NCERT Exemplar Class 12 Maths Solutions Chapter 9 Differential Equations 2021

NCERT Exemplar Class 12 Maths Solutions Chapter 9 Differential Equations

Edited By Komal Miglani | Updated on Mar 30, 2025 11:24 PM IST | #CBSE Class 12th

When considering an example where you are trying to understand how a particular physical quantity changes as a function of time. For instance, how quickly a car accelerates, how quickly a drug is eliminated from the blood, or how quickly a population is growing or declining. Each of these real-world problems can be modelled mathematically with a study known as Differential Equations. The study of Differential Equations provides the tools to analyze these dynamic changes through modelling. Differential Equations will describe how to relate a function to its rate of change over various phenomena in the domains of physics, biology, economics, and engineering. NCERT Exemplar for Class 12 Chapter 9 will cover the fundamental concepts of differential equations, including first-order and second-order equations and solving and applying them. Students will learn methods such as the separation of variables, homogeneous and non-homogeneous, and integrating factors.

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To understand this chapter, there has to be practice. Having an idea of how these equations are being solved and knowing how to solve them in real circumstances will improve the student's understanding of the concepts from theory and will make them more adept at problem-solving. Students can get help from the NCERT Class 12 Maths Solutions if they need additional context or explanation.

NCERT Exemplar Class 12 Maths Solutions Chapter 9 Differential Equations

Class 12 Maths Chapter 9 Solutions Exercise: 9.1
Page number: 193-202
Total questions: 77

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Question:1

Find the solution of $\frac{d y}{d x} = 2^{y - x}$

Answer:

Given
$\frac{d y}{d x} = 2^{y - x}$
To find: Solution of the given differential equation
Rewrite the equation as,
$\frac{d y}{2^{y}} = \frac{d x}{2^{x}}$
Integrating on both sides,
$\int \frac{d y}{2^{y}} = \int \frac{d x}{2^{x}} (\int \frac{d x}{a^{x}} = - \frac{a^{- x}}{I n a})$
Formula:
$- \frac{2^{- y}}{I n 2} = - \frac{2^{- x}}{I n 2} + c$
Here c is some arbitrary constant
$2^{- x} - 2^{- y} = c I n 2$

$\Rightarrow 2^{- x} - 2^{- y} = d$
d is also some arbitrary constant = c In2

Question:2

Find the differential equation of all non-vertical lines in a plane.

Answer:

To find: Differential equation of all non vertical lines

The general form of equation of line is given by y=mx+c where, m is the slope of the line

The slope of the line cannot be $\frac{π}{2}$ or $\frac{3 π}{2}$ for the given condition because if it is so, the line will become perpendicular without any necessity.
So,
$m \neq \frac{π}{2}, m \neq \frac{3 π}{2}$
Differentiate the general form of equation of the line
$\frac{d y}{d x} = m$
Formula:
$\frac{d (a x)}{d x} = a$
Differentiating it again it becomes,
$\frac{d^{2} y}{d x^{2}} = 0$
Thus we get the differential equation of all non-vertical lines.

Question:3

Given that $\frac{d y}{d x} = e^{- 2 y}$ and y = 0 when x = 5. Find the value of x when y = 3.

Answer:

Given:
$\frac{d y}{d x} = e^{- 2 y}$
$(5, 0)$ is a solution of this equation
To find: Solution of the given differential equation
Rewriting the equation.
$\frac{d y}{e^{- 2 y}} = d x$
Integrate on both the sides,
$\int \frac{d y}{e^{- 2 y}} = \int d x$

$\Rightarrow \int e^{2 y} d y = \int d x$
Formula:
$\int e^{a x} d x = \frac{1}{a} e^{a x}$

$\frac{e^{2 y}}{2} = x + c$
Given $(5, 0)$ is a solution so to get c, satisfying these values
$\frac{1}{2} = 5 + c c = - \frac{9}{2}$
Hence the solution is
e^2y=2x + 9
when y=3,
e²⁽³⁾ =2x + 9
$\Rightarrow$ e⁶=2x + 9
$\Rightarrow$ e⁶+ 9=2x
$\Rightarrow x = \frac{e^{6} + 9}{2}$

Question:4

Solve the differential equation $(x^{2} - 1) \frac{d y}{d x} + 2 x y = \frac{1}{x^{2} - 1}$

Answer:

Given:
$(x^{2} - 1) \frac{d y}{d x} + 2 x y = \frac{1}{x^{2} - 1}$
To find the solution of the given differential equation
Rewriting the equations as,
$\frac{d y}{d x} + \frac{2 x y}{(x^{2} - 1)} = \frac{1}{{(x^{2} - 1)}^{2}}$
It is a first-order linear differential equation Compare it with,
$\frac{d y}{d x} + p (x) y = q (x)$

$p (x) = \frac{2 x}{(x^{2} - 1)}$ and $q (x) = \frac{1}{{(x^{2} - 1)}^{2}}$
Calculate the Integrating Factor,
$I F = e^{\int p (x) d x}$

$\Rightarrow I F = e^{\int \frac{2 x}{x^{2} - 1} d x}$

$\Rightarrow I F = e^{l n (x^{2} - 1)}$

$\Rightarrow I F = x^{2} - 1$

$\int \frac{2 x}{(x^{2} - 1)} d x = \int \frac{(x + 1) + (x - 1)}{(x^{2} - 1)} d x$

$= \int \frac{d x}{x - 1} + \int \frac{d x}{x + 1} = l n (x + 1) + l n (x - 1)$

Hence, the solution of the differential equation is given by,
$y . (I F) = \int q (x) . (I F) d x$

$\Rightarrow y (x^{2} - 1) = \int \frac{1}{(x^{2} - 1)^{2}} (x^{2} - 1) d x$

$\Rightarrow y (x^{2} - 1) = \int \frac{1}{(x^{2} - 1)} d x$

Formula: $\int \frac{1}{(x^{2} - 1)} d x = \frac{1}{2} \log (\frac{x - 1}{x + 1}) \frac{1}{(x^{2} - 1)} d x = \frac{1}{2} \log (\frac{x - 1}{x + 1}) + c$

Question:5

Solve the differential equation $\frac{d y}{d x} + 2 x y = y$

Answer:

$\frac{d y}{d x} + 2 x y = y$
To find: Solution of the given differential equation
$\int \frac{1}{x} d x = l n x + c$

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c$
Rewriting the given equation as,
$\frac{d y}{d x} = y (1 - 2 x)$

$\Rightarrow \frac{d y}{y} = (1 - 2 x) d x$
Integrate on both the sides,
$\int \frac{d y}{y} = \int (1 - 2 x) d x$

$\Rightarrow \ln y = (x - x^{2}) + \log c$

$\Rightarrow \ln y - \ln c = x - x^{2}$

$\Rightarrow \ln \frac{y}{c} = x - x^{2}$

$\Rightarrow \frac{y}{c} = e^{x - x^{2}}$

$\Rightarrow y = c e^{x - x^{2}}$

Question:6

Find the general solution of $\frac{d y}{d x} + a y = e^{m x}$

Answer:

Given:
$\frac{d y}{d x} + a y = e^{m x}$
It is a first order differential equation. Comparing it with,
$\frac{d y}{d x} + p (x) y = q (x)$
$P (x) = a$
$Q (x) = e^{x m}$
Calculating Integrating Factor
$I F = e^{\int p (x) d x}$

$I F = e^{\int a d x}$

$I F = e^{a x}$
Hence the solution of the given differential equation is ,
$y (I F) = \int q (x) . (I F) d x$

$\Rightarrow y . (e^{a x}) = \int e^{m x} e^{a x} d x$

$\Rightarrow y . (e^{a x}) = \int e^{(m + a) x} d x$

$\Rightarrow y . (e^{a x}) = \frac{(e^{(m + a) x})}{m + a} + c$

Question:7

Solve the differential equation $\frac{d y}{d x} + 1 = e^{x + y}$

Answer:

$\frac{d y}{d x} + 1 = e^{x + y}$
To find: Solution of the given differential equation
Assume $x + y = t$
Differentiate on both sides with respect to x
$1 + \frac{d y}{d x} = \frac{d t}{d x}$
Substitute
$\frac{d y}{d x} + 1 = e^{x + y}$ in the above equation
$e^{x + y} = \frac{d t}{d x} e^{t} = \frac{d t}{d x}$
Rewriting the equation,
$d x = e^{- t} d t$
Integrate on both the sides,
$\int d x = \int e^{- t} d t$

formula: $\int e^{x} d x = e^{x} x = - e^{- t} + c$
Substituting $t = x + y$

$x = - e^{- (x + y)} + c$ is the solution of the differential equation.

Question:8

Solve: $y d x - x d y = x^{2} y d x$

Answer:

Given:
$y d x - x d y = x^{2} y d x$
To find: solution of the differential equation
Rewriting the given equation,
$\int \frac{1 - x^{2}}{x} d x = \int \frac{d y}{y}$

$\Rightarrow \int \frac{1}{x} - x d x = \int \frac{d y}{y}$

$\Rightarrow \ln x - \frac{x^{2}}{2} + \ln c = \ln y$

$\Rightarrow - \frac{x^{2}}{2} = \ln y - \ln x - \ln c$

$\Rightarrow - \frac{x^{2}}{2} = \ln \frac{y}{c x}$

$\Rightarrow y = c x e^{- \frac{x^{2}}{2}}$

Question:9

Solve the differential equation $\frac{d y}{d x} = 1 + x + y^{2} + x y^{2}$ when y = 0, x = 0.

Answer:

Given:
$\frac{d y}{d x} = (1 + x) (1 + y^{2})$ and (0,0) is solution of the equation
To find: solution of the differential equation
Rewriting the given equation as,
$\frac{d y}{1 + y^{2}} = (1 + x) d x$
Integrating on both the sides
$\int \frac{d y}{1 + y^{2}} = \int (1 + x) d x$

$\tan^{- 1} y = x + \frac{x^{2}}{2} + c$
Substitute(0,0) to find c’s value
$0 + 0 = c$
$c = 0$
Hence, the solution is
$\tan^{- 1} y = x + \frac{x^{2}}{2}$

$\Rightarrow y = \tan (x + \frac{x^{2}}{2})$

Question:10

Find the general solution of $(x + 2 y^{3}) \frac{d y}{d x} = y$

Answer:

Given:
$(x + 2 y^{3}) \frac{d y}{d x} = y$
To find: Solution of the differential equation
Rewriting the equation as
$\frac{d x}{d y} = \frac{(x + 2 y^{3})}{y}$

$\Rightarrow \frac{d x}{d y} = \frac{x}{y} + 2 y^{2}$

$\Rightarrow \frac{d x}{d y} - \frac{x}{y} = 2 y^{2}$
It is a first order linear differential equation
Comparing it with
$\frac{d x}{d y} + p (y) x = q (y)$

$p (y) = - \frac{1}{y}$ and $q (y) = 2 y^{2}$
Calculation the integrating factor,
$I F = e^{\int p (y) d y}$

$\Rightarrow I F = e^{\int - \frac{1}{y} d y}$

$\Rightarrow I F = e^{- \ln y} = \frac{1}{y}$

Therefore, the solution of the differential equation is
$x . (I F) = \int q (y) . (I F) d y$

$\Rightarrow \frac{x}{y} = \int 2 y d y$

$\Rightarrow \frac{x}{y} = y^{2} + c$

$\Rightarrow x = y^{3} + c y$

Question:11

If y(x) is a solution of $\frac{2 + \sin x}{1 + y} \frac{d y}{d x} = - \cos x$ and y(0) = 1, then find the value of $y = \frac{π}{2}$

Answer:

Given:

$\frac{2 + \sin x}{1 + y} \frac{d y}{d x} = - \cos x$

To find: Solution of the differential equation

Rewriting the given equation as,

$\frac{d y}{1 + y} = \frac{- \cos x d x}{2 + \sin x}$

Integrating on both sides,

$\int \frac{d y}{1 + y} = \int \frac{- \cos x d x}{2 + \sin x}$

$\Rightarrow \ln (1 + y) = \int \frac{- \cos x d x}{2 + \sin x}$

Let $\sin x = t$ and $\cos x d x = d t$
$\Rightarrow \ln (1 + y) = \int \frac{- dt}{2 + t}$

$\Rightarrow \ln (1 + y) = - \ln (2 + t) + \log c$

$\Rightarrow \ln (1 + y) + \ln (2 + \sin x) = \log c$

$\Rightarrow (1 + y) (2 + \sin x) = c$

When $x = 0$ and $y = 1$

$\begin{array}{l} 1 = \frac{c}{2 + \sin 0} - 1 \\ 1 + 1 = \frac{c}{2} \end{array}$

$c = 4$

$\begin{array}{l} \Rightarrow y = \frac{4}{2 + \sin x} - 1 \\ If x = π / 2 then, \\ y = \frac{4}{2 + \sin \frac{π}{2}} - 1 \end{array}$

$\begin{array}{l} y = \frac{4}{2 + 1} - 1 \\ y = \frac{4}{3} - 1 \\ y = \frac{1}{3} \end{array}$

Question:12

If y(t) is a solution of $(1 + t) \frac{d y}{d t} - t y = 1$ and y(0) = –1, then show that $y (1) = - \frac{1}{2}$

Answer:

$(1 + t) \frac{d y}{d t} - t y = 1$ and $(0, - 1)$ is a solution
To find: Solution for the differential equation
Rewriting the given equation as,
$\frac{d y}{d t} - \frac{t y}{1 + t} = \frac{1}{1 + t}$
It is a first order linear differential equation
Comparing it with,
$\frac{d y}{d t} - p (t) y = q (t)$

$p (t) = - \frac{t}{1 + t}$ and $q (t) = \frac{1}{1 + t}$
Calculation Integrating Factor
$I F = e^{\int \frac{t}{1 + t} d t}$

$I F = e^{\int \frac{- t + 1}{1 + t} d t}$

$I F = e^{\int - 1 + \frac{1}{1 + t} d t}$

$I F = e^{- t + \ln (1 + t)} = e^{- t} (1 + t)$
Hence the solution for the differential equation is,
$y . (I F) = \int q (t) . (I F) d t$

$\Rightarrow y (e^{- t} (1 + t)) = \int e^{- t} (1 + t) \frac{1}{1 + t} d t$

$\Rightarrow y (e^{- t} (1 + t)) = \int e^{- t} d t$

$\Rightarrow y (e^{- t} (1 + t)) = - e^{- t} + c$
Substitution (0,-1) to find the value of c
$- 1 = - 1 + c$

$c = 0$

$y (e^{- t} (1 + t)) = - e^{- t}$
The solution therefore y(1) is
$y (1) = - \frac{1}{1 + 1} = - \frac{1}{2}$

Question:13

Form the differential equation having $y = (\sin^{- 1} x)^{2} + A \cos^{- 1} x + B$ , where A and B are arbitrary constant, as its general solution.

Answer:

Given:

$y = (\sin^{- 1} x)^{2} + A \cos^{- 1} x + B$
To find: Solution of the differential equation
Differentiating on both the sides,
$\frac{d y}{d x} = 2 \sin^{- 1} x \frac{1}{\sqrt{1 - x^{2}}} - A \frac{1}{\sqrt{1 - x^{2}}}$

$\Rightarrow \sqrt{1 - x^{2}} \frac{d y}{d x} = 2 \sin^{- 1} x - A$

Differntiate again on both the sides
$\sqrt{1 - x^{2}} \frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} \frac{x}{\sqrt{1 - x^{2}}} = 2 \frac{1}{\sqrt{1 - x^{2}}}$

$\Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = 2$
Hence the solution is
$(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - 2 = 0$

Question:14

Form the differential equation of all circles which pass through origin and whose centres lie on y-axis.

Answer:

To find: Differential equations of all circles which pass though origin and centre lies on x axis
Assume a point (0,k) on y-axis
Radius of the circle is
$\sqrt{0^{2} + k^{2}} = k$
General form of the equation of circle is,
$(x - a)^{2} + (y - b)^{2} = r^{2}$
Here a, c is the center and r is the radius.
Substituting the values in the above equation,
$(x - 0)^{2} + (y - b)^{2} = k^{2} x^{2} + y^{2} - 2 y k = 0$
Differentiate the equation with respect to x
$2 x + 2 y \frac{d y}{d x} - 2 k \frac{d y}{d x} = 0. . . . . . . (i)$

$k = \frac{x + y \frac{d y}{d x}}{\frac{d y}{d x}}$
Substituting the value of k in (i)
$x^{2} + y^{2} - 2 y \frac{x + y \frac{d y}{d x}}{\frac{d y}{d x}} = 0$

$\Rightarrow (x^{2} + y^{2}) \frac{d y}{d x} - 2 y x + 2 y^{2} \frac{d y}{d x} = 0$

$\Rightarrow (x^{2} - y^{2}) \frac{d y}{d x} - 2 y x = 0$

Question:15

Find the equation of a curve passing through origin and satisfying the differential equation $(1 + x^{2}) \frac{d y}{d x} + 2 x y = 4 x^{2}$

Answer:

Given:
$(1 + x^{2}) \frac{d y}{d x} + 2 x y = 4 x^{2}$ and $(0, 0)$ is a solution to the curve
To find: Equation of the curve satisfying the differential equation
Rewrite the given equation
$\frac{d y}{d x} + \frac{2 x y}{(1 + x^{2})} = \frac{4 x^{2}}{1 + x^{2}}$
Comparing with
$\frac{d y}{d x} + p (x) y = q (x)$

$p (x) = \frac{2 x}{1 + x^{2}}$ and $q (x) = \frac{4 x^{2}}{1 + x^{2}}$
Calculating Integrating Factor
$I F = e^{\int p (x) d x}$

$I F = e^{\int \frac{2 x}{1 + x^{2}} d x}$
Calculating
$\int \frac{2 x}{1 + x^{2}} d x$
Assume
$1 + x^{2} = t$

$\Rightarrow 2 x d x = d t$

Substituting $t = 1 + x^{2}$

$\int \frac{2 x}{1 + x^{2}} d x = \ln (1 + x^{2})$

$I F = e^{\ln (1 + x^{2})} = (1 + x^{2})$
Hence the solution is
$y . (I F) = \int q (x) . (I F) d x$

$y (1 + x^{2}) = \int \frac{4 x^{2}}{1 + x^{2}} (1 + x^{2}) d x$

$y (1 + x^{2}) = \frac{4}{3} x^{3} + c$
Satisfying $(0, 0)$ in the equation of the curve to find the value of c
0+0=c
c=0
Therefore equation of the curve is
$y (1 + x^{2}) = \frac{4}{3} x^{3} y = \frac{4 x^{2}}{3 (1 + x^{2})}$

Question:16

solve $x^{2} \frac{d y}{d x} = x^{2} + x y + y^{2}$

Answer:

Given:
$x^{2} \frac{d y}{d x} = x^{2} + y^{2} + x y$
To find: solution for the differential equation
Rewriting the given equation as
$\frac{d y}{d x} = 1 + \frac{y^{2}}{x^{2}} + \frac{y}{x}$
Clearly it is a homogenous equation
Assume $y = V x$
Differentiate on both sides
$\frac{d y}{d x} = V + x \frac{d V}{d x}$
Substituting dy/dx in the equation
$1 + \frac{y^{2}}{x^{2}} + \frac{y}{x} = V + x \frac{d V}{d x}$

$\Rightarrow 1 + V^{2} + V = V + x \frac{d V}{d x}$

$\Rightarrow 1 + V^{2} = x \frac{d V}{d x}$

$\Rightarrow \frac{d x}{x} = \frac{d V}{1 + V^{2}}$
Integrating on both the sides
$\int \frac{d x}{x} = \int \frac{d V}{1 + V^{2}}$

$\Rightarrow \ln x = \tan^{- 1} V + c$

Substitute $V = \frac{y}{x}$

$\ln x = \tan^{- 1} \frac{y}{x} + c$
is the solution for the differential equation.

Question:17

Find the general solution of the differential equation $(1 + y^{2}) + (x - e^{\tan^{- 1} y}) \frac{d y}{d x} = 0$

Answer:

Given
$(1 + y^{2}) + (x - e^{\tan^{- 1} y}) \frac{d y}{d x} = 0$
To find: Solution of the given differential equation
Rewrite the given equation as,
$(1 + y^{2}) \frac{d x}{d y} + x - e^{\tan^{- 1} y} = 0$

$\Rightarrow \frac{d x}{d y} + \frac{x}{(1 + y^{2})} = \frac{e^{\tan^{- 1} y}}{(1 + y^{2})}$
It is a first order differential equation
Comparing it with
$\frac{d x}{d y} + p (y) x = q (y)$

$p (y) = \frac{1}{(1 + y^{2})}$ and $q (y) = \frac{e^{\tan^{- 1} y}}{(1 + y^{2})}$
Calculating Integrating Factor
$I F = e^{\int p (y) d y}$

$I F = e^{\int \frac{1}{1 + y^{2}} d y}$

$I F = e^{\tan^{- 1} y}$
Hence the solution of the given differential equation is
$x . (I F) = \int q (y) . (I F) d y$

$\Rightarrow x . (e^{\tan^{- 1} y}) = \int \frac{e^{\tan^{- 1} y}}{(1 + y^{2})} (e^{\tan^{- 1} y}) d y$

Assume $(e^{\tan^{- 1} y}) = t$
Differentiate on both the sides
$\frac{e^{\tan^{- 1} y}}{(1 + y^{2})} d y = d t$

$x . t = \int t d t$

$x . t = \frac{t^{2}}{2} + c$

Substituting $t$

$x . (e^{\tan^{- 1} y}) = \frac{e^{2 \tan^{- 1} y}}{2} + c$

Question:18

Find the general solution of $y^{2} d x + (x^{2} - x y + y^{2}) d y = 0.$

Answer:

Given:
$y^{2} d x + (x^{2} - x y + y^{2}) d y = 0$
To find: Solution for the given differential equation
Rewrite the given equation
$y^{2} \frac{d x}{d y} = x y - x^{2} - y^{2}$

$\frac{d x}{d y} = \frac{x}{y} - 1 - \frac{x^{2}}{y^{2}}$
It is a homogenous differential equation
Assume $x = v y$
Differentiating on both the sides
$\frac{d x}{d y} = v + y \frac{d v}{d y}$
Substitute dy/dx in the given equation
$v + y \frac{d v}{d y} = \frac{x}{y} - 1 - \frac{x^{2}}{y^{2}}$
Substitute v=x/y
$v + y \frac{d v}{d y} = v - 1 - v^{2}$

$y \frac{d v}{d y} = - 1 - v^{2}$

$\frac{d v}{1 + v^{2}} = - \frac{d y}{y}$
Integrating on both the sides
$\int \frac{d v}{1 + v^{2}} = - \int \frac{d y}{y}$

$\tan^{- 1} v = - \ln y + c$

Substituting $v = \frac{x}{y}$

$\tan^{- 1} \frac{x}{y} = - \ln y + c$

Question:19

Solve: (x + y) (dx – dy) = dx + dy. [Hint: Substitute x + y = z after separating dx and dy].

Answer:

Given:
$(x + y) (d x - d y) = d x + d y$
To find: Solution of the given differential equation
Rewriting the given equation
$(x + y) (1 - \frac{d y}{d x}) = (1 + \frac{d y}{d x})$
Assume $x + y = z$
Differentiate on both sides with respect to x
$1 + \frac{d y}{d x} = \frac{d z}{d x}$
Substituting the values in the equation
$z (1 - \frac{d z}{d x} + 1) = \frac{d z}{d x}$

$\Rightarrow 2 z - z \frac{d z}{d x} - \frac{d z}{d x} = 0$

$\Rightarrow 2 z = (z + 1) \frac{d z}{d x}$

$\Rightarrow d x = (\frac{1}{2} + \frac{1}{2 z}) d z$
Integrate on both the sides
$\int d x = \int (\frac{1}{2} + \frac{1}{2 z}) d z$

$x = \frac{z}{2} + \frac{1}{2} \ln z + c$
Substitute $v = x y$
$x = \frac{x + y}{2} + \frac{1}{2} \ln (x + y) + \ln c$

$\Rightarrow x - y - \ln (x + y) - \ln c = 0$

$\Rightarrow \ln (x + y) + \ln c = x - y$

$\Rightarrow \ln c (x + y) = x - y$

$\Rightarrow c (x + y) = e^{x - y}$

$\Rightarrow x + y = 1 / c (e^{x - y})$

$\Rightarrow x + y = d (e^{x - y})$ where $d = \frac{1}{c}$

Question:20

Solve: $2 (y + 3) - x y \frac{d y}{d x} = 0$ given that y (1) = –2

Answer:

Given:
$2 (y + 3) - x y \frac{d y}{d x} = 0$
To Find: Solution of the differential equation
$2 \frac{d x}{x} = \frac{y d y}{y + 3}$

$\Rightarrow 2 \frac{d x}{x} = \frac{[(y + 3) - 3] d y}{y + 3}$

$\Rightarrow 2 \frac{d x}{x} = d y - \frac{3 d y}{y + 3}$
Integrating on both sides
$\int 2 \frac{d x}{x} = \int d y - \int \frac{3 d y}{y + 3}$

$\Rightarrow 2 \ln x = y - 3 \ln (y + 3) + c$
Substitute $(- 2, 1)$ to find value of c
$0 = - 2 + c$

$\Rightarrow c = 2$

$2 \ln x = y - 3 \ln (y + 3) + 2$

$\Rightarrow 2 \ln x + 3 \ln (y + 3) = y + 2$

$\Rightarrow \ln x^{2} + \ln (y + 3)^{3} = y + 2$

$\Rightarrow x^{2} (y + 3)^{3} = e^{y + 2}$

Question:21

Solve the differential equation $d y = \cos x (2 - y c o s e c x) d x$ given that y = 2 when $x = \frac{π}{2}$

Answer:

Given:
$d y = \cos x (2 - y c o s e c x) d x$
$(\frac{π}{2}, 2)$ is a solution of the given differential equation
Rewriting the given equation
$\frac{d y}{d x} = 2 \cos x - y \cot x$

$\Rightarrow \frac{d y}{d x} + y \cot x = 2 \cos x$
It is a first order differential equation
$p (x) = \cot x$ and $q (x) = 2 \cos x$
Calculate integrating factor
$I F = e^{\int p (x) d x}$

$\Rightarrow I F = e^{\int \cot x d x}$

$\Rightarrow I F = e^{\ln \sin x}$

$\Rightarrow I F = \sin x$
Therefore, the solution of the differential equation is
$y . (I F) = \int q (x) . (I F) d x$

$\Rightarrow y \sin x = 2 \int \cos x \sin x d x$

$\Rightarrow y \sin x = \int \sin 2 x d x$

$\Rightarrow y \sin x = - \frac{1}{2} \cos 2 x + c$
Substituting $(\frac{π}{2}, 2)$ to find the value of c
$2 = \frac{1}{2} + c$

$\Rightarrow c = \frac{3}{2}$
Hence the solution is
$y \sin x = - \frac{1}{2} \cos 2 x + \frac{3}{2}$

Question:22

Form the differential equation by eliminating A and B in Ax² + By² = 1

Answer:

Given :
Ax²+By²=1
To find: Solution of the differential equation

Differentiate with respect to x
$2 A x + 2 B y \frac{d y}{d x} = 0$

$\Rightarrow A x + B y \frac{d y}{d x} = 0. . . . (i)$

$\Rightarrow \frac{d y}{d x} = - \frac{A x}{B y} . . . . . (i i)$
Differentiate the curve (i) again to get,
$A + B (\frac{d y}{d x} \frac{d y}{d x} + y \frac{d^{2} y}{d x^{2}}) = 0$

$\Rightarrow - \frac{A}{B} = ({(\frac{d y}{d x})}^{2} + y \frac{d^{2} y}{d x^{2}})$
Substituting this in eq(i)
$\frac{d y}{d x} = - \frac{x}{y} ({(\frac{d y}{d x})}^{2} + y \frac{d^{2} y}{d x^{2}})$

$\Rightarrow y \frac{d y}{d x} = - x {(\frac{d y}{d x})}^{2} - x y \frac{d^{2} y}{d x^{2}}$

$\Rightarrow y \frac{d y}{d x} + x {(\frac{d y}{d x})}^{2} + x y \frac{d^{2} y}{d x^{2}} = 0$

Question:23

Solve the differential equation (1+ y²) tan^-1x dx + 2y (1 + x²) dy = 0

Answer:

Given:
$(1 + y^{2}) \tan^{- 1} x d x + 2 y (1 + x^{2}) d y = 0$
To find: Solution for differential equation that s given
Rewriting the given equation as.
$(1 + y^{2}) \tan^{- 1} x = - 2 y (1 + x^{2}) \frac{d y}{d x}$

$\Rightarrow \frac{\tan^{- 1} x}{(1 + x^{2})} d x = - \frac{2 y}{(1 + y^{2})} d y$
Integrate on the both sides
$\int \frac{\tan^{- 1} x}{(1 + x^{2})} d x = - \int \frac{2 y}{(1 + y^{2})} d y$
For LHS
Assume tan^-1 =t
$\frac{1}{1 + x^{2}} d x = d t$
For RHS
Assume 1+y²=z
$2 y d y = d z$
Substituting and integrating on both the sides
$\int t d t = - \int \frac{d z}{z}$

$\Rightarrow \frac{t^{2}}{2} = - \ln z + c$
Substitute for t and z
Solution for the differential equation is
$\frac{{(\tan^{- 1} x)}^{2}}{2} = - \ln (1 + y^{2}) + c$

$\Rightarrow \frac{{(\tan^{- 1} x)}^{2}}{2} + \ln (1 + y^{2}) = c$

Question:24

Find the differential equation of system of concentric circles with centre (1, 2).

Answer:

To find: Differential equation of concentric circles whose center is (1,2)
Equation of the curve is given by
(x-a)²+(y-b)²=k²
Where (a,b) is the center and k, radius.
Subsitute the values now,
(x-1)²+(y-2)²=k²
Differentiate with respect to x
$2 (x - 1) + 2 (y - 2) \frac{d y}{d x} = 0$

$\Rightarrow (x - 1) + (y - 2) \frac{d y}{d x} = 0$

$\Rightarrow \frac{d y}{d x} = - \frac{(x - 1)}{(y - 2)}$

Question:25

Solve: $y + \frac{d y}{d x} (x y) = x (\sin x + \log x)$

Answer:

$y + \frac{d y}{d x} (x y) = x (\sin x + \log x)$
Now dx/dy (xy) refers to the differentiation of xy with respect to x
Using product rule
$\Rightarrow \frac{d}{dx} (xy) = y + x \frac{dy}{dx}$
When we put it back originally in the differential equation given,
$\Rightarrow y + y + x \frac{d y}{d x} = x (\sin x + \log x)$

$\Rightarrow x \frac{d y}{d x} + 2 y = x (\sin x + \log x)$
Divide by $x$
$\Rightarrow \frac{d y}{d x} + (\frac{2}{x}) y = \sin x + \log x$
Compare $\frac{dy}{dx} + (\frac{2}{x}) y = \sin x + \log_{x}$ with $\frac{dy}{dx} + Py = Q$
We get
$P = \frac{2}{x} and Q = \sin x + \log x$
The above equation is a linear differential equation with P and Q as functions of x
The first to find the solution of a linear differential equation is to find the integrating factor. $\Rightarrow IF = e^{[Pdx}$
$\Rightarrow IF = e^{2 \int \frac{1}{x} dx} \Rightarrow IF = e^{2 \log x}$

$\Rightarrow IF = e^{\log x^{2}} = x^{2} \Rightarrow IF = x^{2}$
The solution of the linear differential equation is
$y (IF) = \int Q (IF) d x + c$
Substituting values for Q and IF
$\Rightarrow y x^{2} = \int (\sin x + \log x) x^{2} d x + c$
$\Rightarrow y x^{2} = \int x^{2} \sin x d x + \int x^{2} \log x d x + c \dots (a)$
Find the integrals individually,
Using uv for integration
$\Rightarrow \int_{U V} d x = u \int v d x - \int (u^{'} j v) d x$

$\Rightarrow \int x^{2} \sin x d x = x^{2} (- \cos x) - \int 2 x (- \cos x) d x$
$\Rightarrow \int x^{2} \sin x d x = - x^{2} \cos x + 2 \int x \cos x d x$
$\Rightarrow \int x^{2} \sin x d x = - x^{2} \cos x + 2 (x \sin x - \int \sin x d x)$
$\Rightarrow \int x^{2} \sin x d x = - x^{2} \cos x + 2 (x \sin x - (- \cos x))$
$\Rightarrow \int x^{2} \sin x d x = - x^{2} \cos x + 2 x \sin x + 2 \cos x \dots (i)$
Now use product rule
$\Rightarrow \int x^{2} \log x d x = \log x (\frac{x^{3}}{3}) - \int (\frac{1}{x}) (\frac{x^{3}}{3}) d x$

$\Rightarrow \int x^{2} \log x d x = (\frac{x^{3}}{3}) \log x - \frac{1}{3} \int x^{2} d x$
$\Rightarrow \int x^{2} \log x d x = (\frac{x^{3}}{3}) \log x - \frac{x^{3}}{9} \dots (i i)$
Substitute (i) and (ii) in (a)
$\Rightarrow {yx}^{2} = - x^{2} \cos x + 2 x \sin x + 2 \cos x + (\frac{x^{3}}{3}) \log x - \frac{x^{3}}{9} + c$
Divide by $x^{2}$
$\Rightarrow y = - \cos x + \frac{2 \sin x}{x} + \frac{2 \cos x}{x^{2}} + (\frac{x}{3}) \log x - \frac{x}{9} + \frac{c}{x^{2}}$

Question:26

Find the general solution of $(1 + \tan y) (d x - d y) + 2 x d y = 0.$

Answer:

$(1 + \tan y) (d x - d y) + 2 x d y = 0$
$\Rightarrow dx - dy + \tan y d x - \tan y dy + 2 xdy = 0$
Divide throughout by dy
$\Rightarrow \frac{d x}{d y} - 1 + \tan y \frac{d x}{d y} - \tan y + 2 x = 0$ $\Rightarrow (1 + \tan y) \frac{d x}{dy} - (1 + \tan y) + 2 x = 0$
Divide by $(1 + \tan y)$
$\Rightarrow \frac{d x}{d y} - 1 + \frac{2 x}{1 + \tan y} = 0$

$\Rightarrow \frac{d x}{d y} + (\frac{2}{1 + t a n y}) x = 1$
Compare $\frac{dx}{dy} + (\frac{2}{1 + tany}) x = 1$ with $\frac{dx}{dy} + Px = Q$
We get
$P = \frac{2}{1 + \tan y}$ and $Q = 1$
This is the linear differential equation with P and Q as functions of x
$\Rightarrow IF = e^{\int Pdy}$

$\Rightarrow IF = e^{\int \frac{2}{1 + \tan y} dy}$
Put $\tan y = \frac{\sin y}{\cos y}$
$\Rightarrow IF = e^{\int \frac{2 \cos y}{\sin y + \cos y}} dy$
Adding and subtracting $\sin y$ in the numerator
$\Rightarrow IF = e^{\int \frac{\cos y + \sin y + \cos y - \sin y}{\sin y + \cos y}} d y$

$\Rightarrow IF = e^{\int (1 + \frac{\cos y - \sin y}{\sin y + \cos y}) d y}$

$\Rightarrow IF = e^{\int 1 d y + \int \frac{\cos y - \sin y}{\sin y + \cos y} d y}$
Consider the integral $\int \frac{\cos y - \sin y}{\sin y + \cos y} d y$
Let $\sin y + \cos y = t$
Differentiate with respect to y
We get
$\frac{dt}{dy} = cosy - \sin y$

$\Rightarrow dt = (\cos y - \sin y) d y$
$\begin{aligned} \Rightarrow \int \frac{\cos y - \sin y}{\sin y + \cos y} d y = \int \frac{1}{t} d t \\ \Rightarrow \int \frac{\cos y - \sin y}{\sin y + \cos y} d y = \log t \\ Resubstitue \\ \Rightarrow \int \frac{\cos y - \sin y}{\sin y + \cos y} d y = \log (\sin y + \cos y) \\ \Rightarrow IF = e^{y + \log (\sin y + \cos y)} \end{aligned}$
$\Rightarrow IF = e^{y} \times e^{log (siny + \cos y)}$

$\Rightarrow IF = e^{y} (\sin y + \cos y)$
The solution of the linear differential equation will be $x (IF) = \int Q (IF) d y + c$
Substitute values for Q and IF
$\Rightarrow x e^{y} (\sin y + \cos y) = \int (1) e^{y} (\sin y + \cos y) d y + c$ $\Rightarrow x e^{y} (\sin y + \cos y) = \int (e^{y} \sin y + e^{y} \cos y) d y + c$
Put $e^{y} \sin y = t$ and differentiate with respect to y
We get
$\frac{d t}{d y} = e^{y} \sin y + e^{y} \cos y$
Which means
$dt = (e^{y} \sin y + e^{y} \cos y) d y + c$
Hence
$\Rightarrow x e^{y} (\sin y + \cos y) = \int d t + c$

$\Rightarrow x e^{y} (\sin y + \cos y) = t + c$
Substitute t again
$x e^{y} (\sin y + \cos y) = e^{y} \sin y + c$
$\Rightarrow x = \frac{\sin y}{\sin y + \cos y} + \frac{c}{e^{y} (\sin y + cosy)}$

Question:27

Solve: $\frac{d y}{d x} = \cos (x + y) + \sin (x + y)$ [Hint: Substitute x + y = z]
Answer:

$\frac{d y}{d x} = \cos (x + y) + \sin (x + y)$
Using the given hint substitute $x + y = z$
$\Rightarrow \frac{d (z - x)}{dx} = \cos z + \sin z$
Differentiate $z - x$ with respect to $x$
$\Rightarrow \frac{d z}{d x} - 1 = \cos z + \sin z$

$\Rightarrow \frac{d z}{d x} = 1 + \cos z + \sin z$

$\Rightarrow \frac{d z}{1 + \cos z + \sin z} = d x$
Integrate
$\Rightarrow \int \frac{d z}{1 + \cos z + \sin z} = \int d x$
As we know
$\cos 2 z = 2 \cos^{2} z - 1$
And $\sin 2 z = 2 \sin z \cos z$
$\Rightarrow \int \frac{d z}{1 + 2 \cos^{2} \frac{z}{2} - 1 + 2 \sin \frac{z}{2} \cos \frac{z}{2}} = x$

$\Rightarrow \int \frac{d z}{2 \cos^{2} \frac{z}{2} + 2 \sin \frac{z}{2} \cos \frac{z}{2}} = x$

$\Rightarrow \int \frac{d z}{2 \cos \frac{z}{2} (\cos \frac{z}{2} + \sin \frac{z}{2})} = x$

$\Rightarrow \int \frac{d z}{2 \cos^{2} \frac{z}{2} (1 + \frac{\sin \frac{z}{2}}{\cos \frac{2}{2}})} = x$
$\int \frac{\sec^{2} \frac{z}{2}}{2 (1 + \tan \frac{z}{2})} d z$
$1 + \tan \frac{z}{2} = t$
Differentiate with respect to z
We get
$\frac{d t}{d z} = \frac{\sec^{2} \frac{z}{2}}{2}$
Hence $\frac{\sec^{2} \frac{z}{2} dz}{2} = dt$
$\Rightarrow \int \frac{d t}{t} = x$
$\log t + c = x$
Again substitute t
$\Rightarrow \log (1 + \tan \frac{z}{2}) + c = x$
Similarly substitute z
$\Rightarrow \log (1 + \tan \frac{x + y}{2}) + c = x$

Question:28

Find the general solution of $\frac{d y}{d x} - 3 y = \sin 2 x$

Answer:

$\frac{dy}{dx} - 3 y = \sin 2 x Compare \frac{dy}{dx} - 3 y = \sin 2 x, and \frac{dy}{dx} + Py = Q$
We get, $P = - 3$ and $Q = \sin 2 x$
The equation is a linear differential equation where P and Q are functions of x
For the solution of the linear differential equation, we need to find Integrating factor,
$\Rightarrow IF = e^{[P dx} \Rightarrow IF = e^{[(- 3) d x}$
$\Rightarrow IF = e^{- 3 x}$
$y (IF) = \int Q (IF) d x + c$
The solution of the linear differential equation is
Substitute values for Q and IF
$\Rightarrow y e^{- 3 x} = \int e^{- 3 x} \sin 2 x d x \dots (1)$
Let $I = \int e^{- 3 x} \sin 2 x d x$
If $u (x)$ and $v (x)$ are two functions, then by integration by parts.
$\int uv = u \int v - \int u^{'} \int v$
$v = \sin 2 x$ and $u = e^{- 2 x}$
After applying the formula we get,
$I = \int e^{- 3 x} \sin 2 x dx = e^{- 3 x} \int \sin 2 x d x - \int {(e^{- 3 x})}^{'} d x \int \sin 2 x d x$

$\Rightarrow I = - e^{- 3 x} \frac{\cos 2 x}{2} + \int 3 e^{- 3 x} \frac{\cos 2 x}{2} d x + c$
Again, applying the above stated rule in $\int 3 e^{- 3 x \frac{\cos 2 x}{2}} we get$
$\Rightarrow I = - e^{- 3 x} \frac{\cos 2 x}{2} - \frac{3}{2} [e^{- 3 x} \frac{\sin 2 x}{2} + \int 3 e^{- 3 x} \frac{\sin 2 x}{2}] + c$
$\Rightarrow I = - e^{- 3 x} \frac{\cos 2 x}{2} - \frac{3}{4} e^{- 3 x} \sin 2 x - \frac{9 I}{4} + c$
$\Rightarrow \frac{13 I}{4} = - e^{- 3 x} \frac{\cos 2 x}{2} - \frac{3}{4} e^{- 3 x} \sin 2 x + c$
$\Rightarrow I = \frac{- 1}{13} e^{- 3 x} (2 \cos 2 x + 3 \sin 2 x) + c$
Put this value in (1) to get
${ye}^{- 3 x} = \int e^{- 3 x} \sin 2 x d x$

$\Rightarrow {ye}^{- 3 x} = \frac{- e^{- 3 x} (2 \cos 2 x + 3 \sin 2 x)}{13} + c$

$\Rightarrow y = - \frac{1}{13} (3 \sin 2 x + 2 \cos 2 x) + c e^{3 x}$

Question:29

Find the equation of a curve passing through (2, 1) if the slope of the tangent to the curve at any point (x, y) is $\frac{x^{2} + y^{2}}{2 x y}$

Answer:

Slope of the tangent is given by $\frac{x^{2} + y^{2}}{2 x y}$
Slope of the tangent of the curve $y = f (x)$ is given by $dy / d x$
$\Rightarrow \frac{dy}{dx} = \frac{x^{2} + y^{2}}{2 xy}$
Put $y = v x$
$\Rightarrow \frac{d (vx)}{dx} = \frac{x^{2} + v^{2} x^{2}}{2 {vx}^{2}}$
Using product rule differentiate $v x$
$\Rightarrow x \frac{dv}{dx} + v = \frac{1 + v^{2}}{2 v}$

$\Rightarrow x \frac{dv}{dx} = \frac{1 + v^{2}}{2 v} - v$

$\Rightarrow x \frac{dv}{dx} = \frac{1 + v^{2} - 2 v^{2}}{2 v}$ $\Rightarrow \frac{d v}{d x} = (\frac{1}{x}) \frac{1 - v^{2}}{2 v}$
$\Rightarrow \frac{2 v}{1 - v^{2}} d v = (\frac{1}{x}) d x$
Integrate
$\Rightarrow \int \frac{2 v d v}{1 - v^{2}} = \int (\frac{1}{x}) d x$
Put $1 - v^{2} = t$
$2 vdv = - dt$
$\Rightarrow \int \frac{- d t}{t} = \log x + c \Rightarrow - \log t = \log x + c$
Resubstitute 1
$\Rightarrow - \log (1 - v^{2}) = \log x + c$
Resubstitute v
$\Rightarrow - \log (1 - \frac{y^{2}}{x^{2}}) = \log x + c \dots (a)$
The curve is passing through (2,1)
Hence (2,1) will satisfy the equation (a)
Put x=1 and y=2 in (a)
$\Rightarrow - \log (1 - \frac{1^{2}}{2^{2}}) = \log 2 + c$

$\Rightarrow - \log (\frac{4 - 1}{4}) = \log 2 + c$

$\Rightarrow - \log (\frac{3}{4}) = \log 2 + c$

$\Rightarrow - \log (\frac{3}{4}) - \log 2 = c$

$\Rightarrow - (\log (\frac{3}{4}) + \log 2) = c$
Use loga+logb=logab
$\Rightarrow - \log \frac{3}{2} = c$
Put c in equation (a)
$\Rightarrow - \log (1 - \frac{y^{2}}{x^{2}}) = \log x - \log \frac{3}{2}$

$\Rightarrow - \log (\frac{x^{2} - y^{2}}{x^{2}}) = \log x - \log \frac{3}{2}$

$\Rightarrow \log {(\frac{x^{2} - y^{2}}{x^{2}})}^{- 1} = \log x - \log \frac{3}{2}$

$\Rightarrow \log (\frac{x^{2}}{x^{2} - y^{2}}) - \log x = - \log \frac{3}{2}$

$\Rightarrow \frac{3 x}{2 (x^{2} - y^{2})} = e^{0}$

$\Rightarrow 3 x = 2 x^{2} - 2 y^{2}$
$\begin{aligned} \Rightarrow 2 y^{2} = 2 x^{2} - 3 x \\ \Rightarrow y = \sqrt{\frac{2 x^{2} - 3 x}{2}} \\ Hence the equation of the curve is & y = \sqrt{\frac{2 x^{2} - 3 x}{2}} \end{aligned}$

Question:30

Find the equation of the curve through the point (1, 0) if the slope of the tangent to the curve any point (x, y) is $\frac{y - 1}{x^{2} + x}$
Answer:

Given: Slope of the tangent is $\frac{y - 1}{x^{2} + x}$
Slope of tangent of a curve $y = f (x)$ is given by $dy / dx$
$\Rightarrow \frac{d y}{d x} = \frac{y - 1}{x^{2} + x} \Rightarrow \frac{d y}{y - 1} = \frac{d x}{x^{2} + x}$
Integrate
$\Rightarrow \int \frac{dy}{y - 1} = \int \frac{dx}{x (x + 1)} \dots (a)$
Use partial fraction for
$\Rightarrow \frac{1}{x (x + 1)} = \frac{A}{x} + \frac{B}{x + 1}$

$\Rightarrow \frac{1}{x (x + 1)} = \frac{A (x + 1) + B x}{x (x + 1)}$
Equate the numerator
$A (x + 1) + B x = 1$
Put $x = 0$
$A = 1$
Put $x = - 1$
$B = - 1$
Hence

$\Rightarrow \frac{1}{x (x + 1)} = \frac{1}{x} + \frac{- 1}{x + 1}$
Hence equation (a) becomes,
$\Rightarrow \int \frac{d y}{y - 1} = \int (\frac{1}{x} - \frac{1}{x + 1}) dx$

$\Rightarrow \int \frac{d y}{y - 1} = \int \frac{1}{x} d x - \int \frac{1}{x + 1} d x$
$\log (y - 1) = \log x - \log (x + 1) + c \dots (b)$
Now it is given that the curve is passing through (1,0)
Hence (1,0) will satisfy the equation (b)
Put x=1 and y=0 in b
When we put y=0 in equation b the result is $\log (- 1)$ which is undefined
hence, we must simplify equation (b) further
$\log (y - 1) - \log x = - \log (x + 1) + c$
using loga-logb=loga/b
$\Rightarrow \log (\frac{y - 1}{x}) (x + 1) = c$
Constant c must be taken as log c to eliminate undefined elements in the
equation.(log cand not any other terms because taking logc completely
eliminates the log terms and we don't have to worry about undefined terms
in the equation)
$\Rightarrow \log (\frac{y - 1}{x}) (x + 1) = \log c$
Eliminate log
$\Rightarrow (\frac{y - 1}{x}) (x + 1) = c \dots (c)$
Substitute x=1 and y=0
$\Rightarrow (\frac{0 - 1}{1}) (1 + 1) = c$
c=-2
put back c=-2 in (c)
$\begin{aligned} \Rightarrow (\frac{y - 1}{x}) (x + 1) = - 2 \\ Hence the equation of the curve is (y - 1) (x + 1) = - 2 x \end{aligned}$

Question:31

Find the equation of a curve passing through origin if the slope of the tangent to the curve at any point (x, y) is equal to the square of the difference of the abscissa and ordinate of the point.

Answer:

Abscissa refers to the x coordinate and ordinate refers to the y coordinate.
Slope of the tangent is the square of the difference of the abscissa and the ordinate.
Difference of the abscissa and ordinate is (x-y) and its square is $(x - y)^{2}$
Hence the Slope of the tangent is $(x - y)^{2}$
$\frac{d y}{d x} = (x - y)^{2}$
$\begin{aligned} Put x - y = z \\ \Rightarrow 1 - \frac{d y}{d x} = \frac{d z}{d x} \\ \Rightarrow 1 - \frac{d z}{d x} = \frac{d y}{d x} = z^{2} \\ \Rightarrow 1 - z^{2} = \frac{d z}{d x} \\ \Rightarrow d x = \frac{d z}{1 - z^{2}} \\ \Rightarrow \int d x = \int \frac{d z}{1 - z^{2}} \end{aligned}$
$x = \frac{1}{2} \log | \frac{1 + z}{1 - z} | + C x = \frac{1}{2} \log | \frac{1 + x - y}{1 - x + y} | + C$
The curve passes through the (0,0)
$0 = \frac{1}{2} \log 1 + C C = 0$
$\Rightarrow e^{2 x} = \frac{1 + x - y}{1 - x + y} \Rightarrow e^{2 x} (1 - x + y) = (1 + x - y)$

Question:32

Find the equation of a curve passing through the point (1, 1). If the tangent drawn at any point P(x,y) on the curve meets the co-ordinate axes at A and B such that P is the mid-point of AB.

Answer:

Points on the y axis and x axis are namely $A (0, a), B (b, 0)$ . The midpoint of AB is $P (x, y)$ .
The x coordinate of the points is given by the addition of the x coordinates of A and B divided by 2.
$\begin{aligned} \Rightarrow x = \frac{b + 0}{2} \\ b = 2 x \\ Similarly, for y coordinate \\ \Rightarrow y = \frac{0 + a}{2} \\ a = 2 y \end{aligned}$
Therefore, the coordinates of A and B are (0,2y) and (2x,0) respectively.
AB is the tangent to curve where P is the point of contact.
Slope of the line given with two points $(x_{1,} y_{1})$ and $(x_{2}, y_{2})$ on it is $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$
Here $(x_{1}, y_{1}) (x_{2}, y_{2})$ are $(0, 2 y) (2 x, 0)$ respectively.
Slope of the tangent AB is
$\frac{0 - 2 y}{2 x - 0}$
Hence the slope of the tangent is -y/x
Slope of the tangent curve is given by,
$\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx} = - \frac{y}{x}$ $\Rightarrow \frac{dy}{y} = - \frac{dx}{x}$
Integrate
$\Rightarrow \int \frac{dy}{y} = - \int \frac{dx}{x}$
$\log y = - \log x + c$

$\log y + \log x = c$
Using $\log a + \log b = \log a b$ .
$\log x y = c$
as given curve is passing through(1,a)
Hence (1,1) will satisfy the equation of the curve(a)
Putting $x = 1, y = 2$ in $(a)$
$\log 1 = c$

$\Rightarrow c = 0$
put c back in (a)
$\log x y = 0$

$\Rightarrow x y = e^{0}$
$\Rightarrow x y = 1$
Hence the equation of the curve is $x y = 1$

Question:33

solve $x \frac{d y}{d x} = y (\log y - \log x + 1)$

Answer:

$x \frac{d y}{d x} = y (\log y - \log x + 1)$
Using loga-logb =loga/b
$\Rightarrow \frac{d y}{d x} = \frac{y}{x} (\log \frac{y}{x} + 1)$
Put $y = v x$
$\Rightarrow \frac{d (v x)}{d x} = \frac{v x}{x} (\log \frac{v x}{x} + 1)$
Differentiate yx with respect to x using product rule
$\Rightarrow \frac{d v}{d x} x + v = v (\log v + 1)$
$\Rightarrow \frac{dv}{dx} x + v = v \log v + v$
$\Rightarrow \frac{d v}{d x} x = v \log v \Rightarrow \frac{d v}{v \log v} = \frac{d x}{x}$
Now Integrate
$\Rightarrow \int \frac{dv}{vlogv} = \int \frac{dx}{x}$
Substitute $\log v = t$
Differentiate with respect to v.
$\frac{d v}{v} = d t$
$\Rightarrow \int \frac{d t}{t} = \log x + c$
$\log t = \log x + \log c$
Resubstitute value of t
log(log v)=log x + logc.
Resubstitute v
$\Rightarrow \log (\log \frac{y}{x}) = \log x + \log c \Rightarrow \log \frac{y}{x} = c x$
Therefore the solution of the differential equation is
$\log \frac{y}{x} = c x$

Question:34

The degree of the differential equation ${(\frac{d^{2} y}{d x^{2}})}^{2} + {(\frac{d y}{d x})}^{2} = x \sin \frac{d y}{d x}$ is:
A. 1
B. 2
C. 3
D. Not defined

Answer:

Degree of differential equation is defined as the highest integer power of the highest order derivative in the equation.
Here’s the differential equation
${(\frac{d^{2} y}{d x^{2}})}^{2} + {(\frac{d y}{d x})}^{2} = x \sin \frac{d y}{d x}$
Now for the degree to exit the differential equation must be a polynomial in some differentials.
Differential means
$\frac{d y}{d x} or \frac{d^{2} y}{d x^{2}} or \dots \frac{d^{n} y}{d x^{n}}$
The given differential equation is not a polynomial because of the term sin dy/dx and therefore degree of such a differential equation is not defined.
Option D is correct.

Question:35

The degree of the differential equation ${[1 + {(\frac{d y}{d x})}^{2}]}^{3 / 2} = \frac{d^{2} y}{d x^{2}}$ is:
A. 4
B. 3/4
C. not defined
D. 2

Answer:

Generally, for a polynomial degree is the highest power.
${(1 + {(\frac{d y}{d x})}^{2})}^{\frac{3}{2}} = \frac{d^{2} y}{d x^{2}}$
Differential equation is Squaring both the sides,
$\Rightarrow {(1 + {(\frac{d y}{d x})}^{2})}^{3} = {(\frac{d^{2} y}{d x^{2}})}^{2}$
Now for the degree to exit the differential equation must be a polynomial in
some differentials.
The given differential equation is polynomial in differential is
$\frac{dy}{dx}$ and $\frac{d^{2} y}{{dx}^{2}}$
Degree of differential equation is the highest integer power of the highest order
derivative in the equation.
Highest derivative is
$\frac{d^{2} y}{d x^{2}}$
There is only one term of the highest order derivative in the equation which is
${(\frac{d^{2} y}{d x^{2}})}^{2}$ Whose power is 2 hence the degree is 2
Option D is correct.

Question:36

The order and degree of the differential equation $\frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{1 / 4} + x^{1 / 5} = 0$ respectively, are
A. 2 and 4
B. 2 and 2
C. 2 and 3
D. 3 and 3

Answer:

The differential equation is
$\frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{1 / 4} + x^{1 / 5} = 0$
Order is defined as the number which represents the highest derivative in a differential equation.
$\frac{d^{2} y}{d x^{2}}$ is the highest derivative in the given equation is second order.

Hence the degree of the equation is 2.
Integer powers on the differentials,
$\Rightarrow {(\frac{d y}{d x})}^{\frac{1}{4}} = - \frac{d^{2} y}{d x^{2}} - x^{\frac{1}{5}} \Rightarrow {(\frac{d y}{d x})}^{\frac{1}{4}} = - (\frac{d^{2} y}{d x^{2}} + x^{\frac{1}{5}})$
Now for the degree to exit the differential equation must be a polynomial in some differentials.
Here differentials mean
$\frac{dy}{dx}$ or $\frac{d^{2} y}{{dx}^{2}}$ or $\dots \frac{d^{n} y}{{dx}^{n}}$
The given differential equation is polynomial in differentials
Degree of differential equation is the highest integer power of the highest
order derivative in the equation.
Observe that
${(\frac{d^{2} y}{d x^{2}} + x^{\frac{1}{5}})}^{4}$
Of differential equation (a) the maximum power $d^{2} y / d x^{2}$ will be 4
Highest order is $d^{2} y / {dx}^{2}$ and highest power is 4.
Degree of the given differential equation is 4.
Hence order is 2 and the degree is 4
Option A is correct.

Question:37

if $y = e^{- x} (A \cos x + B \sin x)$ then y is a solution of

$A . \frac{d^{2} y}{{dx}^{2}} + 2 \frac{dy}{dx} = 0$

$B . \frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 2 y = 0$

$C . \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} + 2 y = 0$

$D . \frac{d^{2} y}{d x^{2}} + 2 y = 0$

Answer:

If $y = f (x)$ is a solution of a differential equation, then differentiating it will give the same differential equation.
Differentiate the differential equation twice. Twice because all the options have order as 2 and also because there are two constants A and B
$y = e^{- x} (A \cos x + B \sin x)$
Differentiating using product rule
$\Rightarrow \frac{dy}{dx} = - e^{- x} (Acosx + B \sin x) + e^{- x} (- A \sin x + B \cos x)$
But $e^{- x} (A \cos x + B \sin x) = y$
$\Rightarrow \frac{dy}{dx} = - y + e^{- x} (- Asinx + Bcosx)$
Differentiating again with respect to x,
$\Rightarrow \frac{d^{2} y}{d x^{2}} = - \frac{d y}{d x} - e^{- x} (- A \sin x + B \cos x) + e^{- x} (- A \cos x - B \sin x)$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - \frac{d y}{d x} - e^{- x} (- A \sin x + B \cos x) - e^{- x} (A \cos x + B \sin x)$
But $e^{- x} (A \cos x + B \sin x) = y$
$\Rightarrow \frac{d^{2} y}{{dx}^{2}} = - \frac{dy}{dx} - e^{- x} (- Asinx + B \cos x) - y$
Also,
$\frac{d y}{d x} = - y + e^{- x} (- A \sin x + B \cos x)$
Means,
$e^{- x} (- A \sin x + B \cos x) = \frac{d y}{d x} + y$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - \frac{d y}{d x} - (\frac{d y}{d x} + y) - y$
$\Rightarrow \frac{d^{2} y}{{dx}^{2}} = - \frac{dy}{dx} - \frac{dy}{dx} - y - y$

$\Rightarrow \frac{d^{2} y}{{dx}^{2}} = - 2 \frac{dy}{dx} - 2 y$

$\Rightarrow \frac{d^{2} y}{{dx}^{2}} + 2 \frac{dy}{dx} + 2 y = 0$

Question:38

The differential equation for $y = A \cos α x + B \sin α x,$ where $A$ and $B$ are arbitrary constants is

$A . \frac{d^{2} y}{d x^{2}} - α^{2} y = 0$

$B .$ $\frac{d^{2} y}{d x^{2}} + α^{2} y = 0$

$C . \frac{d^{2} y}{d x^{2}} + α y = 0$

$D . \frac{d^{2} y}{d x^{2}} - α y = 0$

Answer:

Let us find the differential equation by differentiating y with respect to x twice
Twice because we have to eliminate two constants $A$ and $B$ .
$y = A \cos α x + B \sin α x$
Differentiating,
$\Rightarrow \frac{dy}{dx} = - A α \sin α x + B α \cos α x$
Differentiating again
$\Rightarrow \frac{d^{2} y}{d x^{2}} = - A α^{2} \cos α x - B α^{2} \sin α x$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - α^{2} (A \cos α x + B \sin α x)$
$But y = A \cos a x + B \sin a x$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - α^{2} y$

$\Rightarrow \frac{d^{2} y}{d x^{2}} + α^{2} y = 0$
Option B is correct.

Question:39

Solution of differential equation xdy – ydx = 0 represents:
A. a rectangular hyperbola
B. parabola whose vertex is at origin
C. straight line passing through origin
D. a circle whose centre is at origin

Answer:

$\begin{aligned} Lets solve the differential equation \\ \begin{matrix} x d y - y d x = 0 \\ x d y = y d x \\ \Rightarrow \frac{d y}{y} = \frac{d x}{x} \\ \log y = \log x + c \\ \log x - \log y = c \\ Using \log a - \log b = \log a / b \end{matrix} \end{aligned}$

$\Rightarrow \log \frac{y}{x} = c \Rightarrow \frac{y}{x} = e^{c} y = x e^{c}$
$e^{c}$ is constant because e is a constant and c is the integration constant let it be denoted as k hence
$e^{c} = k$

$y = k x$
$y = kx$ is the equation of the straight line and (0,0) satisfies the equation.
Option C is correct.

Question:40

Integrating factor of the differential equation $\cos x \frac{d y}{d x} + y \sin x = 1$ is:
A. cosx
B. tanx
C. sec x
D. sinx

Answer:

Differential equation is
$\cos x \frac{d y}{d x} + y \sin x = 1$

$\Rightarrow \frac{d y}{d x} + \frac{y \sin x}{\cos x} = \frac{1}{\cos x}$

$\Rightarrow \frac{d y}{d x} + (\tan x) y = \sec x$
Compare
$\frac{d y}{d x} + (\tan x) y = \sec x$
With
$\frac{d y}{d x} + P y = Q^{'}$ we get, $P = \tan x$ and $Q = \sec x$
The IF integrating factor is given by
$\Rightarrow e^{\int Pdx} = e^{\int \frac{\sin x}{\cos x} dx}$
Substitute $\cos x = t$ hence
$\frac{dt}{dx} = - \sin x$

$\Rightarrow \sin x d x = - d t$
Resubstitute the value of t
$\Rightarrow e^{\int Pdx} = e^{\log (\cos x)^{- 1}}$
$\Rightarrow e^{\int Pdx} = e^{\log (\cos x)^{- 1}}$

$\Rightarrow e^{\int Pdx} = e^{\log \sec x} = \sec x$
Hence IF is sec x
Option C is correct.

Question:41

Solution of the differential equation $\tan y \sec^{2} x d x + \tan x \sec^{2} y d y = 0$ is:
A. tanx + tany = k
B. tanx – tan y = k
C. $\frac{\tan x}{\tan y} = k$
D. tanx . tany = k

Answer:

The given differential equation is
$\tan y \sec^{2} x d x + \tan x \sec^{2} y d y = 0$
Divide it by tanx tany
$\Rightarrow \frac{\sec^{2} x}{\tan x} d x + \frac{\sec^{2} y}{tany} d y = 0$
Integrate
$\Rightarrow \int \frac{\sec^{2} x}{\tan x} d x + \int \frac{\sec^{2} y}{\tan y} d y = 0$
Put $\tan x = t$ hence,
$\sec^{2} x d x = d t$
Put tany =z hence
$\frac{d z}{d y} = \sec^{2} y$
That is $\sec^{2} y d y = d t$
$\Rightarrow \int \frac{d t}{t} + \int \frac{d z}{z} = 0$
$\log t + \log z + c = 0$
Resubstitue t and z
$\log (\tan x) + \log (\tan y) + c = 0$
Using $\log a + \log b = \log ab$
$\log (\tan x \tan y) = - c$
$\tan x \tan y = e^{- c}$
$e^{- c}$ is constant because e is a constant and c is the integration constant let it be denoted as $e^{- c} = k$
$\tan x \tan y = k$
Option D is correct.

Question:42

Family $y = A x + A^{3}$ of curves is represented by the differential equation of degree:
A. 1
B. 2
C. 3
D. 4

Answer:

$y = A x + A^{3}$
let us find the differential equation representing it so we have to eliminate
the constant A
Differentiate with respect to x
$\Rightarrow \frac{dy}{dx} = A$
Put back value of A in y
$\Rightarrow y = \frac{dy}{dx} x + {(\frac{dy}{dx})}^{3}$
Now for the degree to exit the differential equation must be a polynomial in some differentials.
Here the differentials mean
$\frac{dy}{dx}$ or $\frac{d^{2} y}{{dx}^{2}}$ or $\dots \frac{d^{n} y}{{dx}^{n}}$
The given differential equation is polynomial in differentials
$\frac{dy}{dx}$
Degree of differential equation is the highest integer power of the highest order derivative in the equation.
Highest derivative is
$\frac{dy}{dx}$
And highest power to it is 3 . Hence degree is 3 .
Option C is correct.

Question:43

Integrating factor of $\frac{x d y}{d x} - y = x^{4} - 3 x$ is:
A. x
B. logx
C. $\frac{1}{x}$
D. –x

Answer:

Given differential equation
$\Rightarrow x \frac{d y}{d x} - y = x^{4} - 3 x$
Divide though by x
$\Rightarrow \frac{d y}{d x} - \frac{y}{x} = x^{3} - 3$

$\Rightarrow \frac{d y}{d x} + (- \frac{1}{x}) y = x^{3} - 3$
Compare
$\frac{d y}{d x} + (- \frac{1}{x}) y = x^{3} - 3$ and $\frac{dy}{dx} + Py = Q$
We get
$P = - \frac{1}{x} and Q = x^{3} - 3$
The IF integrating factor is given by
$\Rightarrow e^{\int Pdx} = e^{- \int \frac{1}{x} d x} \Rightarrow e^{\int Pdx} = e^{- \log x}$
$\begin{aligned} \Rightarrow e^{\int P d x} = e^{\log x^{- 1}} \\ \Rightarrow e^{\int Pdx} = e^{\log_{x}^{1}} \\ Hence the IF integrating factor is & \Rightarrow e^{\int Pdx} = \frac{1}{x} \end{aligned}$
Option C is correct.

Question:44

Solution of $\frac{d y}{d x} - y = 1, y (0) = 1$ is given by
A. $x y = - e^{x}$

B. $x y = - e^{- x}$
C. $x y = - 1$
D. $y = 2 e^{x} - 1$

Answer:

$\frac{d y}{d x} - y = 1$

$\Rightarrow \frac{d y}{d x} = 1 + y$

$\Rightarrow \frac{dy}{1 + y} = dx$
Integrate
$\Rightarrow \int \frac{d y}{1 + y} = \int d x$

$\Rightarrow \log (1 + x) = x + c$

now it is given that y(0)=1 which means when x=0, y=1 hence substitute x=0 and y=0 in (a)
$\log (1 + y) = x + c$ $

$\log (1 + 1) = 0 + c$

$c = \log 2$
put $c = \log 2$ back in (a)
$\log (1 + y) = x + \log 2$

$\Rightarrow \log (1 + y) - \log 2 = x$
Using $\log a - \log b = \log a / b$
$\Rightarrow \log \frac{1 + y}{2} = x$

$\Rightarrow \frac{1 + y}{2} = e^{x}$

$\Rightarrow 1 + y = 2 e^{x}$

$\Rightarrow y = 2 e^{x} - 1$
Hence solution of differential equation is $y = 2 e^{x} - 1$
Option D is correct.

Question:45

The number of solutions of $\frac{d y}{d x} = \frac{y + 1}{x - 1}$ when y(1) = 2 is:
A. none
B. one
C. two
D. infinite

Answer:

$\begin{aligned} \begin{matrix} \frac{d y}{d x} = \frac{y + 1}{x - 1} \\ \Rightarrow \frac{d y}{y + 1} = \frac{d x}{x - 1} \end{matrix} \\ Integrate \\ \Rightarrow \int \frac{dy}{y + 1} = \int \frac{d x}{x - 1} \end{aligned}$
$\log (y + 1) = \log (x - 1) - \log c$

$\Rightarrow \log (y + 1) + \log c = \log (x - 1)$
Using $\log a + \log b = \log ab$
$\log_{0} c (y + 1) = \log (x - 1)$

$\Rightarrow \frac{x - 1}{y + 1} = c \dots (a)$
Now as given y(1)=2 which means when $x = 1, y = 2$ Substitute x=1 and y=2 in (a)
$\Rightarrow \frac{1 - 1}{2 + 1} = c$

$\Rightarrow c = 0$

$\Rightarrow \frac{x - 1}{y + 1} = 0$

$\Rightarrow x - 1 = 0$
So only one solution exists.
Option B is correct.

Question:46

Which of the following is a second order differential equation?

$A . {(y^{'})}^{2} + x = y^{2}$

$B . y^{'} y^{''} + y = \sin x$

$C . y^{'''} + {(y^{''})}^{2} + y = 0$

$D . y^{'} = y^{2}$

Answer:

Order is defined as the number which defines the highest derivative in a differential equation
Second order means the order should be 2 which means the highest
derivative in the equation should be $\frac{d^{2} y}{{dx}^{2}}$

Let's examine each of the option given
A. ${(y^{'})}^{2} + x = y^{2}$
The highest order derivative is $y^{'}$ is in first order.
B. $y^{'} y^{''} + y = \sin x$
The highest order derivative is $y^{''}$ is in second order
C. $y^{'''} + {(y^{''})}^{2} + y = 0$
The highest order derivative is $y^{'''}$ is in third order
D. $y^{'} = y^{2}$
The highest order derivative is $y^{'}$ is in first order
Option B is correct.

Question:47

Integrating factor of the differential equation $(1 - x^{2}) \frac{d y}{d x} - x y = 1$ is:
A. -x
B. $\frac{x}{1 + x^{2}}$
C. $\sqrt{1 - x^{2}}$
D. $\frac{1}{2} \log (1 - x^{2})$
Answer:

$(1 - x^{2}) \frac{d y}{d x} - x y = 1$
Divide through by $(1 - x^{2})$
$\Rightarrow \frac{d y}{d x} - \frac{x y}{1 - x^{2}} = \frac{1}{1 - x^{2}}$

$\Rightarrow \frac{d y}{d x} + (\frac{- x}{1 - x^{2}}) y = \frac{1}{1 - x^{2}}$
Compare
$\frac{d y}{d x} + (\frac{- x}{1 - x^{2}}) y = \frac{1}{1 - x^{2}} a n d \frac{d y}{d x} + P y = Q$
We get
$P = \frac{- x}{1 - x^{2}}, Q = \frac{1}{1 - x^{2}}$
The IF factor is given by
$\Rightarrow e^{\int Pdx} = e^{\int \frac{- x}{1 - x^{2}} d x}$
Substitute $1 - x^{2} = t$ hence
$\frac{d t}{d x} = - 2 x$
Which means
$- x d x = \frac{d t}{2}$

$\Rightarrow e^{\int Pdx} = e^{\int \frac{dt}{2 t}}$

$\Rightarrow e^{\int Pdx} = e^{\frac{1}{2} \int \frac{dt}{t}}$ $\Rightarrow e^{\int Pdx} = e^{\frac{1}{2} \log t}$ $\Rightarrow e^{\int P d x} = e^{\log t^{\frac{1}{2}}}$

$\Rightarrow e^{\int P d x} = e^{\log \sqrt{t}}$

$\Rightarrow e^{\int P d x} = \sqrt{t}$
Resubstitute
$\Rightarrow e^{\int Pdx} = \sqrt{1 - x^{2}}$
Hence the IF integrating factor is $\sqrt{1 - x^{2}}$
Option C is correct.

Question:48

$\tan^{- 1} x + \tan^{- 1} y = c$ is the general solution of the differential equation:

A. $\frac{d y}{d x} = \frac{1 + y^{2}}{1 + x^{2}}$
B. $\frac{dy}{dx} = \frac{1 + x^{2}}{1 + y^{2}}$
C. $(1 + x^{2}) d y + (1 + y^{2}) d x = 0$
D. $(1 + x^{2}) d x + (1 + y^{2}) d y = 0$

Answer:

If $y = f (x)$ is a solution of differential equation then differentiating it will give the same differential equation.
To find the differential equation differentiate with respect to x.
$\tan^{- 1} x + \tan^{- 1} y = c$
$\Rightarrow \frac{1}{1 + x^{2}} + \frac{1}{1 + y^{2}} (\frac{d y}{d x}) = 0 (1 + y^{2}) d x + (1 + x^{2}) d y = 0$
Option C is correct.

Question:49

The differential equation $y \frac{dy}{dx} + x = c$ represents:
A. Family of hyperbolas
B. Family of parabolas
C. Family of ellipses
D. Family of circles

Answer:

$y \frac{d y}{d x} + x = c$

$\Rightarrow y \frac{d y}{d x} = c - x$

$\Rightarrow y d y = (c - x) d x$
Integrate
$y d y = (c - x) d x$

$\Rightarrow \int y d y = \int (c - x) d x$

$\Rightarrow \int y d y = \int c d x - \int x d x$

$\Rightarrow \frac{y^{2}}{2} = c x - \frac{x^{2}}{2} + k$
k is the integration constant
$\Rightarrow \frac{y^{2}}{2} + \frac{x^{2}}{2} = c x + k$

$\Rightarrow \frac{y^{2} + x^{2}}{2} = c x + k$
This is the equation of circle because there is no ‘xy’ term and $x^{2}$ and $y^{2}$ have the same coefficient.
This equation represents the family of circles because for different values of c and k we will get different circles.
Option D is correct.

Question:50

The general solution of $e^{x} \cos y d x - e^{x} \sin y d y = 0$ is:
A. $e^{x} \cos y = k$
B. $e^{x} \sin y = k$
C. $e^{x} = k \cos y$
D. $e^{x} = k \sin y$

Answer:

$e^{x} \cos y d x - e^{x} \sin y d y = 0$

$\Rightarrow e^{x} \cos y d x = e^{x} \sin y d y$

$\Rightarrow d x = \frac{\sin y}{\cos y} d y$
Integrate
$\Rightarrow \int dx = \int \frac{\sin y}{\cos y} d y$
substitute cosy =t hence
$\frac{dt}{dy} = - \sin y$
Which means $\sin y d y = - d t$
$\Rightarrow x = \int \frac{- d t}{t}$

$x = - \log t + c$

$x + c = \log (\cos y)^{- 1}$
$\Rightarrow x + c = \log \frac{1}{\cos y}$

$x + c = \log (\sec y)$

$e^{x + c} = \sec y$

$e^{x} e^{c} = \sec y$

$\Rightarrow e^{x} = \frac{1}{e^{c} cosy}$

$e^{x} \cos y = e^{- c}$

$e^{- c}$ is constant because e is a constant and $c$ is the integration constant let us it denote as $k$ hence $e^{- c} = k$
$e^{x} \cos y = k$
Option A is correct.

Question:51

The degree of the differential equation $\frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{3} + 6 y^{5} = 0$ is
(a) 1
(b) 2
(c) 3
(d) 5

Answer:

The answer is the option (a) 1 as the degree of a differential equation is the highest exponent of the order derivative.

Question:52

The solution of $\frac{d y}{d x} + y = e^{- x}, y (0) = 0$ is
(a) $y = e^{x} (x - 1)$
(b) $y = x e^{- x}$
(c) $y = x e^{- x} + 1$
(d) $y = (x + 1) e^{- x}$

Answer:

The answer is the option (b) $y = x e^{- x}$
Explanation: -
This is a linear differential equation.
On comparing it with $\frac{d y}{d x} + P y = Q$ , we get
$P = 1, Q = e^{- x}$
$IF = e^{[P d x} e^{\int d x} = e^{x}$
So, the general solution is:
$y \cdot e^{x} = \int e^{- x} e^{x} d x + C$

$\Rightarrow y \cdot e^{x} = \int d x + C$

$\Rightarrow y \cdot e^{x} = x + C$
Given that when x=0 and y=0
$\Rightarrow$ $0 = 0 + C$

$\Rightarrow$ $C = 0$
Eq. (i) becomes $y \cdot e^{x} = x$
$\Rightarrow y = x e^{- x}$

Question:53

Integrating factor of the differential equation $dy / dx + y \tan x - \sec x = 0$
(a) $\cos x$
(b) $\sec x$
(c) $e^{\cos x}$
(d) $e^{\sec x}$

Answer:

The answer is the option (b) Sec x
Explanation: -
$On comparing it with \frac{d y}{d x} + P y = Q, we get$
$P = \tan x, Q = \sec x \cdot$

$I.F. = e^{\int P d x} = e^{\int \tan x d x} = e^{(\log s e c x)} = \sec x$

Question:54

The solution of the differential equation $\frac{d y}{d x} = \frac{1 + y^{2}}{1 + x^{2}}$ is
(a) $y = \tan^{- 1} x$
(b) $y - x = k (1 + x y)$
(c) $x = \tan^{- 1} y$
(d) $\tan (x y) = k$

Answer:

The answer is the option (b) $y - x = k (1 + x y)$
Explanation: -
$\begin{aligned} \Rightarrow \frac{d y}{1 + y^{2}} = \frac{d x}{1 + x^{2}} \\ On integrating both sides, we get \\ \tan^{- 1} y = \tan^{- 1} x + C \\ \Rightarrow \tan^{- 1} y - \tan^{- 1} x = C \\ \Rightarrow \tan^{- 1} (\frac{y - x}{1 + x y}) = C \\ \Rightarrow \frac{y - x}{1 + x y} = \tan C \\ \Rightarrow y - x = \tan C (1 + x y) \\ \Rightarrow y - x = k (1 + x y), where, k = \tan C \end{aligned}$

Question:55

The integrating factor of the differential equation $\frac{d y}{d x} + y = \frac{1 + y}{x}$ is
(a) $\frac{x}{e^{x}}$
(b) $\frac{e^{x}}{x}$

Answer:

The answer is the option (b) $\frac{e^{x}}{x}$
Explanation: -
$\Rightarrow$ $\frac{d y}{d x} = \frac{1}{x} + \frac{y (1 - x)}{x}$
$\Rightarrow \frac{d y}{d x} - (\frac{1 - x}{x}) y = \frac{1}{x}$
This is a linear differential equation.
On comparing it with $\frac{d y}{d x} + P y = Q,$ we get
$P = \frac{- (1 - x)}{x}, Q = \frac{1}{x}$

$IF, = \int P d x = e^{- \int \frac{1 - x}{x} d x} = \frac{e^{x}}{x}$

Question:56

$y = a e^{m x} + b e^{- m x}$ satisfies which of the following differential equation.

$a . \frac{d y}{d x} + m y = 0$

$b . \frac{d y}{d x} - m y = 0$

$c . \frac{d^{2} y}{d x^{2}} - m^{2} y = 0$

$d . \frac{d^{2} y}{d x^{2}} + m^{2} y = 0$

Answer:

Given $y = a e^{m x} + b e^{- m x}$
upon differentiation, we get $\frac{d y}{d x} = a . m e^{m x} - b . m e^{- m x}$
after differentiation again we get
$\frac{d^{2} y}{d x^{2}} = a m^{2} e^{m x} - b m^{2} e^{- m x}$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = m^{2} (a e^{m x} - b e^{- m x})$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = m^{2} y \Rightarrow \frac{d^{2} y}{d x^{2}} - m^{2} y = 0$
Option c is correct.

Question:57

The solution of the differential equation $\cos x \sin y d x + \sin x \cos y d y = 0$ is

$(a) \frac{\sin x}{\sin y} = c$

$(b) \sin x \sin y = c$

$(c) \sin x + \sin y = c$

$(d) \cos x \cos y = c$

Answer:

$\cos x \sin y d x + \sin x \cos y d y = 0$

$\Rightarrow \sin x \cos y d y = - \cos x \sin y d x$

$\Rightarrow \frac{\cos y}{s i n y} d y = - \frac{\cos x}{s i n x} d x$

$\Rightarrow \cot y d y = - \cot x d x$
Upon integration of both sides,
$\Rightarrow \int \cot y d y = - \int \cot x d x$

$\Rightarrow \log | \sin y | = - \log | \sin x | + \log c$

$\Rightarrow \log | \sin y | + \log | \sin x | = \log c$

$\Rightarrow \log | \sin y . \sin x | = \log c$

$\Rightarrow \sin y \sin x = c$

Question:58

The solution of $x \frac{d y}{d x} + y = e^{x}$ is
(a) $y = \frac{e^{x}}{x} + \frac{k}{x}$
(b) $y = x e^{x} + c x$
(c) $y = x e^{x} + k$
(d) $x = \frac{e^{y}}{y} + \frac{k}{y}$

Answer:

The answer is the option (a) $y = \frac{e^{x}}{x} + \frac{k}{x}$
Explanation: -
$\Rightarrow \frac{d y}{d x} + \frac{y}{x} = \frac{e^{x}}{x}$
This is a linear differential equation. Dn comparing it with dy/dx+Py=Q, we get
$P = \frac{1}{x} and Q = \frac{e^{x}}{x}$
$∴ IF = e^{\int \frac{1}{x} d x} = e^{(\log x)} = x$
So, the general solution is:
$y \cdot x = \int \frac{e^{x}}{x} x d x$

$\Rightarrow y \cdot x = \int e^{x} d x$

$\Rightarrow y \cdot x = e^{x} + k$

$\Rightarrow$ $y = \frac{e^{x}}{x} + \frac{k}{x}$

Question:59

The differential equation of the family of curves $x^{2} + y^{2} - 2 a y = 0,$ where a is arbitrary constant, is
(a) $(x^{2} - y^{2}) \frac{d y}{d x} = 2 x y$
(b) $2 (x^{2} + y^{2}) \frac{d y}{d x} = x y$
(c) $2 (x^{2} - y^{2}) \frac{d y}{d x} = x y$
(d) $(x^{2} + y^{2}) \frac{d y}{d x} = 2 x y$

Answer:

The answer is the option (a) $(x^{2} - y^{2}) \frac{d y}{d x} = 2 x y$
Given:
$x^{2} + y^{2} - 2 a y = 0. . . . . (i)$
$2 x + 2 y \frac{d y}{d x} - 2 a \frac{d y}{d x} = 0$

$\Rightarrow a = \frac{x + y \frac{d y}{d x}}{\frac{d y}{d x}}$
Put the value of 'a' in Eq. (i) $x^{2} + y^{2} - 2 y \frac{x + y \frac{d y}{d x}}{\frac{d y}{d x}} = 0$

$\Rightarrow (x^{2} + y^{2}) \frac{d y}{d x} - 2 x y - 2 y^{2} \frac{d y}{d x} = 0$

$\Rightarrow (x^{2} - y^{2}) \frac{d y}{d x} - 2 x y = 0$

Question:60

Family y = Ax + A³ of curves will correspond to a differential equation of order ,
(a) 3 (b) 2 (c) 1 (d) not defined.

Answer:

The answer is the option (c) 1.
Explanation: -
$\Rightarrow \frac{d y}{d x} = A$
Putting the value of A in Eq. (i), we gt
$y = x \frac{d y}{d x} + {(\frac{d y}{d x})}^{3}$
$∴$ Order $= 1$

Question:61

The general solution of $\frac{d y}{d x} = 2 x e^{x^{2} - y}$ is
(a) $e^{x^{2} - y} = c$
(b) $e^{- y} + e^{x^{2}}$
(c) $e^{y} = e^{x^{2}} + c$
(d) $e^{x^{2} + y} = c$

Answer:

The answer is the option (c)
Explanation: -
$\begin{aligned} \begin{array}{ll} \Rightarrow & e^{y} \frac{d y}{d x} = 2 x e^{x^{2}} \\ \Rightarrow & \int e^{y} d y = 2 \int x e^{x^{2}} d x \end{array} \\ Put x^{2} = t in R . H.S. integral, we get \\ 2 x d x = d t \\ \begin{array}{ll} \Rightarrow & \int e^{y} d y = \int e^{t} d t \\ \Rightarrow & e^{y} = e^{t} + C \\ \Rightarrow & e^{y} = e^{x^{2}} + C \end{array} \end{aligned}$

Question:62

The curve for which the slope of the tangent at any point is equal to the ratio of the abscissa to the ordinate of the point is
(a) an ellipse (b) parabola (c) circle (d) rectangular hyperbola

Answer:

The answer is the option (d) Rectangular Hyperbola
Explanation: -
According to the question, $\frac{d y}{d x} = \frac{x}{y}$

$\Rightarrow y d y = x d x$
On integrating both sides, we get
$\frac{y^{2}}{2} = \frac{x^{2}}{2} + C$
$\Rightarrow y^{2} - x^{2} = 2 C,$ which is an equation of rectangular hyperbola.

Question:63

The general solution of differential equation $\frac{d y}{d x} = e^{\frac{x^{2}}{2}} + x y$ is
(a) $y = C e^{- x^{2} / 2}$
(b) $y = C e^{x^{2} / 2}$
(c) $y = (x + C) e^{x^{2} / 2}$
(d) $y = (C - x) e^{x^{2} / 2}$

Answer:

The answer is the option (c)
Explanation: -
$\Rightarrow \frac{d y}{d x} - x y = e^{\frac{x^{2}}{2}}$
This is a linear differential equation. On comparing it with $\frac{d y}{d x} + P y = Q,$ we get
$P = - x, O = e^{x^{2} / 2}$
$∴ I . F . = e^{\int - x d x} = e^{- x^{2} / 2}$
So, the general solution is:
$∴ y \cdot e^{- x^{2} / 2} = \int e^{- x^{2} / 2} e^{x^{2} / 2} d x + C$

$\Rightarrow y e^{- x^{2} / 2} = \int 1 d x + C$

$\Rightarrow y e^{- x^{2} / 2} = x + C$

$\Rightarrow y = (x + C) e^{x^{2} / 2}$

Question:64

The solution of equation $(2 y - 1) d x - (2 x + 3) d y = 0$ is
(a) $\frac{2 x - 1}{2 y + 3} = k$
(b) $\frac{2 y + 1}{2 x - 3} = k$
(c) $\frac{2 x + 3}{2 y - 1} = k$
(d) $\frac{2 x - 1}{2 y - 1} = k$
Answer:

The answer is the option (c)
Explanation: -
$\begin{aligned} \begin{array}{ll} \Rightarrow & (2 y - 1) d x = (2 x + 3) d y \\ \Rightarrow & \frac{d x}{2 x + 3} = \frac{d y}{2 y - 1} \end{array} \\ On integrating both sides, we get \\ \frac{1}{2} \log (2 x + 3) = \frac{1}{2} \log (2 y - 1) + \log C \\ \begin{array}{ll} \Rightarrow & [\log (2 x + 3) - \log (2 y - 1)] = 2 \log C \\ \Rightarrow & \log (\frac{2 x + 3}{2 y - 1}) = \log C^{2} \\ \Rightarrow & \frac{2 x + 3}{2 y - 1} = C^{2} \\ \Rightarrow & \frac{2 x + 3}{2 y - 1} = k, where K = C^{2} \end{array} \end{aligned}$

Question:65

The differential equation for which $y = a \cos x + b \sin x$ is a solution, is
(a) $\frac{d^{2} y}{d x^{2}} + y = 0$
(b) $\frac{d^{2} y}{d x^{2}} - y = 0$
(c) $\frac{d^{2}}{d x^{2}} + (a + b) y = 0$
(d) $\frac{d^{2} y}{d x^{2}} + (a - b) y = 0$

Answer:

The answer is the option (a) $\frac{d^{2} y}{d x^{2}} + y = 0$
Explanation: -
On differentiating both sides w.r.t. x, we get $\frac{d y}{d x} = - a \sin x + b \cos x$
Again, differentiating w.r.t. x, we get $\frac{d^{2} y}{d x^{2}} = - a \cos x - b \sin x$
$\Rightarrow \frac{d^{2} y}{d x^{2}} = - y$

$\Rightarrow \frac{d^{2} y}{d x^{2}} + y = 0$

Question:66

The solution of $\frac{d y}{d x} + y = e^{- x}, y (0) = 0$ is
(a) $y = e^{- x} (x - 1)$
(b) $y = x e^{x}$
(c) $y = x e^{- x} + 1$
(d) $y = x e^{- x}$

Answer:

The answer is the option (d) $y = x e^{- x}$
Explanation: -
$\frac{d y}{d x} + y = e^{- x}, y (0) = 0$
Here, $P = 1$ and $Q = e^{- x}$

$I . F . = e^{\int ldx} = e^{x}$
$∴$ The general solution is $y \cdot e^{x} = \int e^{- x} \cdot e^{x} d x + C$

$\Rightarrow y e^{x} = \int d x + C$

$\Rightarrow$ $y e^{x} = x + C$
Given, when x=0 and y=0
$\Rightarrow$ $0 = 0 + C$

$\Rightarrow C = 0$
Eq. (i) reduces to $y \cdot e^{x} = x$ or $y = x e^{- x}$

Question:67

The order and degree of the differential equation
${(\frac{d^{3} y}{d x^{3}})}^{2} - 3 \frac{d^{2} y}{d x^{2}} + 2 {(\frac{d y}{d x})}^{4} = y^{4} are$
(a) 1,4
(b) 3,4
(c) 2,4
(d) 3,2

Answer:

Ans: - The answer is the option (d) 3, 2

Question:68

The order and degree of the differential equation $[1 + {(\frac{d y}{d x})}^{2}] = \frac{d^{2} y}{d x^{2}}$ are
(a) $2, \frac{3}{2}$
(b) 2,3
(c) 2,1
(d) 3,4

Answer:

Ans: -
The answer is the option (c) 2, 1.

Question:69

The differential equation of family of curves $y^{2} = 4 a (x + a)$ is

(a) $y^{2} = 4 \frac{d y}{d x} (x + \frac{d y}{d x})$
(b) $2 y \frac{d y}{d x} = 4 a$
(c) $\frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} = 0$
(d) $2 x \frac{d y}{d x} + y {(\frac{d y}{d x})}^{2} - y = 0$

Answer:

Ans: - The answer is the option (d) $2 x \frac{d y}{d x} + y {(\frac{d y}{d x})}^{2} - y = 0$
Explanation: -
$y^{2} = 4 a (x + a)$
On differentiating both sides w.r.t. x, we get
$2 y \frac{d y}{d x} = 4 a$
$\Rightarrow \frac{1}{2} y \frac{d y}{d x} = a$
On putting the value of a in Eq. (i), we get
$y^{2} = 2 y \frac{d y}{d x} (x + \frac{1}{2} y \frac{d y}{d x})$
$\Rightarrow y^{2} = 2 x y \frac{d y}{d x} + y^{2} {(\frac{d y}{d x})}^{2}$
$\Rightarrow 2 x \frac{d y}{d x} + y {(\frac{d y}{d x})}^{2} - y = 0$

Question:70

Which of the following is the general solution of $\frac{d^{2} y}{d x^{2}} - 2 (\frac{d y}{d x}) + y = 0$ ?
(a) $y = (A x + B) e^{x}$
(b) $y = (A x + B) e^{- x}$
(c) $y = A e^{x} + B e^{- x}$
(d) $y = A \cos x + B \sin x$

Answer:

Ans: -
The answer is the option (a) $y = (A x + B) e^{x}$
Explanation: -
$\begin{aligned} \frac{d y}{d x} & = (A x + B) e^{x} + A e^{x} = (A x + A + B) e^{x} \\ \Rightarrow & \frac{d^{2} y}{d x^{2}} = (A x + A + B) e^{x} + A e^{x} = (A x + 2 A + B) e^{x} \\ ∴ & \frac{d^{2} y}{d x^{2}} - 2 (\frac{d y}{d x}) + y = (A x + 2 A + B) e^{x} - 2 (A x + A + B) e^{x} + (A x + B) e^{x} = 0 \end{aligned}$

Question:71

General solution of $\frac{d y}{d x} + y \tan x = \sec x$ is
(a) $y \sec x = \tan x + C$
(b) $y \tan x = \sec x + C$
(c) $\tan x = y \tan x + C$
(d) $x \sec x = \tan y + C$

Answer:

Ans: - The answer is the option (a) y sec x = tan x + C
Explanation: -
Here, $P = \tan x, Q = \sec x$
$∴$ I.F. $= e^{\int \tan x d x} = e^{\int \log \sec x} = \sec x$

The general solution is $y \sec x = \int \sec x \cdot \sec x + C$

$\Rightarrow$ $y \sec x = \int \sec^{2} x d x + C$

$\Rightarrow$ $y \sec x = \tan x + C$

Question:72

Solution of the differential equation $\frac{d y}{d x} + \frac{1}{x} y = \sin x$ is
(a) $x (y + \cos x) = \sin x + C$
(b) $x (y - \cos x) = \sin x + C$
(c) $x y \cos x = \sin x + C$
(d) $x (y + \cos x) = \cos x + C$

Answer:

Ans: - The answer is the option (a) $x (y + \cos x) = \sin x + C$
Explanation: -
$\frac{d y}{d x} + \frac{1}{x} y = \sin x$
Here, $P = \frac{1}{x}$ and $Q = \sin x$
$∴ I . F, = e^{\int \frac{1}{x} d x} = e^{\log x} = x$

$∴$ The general solution is $y \cdot x = \int x \cdot \sin x d x + C$
$\begin{aligned} = - x \cos x - \int - \cos x d x \\ = - x \cos x + \sin x + c \\ \Rightarrow x (y + \cos x) & = \sin x + C \end{aligned}$

Question:73

The general solution of differential equation $(e^{x} + 1) y d y = (y + 1) e^{x} d x$ is
(a) $(y + 1) = k (e^{x} + 1)$
(b) $y + 1 = e^{x} + 1 + k$
(c) $y = \log {k (y + 1) (e^{x} + 1)}$
(d) $y = \log {\frac{e^{x} + 1}{y + 1}} + k$

Answer:

Ans: - The answer is the option (c) $y = \log {k (y + 1) (e^{x} + 1)}$
Explanation: -
$(e^{x} + 1) y d y = (y + 1) e^{x} d x$
$\Rightarrow \frac{y d y}{y + 1} = \frac{e^{x}}{e^{x} + 1} d x$

$\Rightarrow \int (1 - \frac{1}{y + 1}) d y = \int \frac{e^{x}}{e^{x} + 1} d x$

$\Rightarrow y - \log (y + 1) = \log (e^{x} + 1) + \log k$

$\Rightarrow y = \log (y + 1) + \log (1 + e^{x}) + \log k$

$\Rightarrow y = \log (k (1 + y) (1 + e^{x}))$

Question:74

The solution of the differential equation $\frac{d y}{d x} = e^{x - y} + x^{2} e^{- y}$ is
(a) $y = e^{x - y} - x^{2} e^{- y} + c$
(b) $e^{y} - e^{x} = \frac{x^{3}}{3} + c$
(c) $e^{x} + e^{y} = \frac{x^{3}}{3} + c$
(d) $e^{x} - e^{y} = \frac{x^{3}}{3} + c$

Answer:

The answer is the option (b) $e^{y} - e^{x} = \frac{x^{3}}{3} + c$
Explaination:
$\frac{d y}{d x} = e^{x - y} + x^{2} e^{- y}$

$\Rightarrow e^{y} d y = (e^{x} + x^{2}) d x$

$\Rightarrow \int e^{y} d y = \int (e^{x} + x^{2}) d x$

$\Rightarrow e^{y} = e^{x} + \frac{x^{3}}{3} + C$

$\Rightarrow e^{y} - e^{x} = \frac{x^{3}}{3} + C$

Question:75

The solution of the differential equation $\frac{d y}{d x} + \frac{2 x y}{1 + x^{2}} = \frac{1}{{(1 + x^{2})}^{2}}$ is
(a) $y (1 + x^{2}) = C + \tan^{- 1} x$
(b) $\frac{y}{1 + x^{2}} = C + \tan^{- 1} x$
(c) $y \log (1 + x^{2}) = C + \tan^{- 1} x$
(d) $y (1 + x^{2}) = C + \sin^{- 1} x$

Answer:

Ans: - The answer is the option (a) $y (1 + x^{2}) = C + \tan^{- 1} x$
Explanation: -
$\frac{d y}{d x} + \frac{2 x y}{1 + x^{2}} = \frac{1}{{(1 + x^{2})}^{2}}$ is
Here, $P = \frac{2 x}{1 + x^{2}}$ and $Q = \frac{1}{{(1 + x^{2})}^{2}}$
$∴$ I.F. $= e^{\int \frac{2 x}{1 + x^{2}} d x} = e^{\log (1 + x^{2})} = 1 + x^{2}$

$∴$ The general solution is
$y (1 + x^{2}) = \int (1 + x^{2}) \frac{1}{{(1 + x^{2})}^{2}} + C$
$\Rightarrow y (1 + x^{2}) = \int \frac{1}{1 + x^{2}} d x + C$

$\Rightarrow$ $y (1 + x^{2}) = \tan^{- 1} x + C$

Question:76

(i)The degree of the differential equation $\frac{d^{2} y}{d x^{2}} + e^{\frac{d y}{d x}} = 0$ is
(ii)The degree of the differential equation $\sqrt{1 + {(\frac{d y}{d x})}^{2}} = x$ is

(iii)The number of arbitrary constants in the general solution of a differential equation of order three is

(iv) $\frac{d y}{d x} + \frac{y}{x \log x} = \frac{1}{x}$ is an equation of the type

(v)General solution of the differential equation of the type $\frac{d y}{d x} + P y = Q$ is given by
(vi)The solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2}$ is
(vii)The solution of $(1 + x^{2}) \frac{d y}{d x} + 2 x y - 4 x^{2} = 0$ is
(viii)The solution of the differential equation $y d x + (x + x y) d y = 0$ is
(ix) General solution of $\frac{d y}{d x} + y = \sin x$ is
(x)The solution of differential equation $\cot y d x = x d y$ is
(xi)The integrating factor of $\frac{d y}{d x} + y = \frac{1 + y}{x}$ is

Answer:

(i) Given differential equation is
$\frac{d^{2} y}{d x^{2}} + e^{\frac{d y}{d x}} = 0$

Degree of this equation is not defined as it cannot be expresses as polynomial of derivatives.
(ii) We have $\sqrt{1 + {(\frac{d y}{d x})}^{2}} = x$
$\Rightarrow 1 + {(\frac{d y}{d x})}^{2} = x^{2}$
So, degree of this equation is two.
(iii) Given that the general solution of a differential equation has three arbitrary constants. So we require three more equations to eliminate these three constants. We can get three more equations by differentiating the given equation three times. So, the order of the differential equation is three.
(iv) We have $\frac{d y}{d x} + \frac{y}{x \log x} = \frac{1}{x}$
The equation is of the type $\frac{d y}{d x} + P y = Q$
Hence it is a linear differential equation.
(v) We have $\frac{d x}{d y} + P_{1} x = Q_{1}$
To solve such equations we multiply both sides by
So we get $e^{\int P_{1} d y} (\frac{d x}{d y} + P_{1} x) = Q_{1} e^{\int P_{1} d y}$
$\Rightarrow \frac{d x}{d y} e^{\int P_{1} d y} + P_{1} e^{\int P_{1} d y} = Q_{1} e^{\int P_{1} d y}$

$\Rightarrow \frac{d}{d y} (x e^{\int P_{1} d y}) = Q_{1} e^{\int P_{1} d y}$ $\Rightarrow \int \frac{d}{d y} (x e^{\int P_{1} d y}) d y = \int Q_{1} e^{P_{1} d y} d y$

$\Rightarrow x e^{\int P_{1} d y} = \int Q_{1} e^{\int P_{1} d y} d y + C$
This is the required solution of the given differential equation.
(vi) We have, $x \frac{d y}{d x} + 2 y = x^{2}$
$\frac{d y}{d x} + \frac{2 y}{x} = x$
This equation of the form $\frac{d y}{d x} + P y = Q$ .
$∴$ I.F. $= e^{\int \frac{2}{x} d x} = e^{2 \log x} = x^{2}$
The general solution is
$y x^{2} = \int x \cdot x^{2} d x + C$
$\Rightarrow y x^{2} = \frac{x^{4}}{4} + C$ \ $\Rightarrow y = \frac{x^{2}}{4} + C x^{- 2}$
(vii) We have $(1 + x^{2}) \frac{d y}{d x} + 2 x y - 4 x^{2} = 0$
$\Rightarrow \frac{d y}{d x} + \frac{2 x}{1 + x^{2}} y = \frac{4 x^{2}}{1 + x^{2}}$
This equation is of the form $\frac{d y}{d x} + P y = Q$ .
$∴ I . F . = e^{\int \frac{2 x}{1 + x^{2}} d x} = e^{\log (1 + x^{2})} = 1 + x^{2}$
So, the general solution is:
$\begin{array}{ll} y \cdot (1 + x^{2}) = \int (1 + x^{2}) \frac{4 x^{2}}{(1 + x^{2})} d x + C \\ \Rightarrow & (1 + x^{2}) y = \int 4 x^{2} d x + C \\ \Rightarrow & (1 + x^{2}) y = \frac{4 x^{3}}{3} + C \\ \Rightarrow & y = \frac{4 x^{3}}{3 (1 + x^{2})} + C {(1 + x^{2})}^{- 1} \end{array}$
(viii) We have, $y d x + (x + x y) d y = 0$

$\Rightarrow y d x + x (1 + y) d y = 0$
$\Rightarrow$ $\frac{d x}{- x} = (\frac{1 + y}{y}) d y$

$\Rightarrow \int \frac{1}{x} d x = - \int (\frac{1}{y} + 1) d y$

$\Rightarrow \log x = - \log y - y + \log C$

$\Rightarrow \log x + \log y - \log C = - y$

$\Rightarrow \log \frac{x y}{C} = - y$

$\Rightarrow \frac{x y}{C} = e^{- y}$
$\Rightarrow x y = C e^{- y}$
(ix) We have, $\frac{d y}{d x} + y = \sin x$
Which is of the form $\frac{d y}{d x} + P y = Q$
$I.F. = e^{\int 1 d x} = e^{x}$
So, the general solution is:
$\begin{array}{ll} y \cdot e^{x} = \int e^{x} \sin x d x + C \\ \Rightarrow & y e^{x} = \frac{1}{2} e^{x} (\sin x - \cos x) + C \\ \Rightarrow & y = \frac{1}{2} (\sin x - \cos x) + C e^{- x} \end{array}$
(x) Given differential equation is $\cot y d x = x d y$
$\Rightarrow \int \frac{1}{x} d x = \int \tan y d y$

$\Rightarrow \log x = \log \sec y + \log C$
$\Rightarrow$ $\log \frac{x}{\sec y} = \log C$

$\Rightarrow \frac{x}{\sec y} = C$

$\Rightarrow$ $x = C \sec y$
(xi) Given differential equation is
$\begin{array}{l} \frac{d y}{d x} + y = \frac{1 + y}{x} \\ \frac{d y}{d x} + y = \frac{1}{x} + \frac{y}{x} \end{array}$
$\Rightarrow \frac{d y}{d x} + y (1 - \frac{1}{x}) = \frac{1}{x}$
Which is a linear differential equation.
$∴ I.F. = e^{\int (1 - \frac{1}{x}) d x} = e^{x - \log x} = e^{x} \cdot e^{- \log x} = \frac{e^{x}}{x}$

Question:77

Answer:

i) Integrating factor of the differential of the form $\frac{d x}{d y} + P_{1} x = Q_{1}$ is given by $e^{\int P_{1} d y}$ .Hence given statement is true.
(ii) Solution of the differential equation of the type $\frac{d x}{d y^{'}} + P_{1} x = Q_{1}$ is given by $x e^{\int P_{1} d y} = \int Q_{1} e^{\int P_{1} d y} d y + C$ .
Hence given statement is true.
iii) Correct substitution for the solution of the differential equation of the type $\frac{d y}{d x} f (x, y),$ where $f (x, y)$ is a homogeneous function of zero degree is $y = v x .$
Hence given statement is true.
(iv) Correct substitution for the solution of the differential equation of the type $\frac{d x}{d y} g (x, y)$ where $g (x, y)$ is a homogeneous function of the degree zero is $x = v y .$
Hence given statement is true.
(V) There is no arbitrary constants in the particular solution of a differential equation. Hence given statement is False.
(vi) In thegiven equation $x^{2} + (y - a)^{2} =$ $a^{2}$ the number of arbitrary constant is one. So the order order will be one.
Hence given statement is False.
(vii) $\frac{d y}{d x} = {(\frac{y}{x})}^{1 / 3}$
$\frac{d y}{y \frac{1}{3}} = (\frac{d x}{x^{\frac{1}{3}}})$
$\int \frac{d y}{y \frac{1}{3}} = \int (\frac{d x}{x^{\frac{1}{3}}})$
$\begin{array}{l} \frac{3}{2} y^{2 / 3} = \frac{3}{2} x^{2 / 3} + C^{'} \\ y^{2 / 3} - x^{2 / 3} = C \end{array}$
Hence the given statement is true.
(viii) $y = e^{x} (A \cos x +$ $B \sin x$ )
$\frac{d y}{d x} = e^{x} (- A \sin x + B \cos x) + e^{x} (A \sin x + B \cos x)$
$\frac{d y}{d x} = e^{x} (- A \sin x + B \cos x) + y$
$\frac{d^{2} y}{d x^{2}} = e^{x} (- A \sin x + B \cos x) + e^{x} (- A \cos x - B \sin x) + \frac{d y}{d x}$
$\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 2 y = 0$ .
Hence the given statement is true.
ix) Given: $\frac{d y}{d x} = \frac{x + 2 y}{x}$
$\frac{d y}{d x} - \frac{2}{x} y = 1$
Compare with $\frac{d y}{d x} + P_{1} y = Q_{1}$
Here $P_{1} = \frac{- 2}{x}$ , $Q_{1} = 1$
$I . F . = e^{\int \frac{- 2}{x} d x} = e^{\log \frac{1}{x^{2}}} = \frac{1}{x^{2}}$
General solution
$y . \frac{1}{x^{2}} = \int \frac{1}{x^{2}} d x$
$\frac{y}{x^{2}} = \frac{- 1}{x} + c$
$y + x = c x^{2}$
Hence the given statement is true.
x) Given: $\frac{x d y}{d x} = y + x \tan (\frac{y}{x})$
$\frac{d y}{d x} = \frac{y}{x} + \tan (\frac{y}{x})$
Let y =vx
$\frac{d y}{d x} = v + x \frac{d v}{d x}$
$v + x \frac{d v}{d x} = v + \tan v$
$x \frac{d v}{d x} = \tan v$
$\int \frac{d v}{\tan v} = \int \frac{d x}{x}$
$\log \sin v = \log x + \log c$
$\sin v = x c$
$\sin \frac{y}{x} = c x$
Hence the given statement is true.
xi) Assume the equation of a non-horizontal line in the plane
$y = m x + c$
$\frac{d y}{d x} = m$
$\frac{d x}{d y} = \frac{1}{m}$
$\begin{aligned} \frac{d^{2} x}{d y^{2}} = 0 \end{aligned}$
Hence the given statement is true.

Main Subtopics of NCERT Exemplar Class 12 Maths Solutions Chapter 9

Below is the list of topics which are covered in Class 12 Maths NCERT exemplar solutions chapter 9

Introductory Concepts
Ordinary differential equations
Order of a differential equation
Degree of a differential equation
General and particular solutions of a differential equation
Formation of a differential equation
Formation of a differential equation whose general solution is given
Formation of a differential equation that will represent a given family of curves
Methods of solving First order, First degree differential equations
Differential equations with separate variables
Homogeneous Differential equations
Linear differential equations

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Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 9

These equations have a variety of uses in academic subjects like Physics, Chemistry, Biology, Geology, etc., which makes it important to acquire detailed learning of these equations.
A thorough understanding of differential equations is needed to solve Newton's laws of motion and cooling, the rate of spread of a pandemic or an epidemic, and also help measure market competition.
If understood well, NCERT exemplar solutions for Class 12 Maths chapter 9 would help you answer real-world questions like - How do you model an antibiotic-resistant bacteria's growth? How do you study ever-changing online purchasing trends? At what rate, a radioactive material decays? What is the trajectory of a biological cell motion? How the suspension system of a car works to give you a smooth ride? These equations are a way to describe many things in the universe and model nearly anything around us. Scientists and geniuses understand the world through differential equations; you can too.

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Other NCERT Exemplar Class 12 Mathematics Chapters

NCERT Exemplar Class 12 Mathematics Chapter 1: Relations and Functions

NCERT Exemplar Class 12 Mathematics Chapter 2: Inverse Trigonometric Functions

NCERT Exemplar Class 12 Mathematics Chapter 3: Matrices

NCERT Exemplar Class 12 Mathematics Chapter 4: determinants

NCERT Exemplar Class 12 Mathematics Chapter 5: Continuity and Differentiability

NCERT Exemplar Class 12 Mathematics Chapter 6: Applications of derivatives

NCERT Exemplar for Class 12 Mathematics Chapter 7: Integrals

NCERT Exemplar Class 12 Mathematics Chapter 8: Applications of Integrals

NCERT Exemplar Class 12 Mathematics Chapter 9: Differential equations

NCERT Exemplar Class 12 Mathematics Chapter 10: Vector Algebra

NCERT Exemplar Class 12 Mathematics Chapter 11: Three-Dimensional Geometry

NCERT Exemplar Class 12 Mathematics Chapter 12: Linear Programming

NCERT Exemplar Class 12 Mathematics Chapter 13: Probability

Important Topics To Cover For Exams From NCERT Exemplar Class 12 Maths Solutions Chapter 9

NCERT exemplar Class 12 Maths chapter 9 solutions revolve around using the mathematical tool of Differentiation, analyses properties like the intervals of increment and decrement, study the local maximum and minimum of functions that are quadratic.
In Class 12 Maths NCERT exemplar solutions chapter 9, you will be introduced to the concepts related to differential equations, types of these equations, general and specific solutions to solve them, the formation and production of these equations, different forms of the equations, and a wide range of applications of modelling real-life situations by applying these equations
In NCERT exemplar Class 12 Maths solutions chapter 9 pdf download, we would also look at the graphical aspects of differential equations, including a family of straight lines and curves, and have a look at the devised solutions and mathematical tools to solve the most complex equations over time.

NCERT Exemplar Class 12 Solutions

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NCERT solutions for class 12 maths - Chapter-wise

NCERT Solutions for Class 12 Maths: Chapter Wise

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions

NCERT solutions for class 12 maths chapter 3 Matrices

NCERT solutions for class 12 maths chapter 4 Determinants

NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability

NCERT solutions for class 12 maths chapter 6 Application of Derivatives

NCERT solutions for class 12 maths chapter 7 Integrals

NCERT solutions for class 12 maths chapter 8 Application of Integrals

NCERT solutions for class 12 maths chapter 9 Differential Equations

NCERT solutions for class 12 maths chapter 10 Vector Algebra

NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry

NCERT solutions for class 12 Maths chapter 12 Linear Programming

NCERT solutions for class 12 maths chapter 13 Probability

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Frequently Asked Questions (FAQs)

1. Are these solutions helpful in competitive exams?

Yes, these NCERT exemplar Class 12 Maths solutions chapter 9 can be highly useful in understanding the way the questions should be solved in entrance exams.

2. How to make use of these NCERT exemplar Class 12 Maths chapter 9 solutions?

These solutions can be used for both getting used to the chapter and its topics and to also get an idea about how to solve questions in exams.

3. What are the basic take away from the Class 12 Maths NCERT exemplar solutions chapter 9?

One can understand how to stepwise solve these questions through NCERT exemplar Class 12 Maths solutions chapter 9 and how the CBSE expects a student to solve in their final paper.

4. Who prepare these solutions to the maths chapter?

We have the best maths teachers onboard to solve the questions as per the students understanding and also CBSE standards. These teachers prepare the NCERT exemplar solutions for Class 12 Maths chapter 9.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Can I change my board from CBSE to CHSE Odisha in Class 12th?

Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

Read Complete Answer

Manasvini Gupta 30 January,2025

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

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Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

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Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer

Ashvadha 12 July,2024

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Exemplar Class 12 Maths Solutions Chapter 9 Differential Equations

NCERT Exemplar Class 12 Maths Solutions Chapter 9 Differential Equations

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Main Subtopics of NCERT Exemplar Class 12 Maths Solutions Chapter 9

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JEE Main Important Mathematics Formulas

Other NCERT Exemplar Class 12 Mathematics Chapters

Important Topics To Cover For Exams From NCERT Exemplar Class 12 Maths Solutions Chapter 9

NCERT Exemplar Class 12 Solutions

NCERT solutions for class 12 maths - Chapter-wise

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