NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.3 - Differential Equations

NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.3 - Differential Equations

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CBSE Class 12th Exam Date:17 Feb' 26 - 17 Feb' 26

Komal MiglaniUpdated on 08 May 2025, 02:26 PM IST

Consider a tank with a hole at the bottom. The exit rate of water is a function of the water level in the tank. As the water level reduces, the flow rate reduces. This can be represented with a differential equation.

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  1. Class 12 Maths Chapter 9 Exercise 9.3 Solutions: Download PDF
  2. NCERT Solutions Class 12 Maths Chapter 9: Exercise 9.3
  3. Topics covered in Chapter 9 Differential Equation: Exercise 9.3
  4. NCERT Solutions Subject Wise
  5. Subject Wise NCERT Exemplar Solutions

The NCERT Solutions for Class 12 Maths Chapter 9 – Exercise 9.3 teach students exactly how to determine solutions to such equations. These are designed by subject matter experts from Careers360 as per the new CBSE 2025–26 syllabus and are described in depth in step-by-step manner. NCERT Chapter 9 Exercise 9.3 is differential equations under the category of variable separable, where the variables can be rearranged such that all the terms of one variable are placed on one side and the remaining terms are placed on the other side. Students will also encounter special cases where the equation needs to be transformed into separable form prior to using the method.

Class 12 Maths Chapter 9 Exercise 9.3 Solutions: Download PDF

This material provides easy solutions to all the questions of Exercise 9.3 of Differential Equations. The PDF can be downloaded by the students for practice and enhancement of the chapter for board and and competitive exams

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NCERT Solutions Class 12 Maths Chapter 9: Exercise 9.3

Question 1: Find the general solution: $\frac{dy}{dx} = \frac{1-\cos x}{1 + \cos x}$

Answer:

Given,

$\frac{dy}{dx} = \frac{1-\cos x}{1 + \cos x}$

$\\ \implies\frac{dy}{dx} = \frac{2sin^2\frac{x}{2}}{2cos^2\frac{x}{2}} = tan^2\frac{x}{2} \\$

$\implies dy = (sec^2\frac{x}{2} - 1)dx$

$\\ \implies \int dy = \int sec^2\frac{x}{2}dx - \int dx \\$

$\implies y = 2tan^{-1}\frac{x}{2} - x + C$

Question 2: Find the general solution: $\frac{dy}{dx} = \sqrt{4-y^2}\ (-2 < y < 2)$

Answer:

Given, in the question

$\frac{dy}{dx} = \sqrt{4-y^2}$

$\\ \implies \frac{dy}{\sqrt{4-y^2}} = dx \\ \implies \int \frac{dy}{\sqrt{4-y^2}} = \int dx$

$\\ (\int \frac{dy}{\sqrt{a^2-y^2}} = sin^{-1}\frac{y}{a})\\$

The required general solution:

$\\ \implies sin^{-1}\frac{y}{2} = x + C$

Question 3: Find the general solution: $\frac{dy}{dx} + y = 1 (y\neq 1)$

Answer:

Given, in the question

$\frac{dy}{dx} + y = 1$

$\\ \implies \frac{dy}{dx} = 1- y \\$

$\implies \int\frac{dy}{1-y} = \int dx$

$(\int\frac{dx}{x} = lnx)$

$\\ \implies -log(1-y) = x + C\ \ (We\ can\ write\ C= log k) \\$

$\implies log k(1-y) = -x \\$

$\implies 1- y = \frac{1}{k}e^{-x} \\$

The required general equation

$\implies y = 1 -\frac{1}{k}e^{-x}$

Question 4: Find the general solution: $\sec^2 x \tan y dx + \sec^2 y \tan x dy = 0$

Answer:

Given,

$\sec^2 x \tan y dx + \sec^2 y \tan x dy = 0$

$\\ \implies \frac{sec^2 y}{tan y}dy = -\frac{sec^2 x}{tan x}dx \\$

$\implies \int \frac{sec^2 y}{tan y}dy = - \int \frac{sec^2 x}{tan x}dx$

Now, let tany = t and tanx = u

$sec^2 y dy = dt\ and\ sec^2 x dx = du$

$\\ \implies \int \frac{dt}{t} = -\int \frac{du}{u} \\$

$\implies log t = -log u +logk \\ \implies t = \frac{1}{ku} \\$

$\implies tany = \frac{1}{ktanx}$

Question 5: Find the general solution:

$(e^x + e^{-x})dy - (e^x - e^{-x})dx = 0$

Answer:

Given, in the question

$(e^x + e^{-x})dy - (e^x - e^{-x})dx = 0$

$\\ \implies dy = \frac{(e^x - e^{-x})}{(e^x + e^{-x})}dx$

Let,

$\\ (e^x + e^{-x}) = t \\$

$\implies (e^x - e^{-x})dx = dt$

$\\ \implies \int dy = \int \frac{dt}{t} \\$

$\implies y = log t + C \\ $

$\implies y = log(e^x + e^{-x}) + C$

This is the general solution

Question 6: Find the general solution: $\frac{dy}{dx} = (1+x^2)(1+y^2)$

Answer:

Given, in the question

$\frac{dy}{dx} = (1+x^2)(1+y^2)$

$\\ \implies \int \frac{dy}{(1+y^2)} = \int (1+x^2)dx$

$(\int \frac{dx}{(1+x^2)} =tan^{-1}x +c)$

$\\ \implies tan^{-1}y = x+\frac{x^3}{3} + C$

Question 7: Find the general solution: $y\log y dx - x dy = 0$

Answer:

Given,

$y\log y dx - x dy = 0$

$\\ \implies \frac{1}{ylog y}dy = \frac{1}{x}dx$

let logy = t

=> 1/ydy = dt

$\\ \implies \int \frac{dt}{t} = \int \frac{1}{x}dx \\$

$\implies \log t = \log x + \log k \\$

$\implies t = kx \\$

$\implies \log y = kx$

This is the general solution

Question 8: Find the general solution: $x^5\frac{dy}{dx} = - y^5$

Answer:

Given, in the question

$x^5\frac{dy}{dx} = - y^5$

$\\ \implies \int \frac{dy}{y^5} = - \int \frac{dx}{x^5} \\$

$\implies \frac{y^{-4}}{-4} = -\frac{x^{-4}}{-4} + C \\$

$\implies \frac{1}{y^4} + \frac{1}{x^4} = C$

This is the required general equation.

Question 9: Find the general solution: $\frac{dy}{dx} = \sin^{-1}x$

Answer:

Given, in the question

$\frac{dy}{dx} = \sin^{-1}x$

$\implies \int dy = \int \sin^{-1}xdx$

Now,

$\int (u.v)dx = u\int vdx - \int(\frac{du}{dx}.\int vdx)dx$

Here, u = $\sin^{-1}x$ and v = 1

$\implies y = \sin^{-1}x .x - \int(\frac{1}{\sqrt{1-x^2}}.x)dx$

$\\ Let\ 1- x^2 = t \\$

$\implies -2xdx = dt \implies xdx = -dt/2$

$\\ \implies y = x\sin^{-1}x+ \int(\frac{dt}{2\sqrt{t}}) \\ $

$\implies y = x\sin^{-1}x + \frac{1}{2}.2\sqrt{t} + C \\$

$\implies y = x\sin^{-1}x + \sqrt{1-x^2} + C$

Question 10: Find the general solution $e^x\tan y dx + (1-e^x)\sec^2 y dy = 0$

Answer:

Given,

$e^x\tan y dx + (1-e^x)\sec^2 y dy = 0$

$\\ \implies e^x\tan y dx = - (1-e^x)\sec^2 y dy \\ $

$\implies \int \frac{\sec^2 y }{\tan y}dy = -\int \frac{e^x }{(1-e^x)}dx$

$\\ let\ tany = t \ and \ 1-e^x = u \\$

$\implies \sec^2 ydy = dt\ and \ -e^xdx = du$

$\\ \therefore \int \frac{dt }{t} = \int \frac{du }{u} \\$

$\implies \log t = \log u + \log k \\$

$\implies t = ku \\$

$\implies \tan y= k (1-e^x)$

Question 11: Find a particular solution satisfying the given condition:

$(x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x; \ y = 1\ \textup{when}\ x = 0$

Answer:

Given, in the question

$
\begin{aligned}
& \left(x^3+x^2+x+1\right) \frac{d y}{d x}=2 x^2+x \\
& \Longrightarrow \int d y=\int \frac{2 x^2+x}{\left(x^3+x^2+x+1\right)} d x \\
& \left(x^3+x^2+x+1\right)=(x+1)\left(x^2+1\right)
\end{aligned}
$


Now,

$
\begin{aligned}
& \Rightarrow \frac{2 x^2+x}{(x+1)\left(x^2+1\right)}=\frac{A}{x+1}+\frac{B x+C}{x^2+1} \\
& \Rightarrow \frac{2 x^2+x}{(x+1)\left(x^2+1\right)}=\frac{A x^2+A(B x+C)(x+1)}{(x+1)\left(x^2+1\right)} \\
& \Rightarrow 2 x^2+x=A x^2+A+B x+C x+C \\
& \Rightarrow 2 x^2+x=(A+B) x^2+(B+C) x+A+C
\end{aligned}
$


Now comparing the coefficients

$
A+B=2 ; B+C=1 ; A+C=0
$


Solving these:

$
\mathrm{A}=\frac{1}{2}, \mathrm{~B}=\frac{3}{2}, \mathrm{C}=-\frac{1}{2}
$


Putting the values of $A, B, C$ :

$
\Rightarrow \frac{2 x^2+x}{(x+1)\left(x^2+1\right)}=\frac{1}{2} \frac{1}{(x+1)}+\frac{1}{2} \frac{3 x-1}{x^2+1}
$


Therefore,

$
\begin{aligned}
& \Rightarrow \int d y=\frac{1}{2} \int \frac{1}{x+1} d x+\frac{1}{2} \int \frac{3 x-1}{x^2+1} d x \\
& \Rightarrow \mathrm{y}=\frac{1}{2} \log (\mathrm{x}+1)+\frac{3}{2} \int \frac{\mathrm{x}}{\mathrm{x}^2+1} \mathrm{dx}-\frac{1}{2} \int \frac{\mathrm{dx}}{\mathrm{x}^2+1} \\
& \Rightarrow \mathrm{y}=\frac{1}{2} \log (\mathrm{x}+1)+\frac{3}{4} \int \frac{2 \mathrm{x}}{\mathrm{x}^2+1} \mathrm{dx}-\frac{1}{2} \tan ^{-1} \mathrm{x}
\end{aligned}
$

Question 12: Find a particular solution satisfying the given condition:

$x(x^2 -1)\frac{dy}{dx} =1;\ y = 0\ \textup{when} \ x = 2$

Answer:

Given, in the question $x\left(x^2-1\right) \frac{d y}{d x}=1$

$
\begin{aligned}
& \Longrightarrow \int d y=\int \frac{d x}{x\left(x^2-1\right)} \\
& \Longrightarrow \int d y=\int \frac{d x}{x(x-1)(x+1)}
\end{aligned}
$


Let,

$
\begin{aligned}
& \Rightarrow \frac{1}{x(x+1)(x-1)}=\frac{A}{x}+\frac{B}{x+1}+\frac{c}{x-1} \\
& \Rightarrow \frac{1}{x(x+1)(x-1)}=\frac{A(x-1)(x+1)+B(x)(x-1)+C(x)(x+1)}{x(x+1)(x-1)} \\
& \Rightarrow \frac{1}{x(x+1)(x-1)}=\frac{(A+B+C) x^2+(B-C) x-A}{x(x+1)(x-1)}
\end{aligned}
$


Now comparing the values of $A, B, C$

$
A+B+C=0 ; B-C=0 ; A=-1
$


Solving these:

$
\mathrm{B}=\frac{1}{2} \text { and } \mathrm{C}=\frac{1}{2}
$


Now putting the values of $A, B, C$,

$
\begin{aligned}
& \Rightarrow \frac{1}{x(x+1)(x-1)}=-\frac{1}{x}+\frac{1}{2}\left(\frac{1}{x+1}\right)+\frac{1}{2}\left(\frac{1}{x-1}\right) \\
& \Rightarrow \int d y=-\int \frac{1}{x} d x+\frac{1}{2} \int\left(\frac{1}{x+1}\right) d x+\frac{1}{2} \int\left(\frac{1}{x-1}\right) d x \\
& \Rightarrow y=-\log x+\frac{1}{2} \log (x+1)+\frac{1}{2} \log (x-1)+\log c \\
& \left.\Rightarrow \mathrm{y}=\frac{1}{2} \log \left[\frac{\mathrm{c}^2(\mathrm{x}-1)(\mathrm{x}+1)}{\mathrm{x}^2}\right\}----\mathrm{iii}\right)
\end{aligned}
$


Given, $\mathrm{y}=0$ when $\mathrm{x}=2$

$
\begin{aligned}
& 0=\frac{1}{2} \log \left[\frac{\mathrm{c}^2(2-1)(2+1)}{4}\right\} \\
& \Rightarrow \log \frac{3 \mathrm{c}^2}{4}=0
\end{aligned}
$


$
\Rightarrow 3 c^2=4
$


Therefore,

$
\begin{aligned}
& \Longrightarrow y=\frac{1}{2} \log \left[\frac{4(x-1)(x+1)}{3 x^2}\right] \\
& \Longrightarrow y=\frac{1}{2} \log \left[\frac{4\left(x^2-1\right)}{3 x^2}\right]
\end{aligned}
$

Question 13: Find a particular solution satisfying the given condition:

$\cos\left(\frac{dy}{dx} \right ) = a\ (a\in R);\ y = 1\ \textup{when}\ x = 0$

Answer:

Given,

$\cos\left(\frac{dy}{dx} \right ) = a$

$\\ \implies \frac{dy}{dx} = \cos^{-1}a \\$

$\implies \int dy = \int\cos^{-1}a\ dx \\$

$\implies y = x\cos^{-1}a + c$

Now, y =1 when x =0

1 = 0 + c

Therefore, c = 1

Putting the value of c:

$\implies y = x\cos^{-1}a + 1$

Question 14: Find a particular solution satisfying the given condition:

$\frac{dy}{dx} = y\tan x; \ y =1\ \textup{when}\ x = 0$

Answer:

Given,

$\frac{dy}{dx} = y\tan x$

$\\ \implies \int \frac{dy}{y} = \int \tan x\ dx \\$

$\implies \log y = \log \sec x + \log k \\$

$\implies y = k\sec x$

Now, y=1 when x =0

1 = ksec0

$\implies$ k = 1

Putting the vlue of k:

y = sec x

Question 15: Find the equation of a curve passing through the point (0, 0) and whose differential equation is $y' = e^x\sin x$ .

Answer:

We first find the general solution of the given differential equation

Given,

$y' = e^x\sin x$

$\\ \implies \int dy = \int e^x\sin xdx$

$\\ Let I = \int e^x\sin xdx \\$

$\implies I = \sin x.e^x - \int(\cos x. e^x)dx \\$

$\implies I = e^x\sin x - [e^x\cos x - \int(-\sin x.e^x)dx] \\$

$\implies 2I = e^x(\sin x - \cos x) \\$

$\implies I = \frac{1}{2}e^x(\sin x - \cos x)$

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

Now, Since the curve passes through (0,0)

y = 0 when x =0

$\\ \therefore 0 = \frac{1}{2}e^0(\sin 0 - \cos 0) + c \\$

$\implies c = \frac{1}{2}$

Putting the value of c, we get:

$\\ \therefore y = \frac{1}{2}e^x(\sin x - \cos x) + \frac{1}{2} \\$

$\implies 2y -1 = e^x(\sin x - \cos x)$

Question 16: For the differential equation $xy\frac{dy}{dx} = (x+2)(y+2)$ , find the solution curve passing through the point (1, –1).

Answer:

We first find the general solution of the given differential equation

Given,

$xy\frac{dy}{dx} = (x+2)(y+2)$

$\\ \implies \int \frac{y}{y+2}dy = \int \frac{x+2}{x}dx \\ $

$\implies \int \frac{(y+2)-2}{y+2}dy = \int (1 + \frac{2}{x})dx \\ $

$\implies \int (1 - \frac{2}{y+2})dy = \int (1 + \frac{2}{x})dx \\$

$\implies y - 2\log (y+2) = x + 2\log x + C$

Now, Since the curve passes through (1,-1)

y = -1 when x = 1

$\\ \therefore -1 - 2\log (-1+2) = 1 + 2\log 1 + C \\$

$\implies -1 -0 = 1 + 0 +C \\ \implies C = -2$

Putting the value of C:

$\\ y - 2\log (y+2) = x + 2\log x + -2 \\$

$\implies y -x + 2 = 2\log x(y+2)$

Question 17: Find the equation of a curve passing through the point $( 0 ,-2)$ given that at any point $(x,y)$ on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Answer:

According to the question,

$y\frac{dy}{dx} =x$

$\\ \implies \int ydy =\int xdx \\$

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

Now, Since the curve passes through (0,-2).

x =0 and y = -2

$\\ \implies \frac{(-2)^2}{2} = \frac{0^2}{2} + c \\ \implies c = 2$

Putting the value of c, we get

$\\ \frac{y^2}{2} = \frac{x^2}{2} + 2 \\ \implies y^2 = x^2 + 4$

Question 18: At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).

Answer:

Slope m of line joining (x,y) and (-4,-3) is $\frac{y+3}{x+4}$

According to the question,

$\\ \frac{dy}{dx} = 2(\frac{y+3}{x+4}) \\$

$\implies \int \frac{dy}{y+3} = 2\int \frac{dx}{x+4} \\$

$\implies \log (y+3) = 2\log (x+4) + \log k \\$

$\implies (y+3) = k(x+4)^2$

Now, Since the curve passes through (-2,1)

x = -2 , y =1

$\\ \implies (1+3) = k(-2+4)^2 \\ \implies k =1$

Putting the value of k, we get

$\\ \implies y+3 = (x+4)^2$

Question 19: The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Answer:

Volume of a sphere, $V = \frac{4}{3}\pi r ^3$

Given, Rate of change is constant.

$\\ \therefore \frac{dV}{dt} = c \\$

$\implies \frac{d}{dt} (\frac{4}{3}\pi r ^3) = c \\ $

$\implies \int d(\frac{4}{3}\pi r ^3) = c\int dt \\$

$\implies \frac{4}{3}\pi r ^3 = ct + k$

Now, at t=0, r=3 and at t=3 , r =6

Putting these value:

$\frac{4}{3}\pi (3) ^3 = c(0) + k \\ \implies k = 36\pi$

Also,

$\frac{4}{3}\pi (6) ^3 = c(3) + 36\pi \\ $

$\implies 3c = 252\pi \\ \implies c = 84\pi$

Putting the value of c and k:

$\\ \frac{4}{3}\pi r ^3 = 84\pi t + 36\pi \\$

$\implies r ^3 = (21 t + 9)(3) = 62t + 27 \\$

$\implies r = \sqrt[3]{62t + 27}$

Question 20: In a bank, principal increases continuously at the rate of r % per year. Find the value of r if Rs 100 double itself in 10 years (log e 2 = 0.6931).

Answer:

Let p be the principal amount and t be the time.

According to question,

$\frac{dp}{dt} = (\frac{r}{100})p$

$\\ \implies \int\frac{dp}{p} = \int (\frac{r}{100})dt \\ $

$\implies \log p = \frac{r}{100}t + C$

$\\ \implies p = e^{\frac{rt}{100} + C}$

Now, at t =0 , p = 100

and at t =10, p = 200

Putting these values,

$\\ \implies 100 = e^{\frac{r(0)}{100} + C} = e^C$

Also,

, $\\ \implies 200 = e^{\frac{r(10)}{100} + C} = e^{\frac{r}{10}}.e^C = e^{\frac{r}{10}}.100 \\$

$\implies e^{\frac{r}{10}} = 2 \\$

$\implies \frac{r}{10} = \ln 2 = 0.6931 \\ $

$\implies r = 6.93$

So value of r = 6.93%

Question 21: In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e 0.5 = 1.648).

Answer:

Let p be the principal amount and t be the time.

According to question,

$\frac{dp}{dt} = (\frac{5}{100})p$

$\\ \implies \int\frac{dp}{p} = \int (\frac{1}{20})dt \\$

$\implies \log p = \frac{1}{20}t + C$

$\\ \implies p = e^{\frac{t}{20} + C}$

Now, at t =0 , p = 1000

Putting these values,

$\\ \implies 1000 = e^{\frac{(0)}{20} + C} = e^C$

Also, At t=10 \,

$\\ \implies p = e^{\frac{(10)}{20} + C} = e^{\frac{1}{2}}.e^C = e^{\frac{1}{2}}.1000 \\$

$\implies p =(1.648)(1000) = 1648$

After 10 years, the total amount would be Rs.1648

Question 22: In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Answer:

Let n be the number of bacteria at any time t.

According to question,

$\frac{dn}{dt} = kn\ \ (k\ is\ a\ constant)$

$\\ \implies \int \frac{dn}{n} = \int kdt \\$

$\implies \log n = kt + C$

Now, at t=0, n = 100000

$\\ \implies \log (100000) = k(0) + C \\$

$\implies C = 5$

Again, at t=2, n= 110000

$\\ \implies \log (110000) = k(2) + 5 \\$

$\implies \log 11 + 4 = 2k + 5 \\$

$\implies 2k = \log 11 -1 =\log \frac{11}{10} \\$

$\implies k = \frac{1}{2}\log \frac{11}{10}$

Using these values, for n= 200000

$\\ \implies \log (200000) = kt + C \\ $

$\implies \log 2 +5 = kt + 5 \\ \implies (\frac{1}{2}\log \frac{11}{10})t = \log 2 $

$\\ \implies t = \frac{2\log 2}{ \log \frac{11}{10}}$

Question 23: The general solution of the differential equation $\frac{dy}{dx} = e^{x+y}$ is

(A) $e^x + e^{-y} = C$

(B) $e^{x }+ e^{y} = C$

(C) $e^{-x }+ e^{y} = C$

(D) $e^{-x }+ e^{-y} = C$

Answer:

Given,

$\frac{dy}{dx} = e^{x+y}$

$\\ \implies\frac{dy}{dx} = e^x.e^y \\ \implies\int \frac{dy}{e^y} = \int e^x.dx \\$

$\implies -e^{-y} = e^x + C \\ \implies e^x + e^{-y} = K\ \ \ \ (Option A)$

Also check -

Topics covered in Chapter 9 Differential Equation: Exercise 9.3


TopicsDescriptionExample
Variable Separable MethodInvolves separating variables x and y on opposite sides of the equation and integrating both sides. Used when a differential equation can be expressed as: $\frac{d y}{d x}=g(x) \cdot h(y)$

Solve: $\frac{d y}{d x}=x y$

$
\begin{aligned}
& \Rightarrow \frac{1}{y} d y=x d x \\
& \Rightarrow \int \frac{1}{y} d y=\int x d x
\end{aligned}
$

Equations Already in Separable FormSome differential equations are already arranged so that dy and dx can be directly separated.$\begin{aligned} & \text { Solve: } \frac{d y}{d x}=\frac{x^2}{1+y^2} \\ & \Rightarrow\left(1+y^2\right) d y=x^2 d x \\ & \Rightarrow \int\left(1+y^2\right) d y=\int x^2 d x \\ & \Rightarrow y+\frac{y^3}{3}=\frac{x^3}{3}+C\end{aligned}$
Implicit SolutionsSolutions are sometimes left in implicit form (not solved for y explicitly). This is valid and acceptable in many cases.From above: $y+\frac{y^3}{3}=\frac{x^3}{3}+C$ is an implicit solution.
Solving Using Initial ConditionsAfter finding the general solution, plug in the given values (initial conditions) to find the constant C for the particular solution.Given: $\frac{d y}{d x}=x y, y=1$ when $x=0$
General: $y=C e^{\frac{x^2}{2}}$
Apply: $1=C e^0 \Rightarrow C=1$
$\Rightarrow$ Final: $y=e^{\frac{x^2}{2}}$
Rearranging to Separable FormIf the equation isn't already in separable form, rearrange terms algebraically to isolate variables on each side.

Solve: $x \frac{d y}{d x}=y$

$
\begin{aligned}
& \Rightarrow \frac{d y}{d x}=\frac{y}{x} \\
& \Rightarrow \frac{1}{y} d y=\frac{1}{x} d x
\end{aligned}
$

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Frequently Asked Questions (FAQs)

Q: Name the subtopics discussed in 9.4
A:

The topics discussed are listed below

Differential equations of variable separable type, homogeneous and linear differential equations. 

Q: What is the order and degree of differential equations handled in topic 9.1?
A:

The differential equations of order 1 and degree 1 are handled.

Q: What type of differential equation is solved in exercise 9.3 Class 12 Maths?
A:

Differential equations of variable separable type are solved in the NCERT solutions for Class 12 Maths chapter 9 exercise 9.3.

Q: How many questions are there in Class 12 Maths chapter 9 exercise 9.3?
A:

Twenty-three questions are explained in the Class 12th Maths chapter 6 exercise 9.3.

Q: In topic 9.4, the number of sub-topics discussed is?
A:

Three subtopics are discussed in topic 9.4.

Q: In session 9.4 of NCERT Class, 12 Maths book how many methods to solve the first-degree first-order differential equations are discussed?
A:

3 methods are discussed. 

Q: How many questions of exercise 9.1 are multiple-choice questions?
A:

One question is of objective type with 4 choices.

Q: How many solved examples are given in topic 9.5.1?
A:

Six example problems and their solutions are given under topic 9.5.1.

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Questions related to CBSE Class 12th

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Have a question related to CBSE Class 12th ?

Hello,

The date of 12 exam is depends on which board you belongs to . You should check the exact date of your exam by visiting the official website of your respective board.

Hope this information is useful to you.

Hello,

Class 12 biology questions papers 2023-2025 are available on cbseacademic.nic.in , and other educational website. You can download PDFs of questions papers with solution for practice. For state boards, visit the official board site or trusted education portal.

Hope this information is useful to you.

Hello Pruthvi,

Taking a drop year to reappear for the Karnataka Common Entrance Test (KCET) is a well-defined process. As a repeater, you are fully eligible to take the exam again to improve your score and secure a better rank for admissions.

The main procedure involves submitting a new application for the KCET through the official Karnataka Examinations Authority (KEA) website when registrations open for the next academic session. You must pay the required application fee and complete all formalities just like any other candidate. A significant advantage for you is that you do not need to retake your 12th board exams. Your previously secured board marks in the qualifying subjects will be used again. Your new KCET rank will be calculated by combining these existing board marks with your new score from the KCET exam. Therefore, your entire focus during this year should be on preparing thoroughly for the KCET to achieve a higher score.

For more details about the KCET Exam preparation, CLICK HERE.

I hope this answer helps you. If you have more queries, feel free to share your questions with us, and we will be happy to assist you.

Thank you, and I wish you all the best in your bright future.

Yes, you can switch from Science in Karnataka State Board to Commerce in CBSE for 12th. You will need a Transfer Certificate from your current school and meet the CBSE school’s admission requirements. Since you haven’t studied Commerce subjects like Accountancy, Economics, and Business Studies, you may need to catch up before or during 12th. Not all CBSE schools accept direct admission to 12th from another board, so some may ask you to join Class 11 first. Make sure to check the school’s rules and plan your subject preparation.



Hello

For the 12th CBSE Hindi Medium board exam, important questions usually come from core chapters like “Madhushala”, “Jhansi ki Rani”, and “Bharat ki Khoj”.
Questions often include essay writing, letter writing, and comprehension passages. Grammar topics like Tenses, Voice Change, and Direct-Indirect Speech are frequently asked.
Students should practice poetry questions on themes and meanings. Important questions also cover summary writing and translation from Hindi to English or vice versa.
Previous years’ question papers help identify commonly asked questions.
Focus on writing practice to improve handwriting and presentation. Time management during exams is key to answering all questions effectively.