RD Sharma Solutions Class 12 Mathematics Chapter 21 FBQ

# RD Sharma Solutions Class 12 Mathematics Chapter 21 FBQ

Edited By Satyajeet Kumar | Updated on Jan 21, 2022 12:17 PM IST

Most class 12 students use the RD Sharma Solution books to clarify their doubts while doing homework. However, when it comes to mathematics, the students feel nervous about solving the sums in new concepts. Chapters like the Differential Equations are challenging for most of the students. The arrival of RD Sharma Class 12th Chapter 21 FBQ solution books has helped many students of this category.

Differential Equation exercise Fill in the blank question 1

1
Hint:
Since, Parabola has symmetric along positive x-axis, its equation is y2 = 4ax
Number of arbitrary constants = Order of differential Equation
Given:
The order of differential equation representing the family of parabola y2 = 4ax is _____
Solution:
We know that order of differential equation = no. of arbitrary constant. Here, no. of arbitrary constant is 1
So, 1 is the arbitrary constant.

Differential Equation exercise Fill in the blank question 2

2
Hint:
The degree of the differential equation is the power of highest order derivative present in the differential equation.
Given:
$\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{3}+\left ( \frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}} \right )^{2}=0$
Solution:
The order of the given equation
$\left ( \frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}} \right )^{2}=2$
Degree is the power of the highest order derivative.
So, the degree in this equation is 2.

Differential Equation exercise Fill in the blank question 3

0
Hint:
In particular solution Number of arbitrary constant = 0.
Given:
The number of arbitrary constants in a particular solution of the differential equation tanx dx + tany dy=0
Solution:
Tanx dx + tany dy=0, any particular solution of a differential equation has no arbitrary constant.

Differential Equation exercise Fill in the blank question 4

x = v y
Hint:
Substituting the equation
Given:
$\frac{\mathrm{d} x}{\mathrm{d} y}=\frac{x^{2}log\left ( \frac{x}{y} \right )-x^{2}}{xy\, log\left ( \frac{x}{y} \right )}$
Solution:
As there are x/y present in equation 3 times. So, it is good to use replacement x=v y.
It will simplify the equation to solve it easily.
∴ The answer is x=v y

Differential Equation exercise Fill in the blank question 5

$\frac{1}{x}$
Hint:
We will use the method of solving linear differential equation.
Given:
$x\frac{\mathrm{d} y}{\mathrm{d} x}-y=sin\, x$
Solution:
The given differential equation can be written as
$\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{y}{x}=\frac{sin\, x}{x}$
The given differential equation is of the form
$\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q$
and therefore by comparison we get;
$P=\frac{-1}{x}, \qquad Q=\frac{sin\, x}{x}$
$I.F.=e^{\int P dx}=e^{\int \frac{-1}{x} dx}=e^{\-log\, x}=\frac{1}{x}$
$\frac{1}{x}$

Differential Equation exercise Fill in the blank question 6

$e^{y}=e^{x}+c$
Hint:
First use the variable separation method then integrating both sides.
Given:
$\frac{\mathrm{d} y}{\mathrm{d} x}=e^{x-y}$
Solution:
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{e^{x}}{e^{y}}$
$\Rightarrow \int e^{y}dy=\int e^{x}dy$
$\Rightarrow e^{y}=e^{x}+c$
$e^{y}=e^{x}+c$

Differential Equation exercise Fill in the blank question 7

$\Rightarrow xy=\frac{x^{2}}{2}+c$
Hint:
Using the form
$\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q$
Given:
$\frac{\mathrm{d} y}{\mathrm{d} x}+\frac{y}{x}=1$
Solution:
Comparing with the given equation with
$\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q$
We get,
$P=\frac{1}{x} \text { and }Q=1$
$I.F=e^{\int P\, dx}=e^{\int\frac{1}{x} dx}=e^{log\, x}=x$
∴The general solution is
\begin{aligned} &y\times I.F=\int Q\times I.F\, dx \\ &\Rightarrow y\times x=\int 1\times x\, dx \\ &\Rightarrow xy=\frac{x^{2}}{2}+c \end{aligned}
\begin{aligned} &\Rightarrow xy=\frac{x^{2}}{2}+c \end{aligned}

Differential Equation exercise Fill in the blank question 8

$\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+y=0$
Hint:
Differentiating the given function w.r.t ‘x’
Given:
$y=Asin\, x+Bcos\, x$
Solution:
Differentiating the given function w.r.t ‘x’ we get,
$\frac{\mathrm{d} y}{\mathrm{d} x}=Acos\, x-Bsin\, x$
and
$\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}=-Asin\, x-Bcos\, x$
$\Rightarrow \frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}+y=0$
is the required differential equation.
$\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}+y=0$

Differential Equation exercise Fill in the blank question 9

$\frac{1}{\sqrt{x}}$
Hint:
Use the concept of linear Differential equation is of the form
$\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q$
Given:
$\left ( \frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}} \right )\frac{\mathrm{d} x}{\mathrm{d} y}=1,\quad x\neq 0$
Solution:
\begin{aligned} &\Rightarrow \left ( \frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}} \right )\frac{\mathrm{d} x}{\mathrm{d} y}=1 \\ &\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\left ( \frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}} \right ) \\ &\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}+\frac{y}{\sqrt{x}}=\frac{e^{-2\sqrt{x}}}{\sqrt{x}} \end{aligned}
The equation in the form
$\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q$
Where
\begin{aligned} &P=\frac{1}{\sqrt{x}} \end{aligned}
\begin{aligned} &P=\frac{1}{\sqrt{x}} \end{aligned}
$\text { Note: In question there will be } \frac{\mathrm{d} x}{\mathrm{d} y} \text { in place of } \frac{\mathrm{d} y}{\mathrm{d} x}.$

Differential Equation exercise Fill in the blank question 10

2
Hint:
Differentiating w.r.t to ‘x’
Given:
The family of ellipses having foci on x-axis and center at the origin, is given by
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Solution:
Differentiating w.r.t ‘x’, we get
\begin{aligned} & \end{aligned}\begin{aligned} &\frac{2x}{a^{2}}+\frac{2y}{b^{2}}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )=0 \\ &\Rightarrow \frac{2y}{b^{2}}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )=-\frac{2x}{a^{2}} \\ &\Rightarrow \frac{y\frac{\mathrm{d} y}{\mathrm{d} x}}{b^{2}}=-\frac{x}{a^{2}} \\ &\Rightarrow \frac{y\frac{\mathrm{d} y}{\mathrm{d} x}}{x}=-\frac{b^{2}}{a^{2}} \end{aligned}
Again by differentiating w.r.t ‘x’ we get
\begin{aligned} &\frac{x\left [ y\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \right ]-\left ( y.\frac{\mathrm{d} y}{\mathrm{d} x} \right )}{x^{2}}=0 \end{aligned}
The required equation is
\begin{aligned} &xy\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+x\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}-y.\frac{\mathrm{d} y}{\mathrm{d} x} \qquad \qquad \dots(i) \end{aligned}
∴ Order of the differential equation is the highest derivative present in the differential equation(i)
\begin{aligned} &\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=2 \end{aligned}

Differential Equation exercise Fill in the blank question 11

1
Hint:
Squaring both sides
Given:
$\sqrt{1+\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}}=\frac{\mathrm{d} y}{\mathrm{d} x}+x$
Solution:
On squaring the given differential equation,
\begin{aligned} &\left ( \sqrt{1+\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}}\right )^{2} =\left ( \frac{\mathrm{d} y}{\mathrm{d} x}+x \right )^{2} \\ &\Rightarrow 1+\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=\left ( \frac{\mathrm{d} y}{\mathrm{d} x}+x \right )^{2} \end{aligned}
Highest order of the given equation is
\begin{aligned} & \frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=2 \end{aligned}
and degree of
\begin{aligned} & \frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}} \, \, is\, \, 1 \end{aligned}
∴ Degree of the differential equation is 1.

Differential Equation exercise Fill in the blank question 12

$\frac{1}{x}$
Hint:
Using the form of linear differential equation
$\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q$
Given:
$x\frac{\mathrm{d} y}{\mathrm{d} x}-y=x\, cos\, x$
Solution:
$x\frac{\mathrm{d} y}{\mathrm{d} x}-y=x\, cos\, x \qquad \qquad \dots(i)$
Divide the eqn (i) by x, we get
$\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{y}{x}= cos\, x \qquad \qquad \dots(ii)$
Comparing eqn (ii) with the standard linear differential equation
$\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q$
We get,
$P=-\frac{1}{x}, \quad Q=cos \, x$
$\therefore I.F=e^{\int P\, dx}=e^{\int\frac{-1}{x} dx}=e^{-log\, x}$
$I.F=e^{log\, x^{-1}}=\frac{1}{x}$
$\frac{1}{x}$

Differential Equation exercise Fill in the blank question 13

Not defined
Hint:
We cannot express polynomial of derivatives
Given:
$\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+e^{\frac{\mathrm{d} y}{\mathrm{d} x}}=0$
Solution:
Degree of this equation is not defined as it cannot be expressed as polynomial of derivatives.
So, the answer is not defined.

Differential Equation exercise Fill in the blank question 14

2
Hint:
Degree is the highest exponent of the highest order derivative.
Given:
$\sqrt{1+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}}=x$
Solution:
The degree is the highest exponent of the highest order derivative and it is 2 in this equation.

Differential Equation exercise Fill in the blank question 15

3
Hint:
Number of arbitrary constants = Order of differential Equation
Given:
The number of arbitrary constants in the general solution of the differential equation of order 3 is_____.
Solution:
As per the given question the numbers of arbitrary constants in third order differential equation are 3.

Differential Equation exercise Fill in the blank question 16

$xe^{\int R\, dy}=\int \left ( Se^{\int R\, dy} \right )dy+c$
Hint:
Using the form
$\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q$
Given:
$\frac{\mathrm{d} y}{\mathrm{d} x}+Rx=S,$
where R and S are functions of y.
Solution:
Integrating factor of given differential equation is
$IF=e^{\int R\, dy}$
General solution is given by:
$xe^{\int R\, dy}=\int (S\, e^{\int R \, dy})dy+c$
$xe^{\int R\, dy}=\int \left ( Se^{\int R\, dy} \right )dy+c$

Differential Equation exercise Fill in the blank question 17

$\frac{e^{x}}{x}$
Hint:
Use the form
$\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q,$
to find the integrating factor.
Given:
$\frac{\mathrm{d} y}{\mathrm{d} x}+y=\frac{1+y}{x}$
Solution:
$\frac{\mathrm{d} y}{\mathrm{d} x}+y-\frac{y}{x}=\frac{1}{x}$
$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}+y\left ( 1-\frac{1}{x} \right )=\frac{1}{x}$
$\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q,$
Where
$P=\left ( 1-\frac{1}{x} \right )$
$I.F=e^{\int P\, dx}=e^{\int(1-\frac{1}{x}) dx}=e^{(x-in\, x)}$
$I.F=e^{x}.e^{-in\, x}=e^{x}.e^{in_{e}\, (x^{-1})}=e^{x}.\frac{1}{x}=\frac{e^{x}}{x}$
$\frac{e^{x}}{x}$

Differential Equation exercise Fill in the blank question 18

$x=C\, sec\, y$
Hint:
Integrating both sides
Given:
$cot \, y\, dx=x\, dy$
Solution:
$cot \, y\, dx=x\, dy$
Separate the variable and integrating both sides, we get
\begin{aligned} &\int \frac{1}{x}dx=\int tan\, y\, dy \end{aligned}
\begin{aligned} &\Rightarrow log\, x=log\left | sec\, y \right |+log\, C \\ &\Rightarrow log\frac{x}{sec\, y}=log \, C \\ &\Rightarrow \frac{x}{sec\, y}= C \\ &\Rightarrow x=C\, sec\, y \end{aligned}
So, the answer is C sec y

Differential Equation exercise Fill in the blank question 19

$y=\frac{1}{4}x^{2}+Cx^{-2}$
Hint:
Substituting the values
Given:
$x\frac{\mathrm{d} y}{\mathrm{d} x}+2y=x^{2}$
Solution:
On dividing the given differential equation by x, we get
$\frac{\mathrm{d} y}{\mathrm{d} x}+2\frac{y}{x}=x$
So, the differential equation is a linear differential equation, and we can observe that the given differential equation is in the form
$\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q$
where,
$P=\frac{2}{x} \, and\, Q=x$
Here we get the integrating factor as
$I.F=e^{\int Pdx}$
By substituting the values,
$=e^{\int \frac{2}{x}dx}$
We get,
\begin{aligned} &=e^{2log\, x} \\ &=e^{log\, x^{2}} \\ &=x^{2} \end{aligned}
Here, the general solution can be written as
\begin{aligned} &Y(I.F)=\int Q(I.F)dx+C \end{aligned}
Substituting the values we get
\begin{aligned} &y(x^{2})=\int x(x)^{2}dx+C \\ &x^{2}y=\frac{x^{4}}{4}+C \end{aligned}
Dividing the entire equation by x2
\begin{aligned} &y=\frac{x^{2}}{4}+\frac{C}{x^{2}} \\ \Rightarrow &y=\frac{1}{4}x^{2}+Cx^{-2} \end{aligned}
\begin{aligned} &y=\frac{1}{4}x^{2}+Cx^{-2} \end{aligned}

Differential Equation exercise Fill in the blank question 20

$xy=Ce^{-y}$
Hint:
First separate the variables and then use simple Integrating to solve the question.
Given:
$y\, dx+(x+xy)dy=0$
Solution:
\begin{aligned} &\Rightarrow y\, dx=-x(1+y)dy \\ &\Rightarrow \frac{dx}{x}=-\left ( \frac{1+y}{y} \right )dy \\ &\Rightarrow \int \frac{1}{x}=-\int \left ( \frac{1}{y}+1 \right )dy \\ &\Rightarrow log\, x=-log\, y-y+log\, C \\ &\Rightarrow log\, x+log\, y-log\, C=-y \\ &\Rightarrow log\frac{xy}{c}=-y \\ &\Rightarrow \frac{xy}{c}=e^{-y} \\ &\Rightarrow xy=Ce^{-y} \end{aligned}
$xy=Ce^{-y}$

Differential Equation exercise Fill in the blank question 21

Order = 1
Given:
The order of differential equation representing the family of circles
x2 + (y-a)2 =a2 Is _________
Hint:
Order = No. of arbitrary constant
Solution:
We know that, order of differential equation = no. of arbitrary constant. Here, no. of arbitrary constant is 1

Differential Equation exercise Fill in the blank question 22

Order =0
Given:
The number of arbitrary constants in the particular solution of a differential equation of order two is ________
Hint:
Here, number of arbitrary constant = order of the equation
Solution:
Particular solution of a differential equation doesn’t contain any arbitrary constant.
So, the number of arbitrary constant in the particular solution of a differential equation = 0

Differential Equation exercise Fill in the blank question 23

$\frac{\mathrm{d} ^{2}x}{\mathrm{d} y^{2}}=0$
Given:
The differential equation of all non-horizontal lines in a plane is ____
Hint:
$y=mx+c \quad (m=slope)$
Solution:
Equation of non-horizontal line in a plane is y = mx + c where ‘m’ is slope.
Since, there are two arbitrary constants. so the order will be two hence we can differentiate the equation of line two times.
Differentiate w.r.t x
\begin{aligned} &\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=m \\ &\Rightarrow \frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=0 \end{aligned}
\begin{aligned} & \frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=0 \end{aligned}

Differential Equation exercise Fill in the blank question 24

$\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=0$
Given:
The differential equation of all non-vertical lines in a plane is ____
Hint:
$y=mx+c \quad (m=slope)$
Solution:
Clearly, equation of all non-vertical lines is y=mx+c
On differentiating w.r.t x we get
$\frac{\mathrm{d} y}{\mathrm{d} x}=m$
and differentiating again We get:
$\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}=0$
$\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}=0$

Differential Equation exercise Fill in the blank question 25

Integrating factor = x2 + 1
Hint:
we will use the linear differential equation to solve the problem.
Given:
The integrating factor of all differential equation
$(x^{2}+1)\frac{\mathrm{d} y}{\mathrm{d} x}+2xy=x^{2}-1$
is _________
Dividing the given equation both sides by x2 +1
Solution:
Dividing the L.H.S by (x2 + 1) we get,
$\frac{\mathrm{d} y}{\mathrm{d} x}+\frac{2x}{x^{2}+1}y=\frac{x^{2}-1}{x^{2}+1} \qquad \qquad \dots(i)$
Clearly, the equation (i) is of the form
$\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q \qquad \qquad \dots(ii)$
Comparing (i) and (ii) we get,
\begin{aligned} &P(x)=\frac{2x}{x^{2}+1} \\ &Q(x)=\frac{x^{2}-1}{x^{2}+1} \end{aligned}
So,
\begin{aligned} &I.F.=e^{\int P(x)dx}=e^{\int \frac{2x}{x^{2}+1}dx} \end{aligned}
Now Put x2 +1 = t ⇒2x dx =dt
\begin{aligned} &I.F.=e^{\int \frac{2x}{x^{2}+1}dx} \\ &\Rightarrow e^{\int \frac{1}{t}dt} \\ &\Rightarrow e^{log\, t}= t \\ &\therefore t=x^{2}+1 \end{aligned}
So, the answer is x2 +1

Differential Equation exercise Fill in the blank question 26

Degree=4
Hint:
Degree = Highest power of the highest order derivative.
Given:
The degree of differential equation
$y=x\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+\left ( \frac{\mathrm{d} x}{\mathrm{d} y} \right )^{2}$
is _______
Solution:
$y=x\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+\left ( \frac{\mathrm{d} x}{\mathrm{d} y} \right )^{2}$
$y=x\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{-2}$
$y=x\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+\frac{1}{\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}}$
$y=\frac{x\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{4}+1}{\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}}$
$y\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}=x\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{4}+1$
So, the degree of above differential equation is 4

Differential Equation exercise Fill in the blank question 27

Order = 2 or 2nd order
Given:
The order of differential equation representing all circles of radius r is ____
Hint:
Here, Order = Highest derivative
Solution:
Any circle with given radius can be written as (x-h)2+(y-k)2=r2
Where (h, k) be the Centre of the circle and radius is constant.
Differentiating both side w.r.t x we get
\begin{aligned} &2(x-h)+2(y-k)\frac{\mathrm{d} y}{\mathrm{d} x}=0 \\ &\Rightarrow 2(y-k)\frac{\mathrm{d} y}{\mathrm{d} x}=-2(x-h) \\ &\Rightarrow (y-k)\frac{\mathrm{d} y}{\mathrm{d} x}=-(x-h) \\ &\Rightarrow y\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}=-1 \\ &\Rightarrow \frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}=-\frac{1}{y} \end{aligned}
Hence order of differential equation will be ‘2’ i.e. 2nd order.
So, the answer is ‘2’ or 2nd order.

Differential Equation exercise Fill in the blank question 28

Degree = 3
Hint:
To find the degree of differential equation representing the family of curves y = Ax + A3,we have to eliminate the constant A.
Given:
The degree of differential equation representing the family of curves y = Ax + A3, where A is arbitrary constant is ______
Solution:
\begin{aligned} &y=Ax + A^{3} \qquad \qquad \dots (i) \\ &\frac{\mathrm{d} y}{\mathrm{d} x}=A\qquad \qquad \dots (ii) \end{aligned}
Using the eqn (ii) in (i), we get
\begin{aligned} &y=\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )x+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{3} \end{aligned}
Hence, Degree = Highest power of the highest order derivative.
So, the degree of the above differential equation is 3.

Differential Equation exercise Fill in the blank question 29

xy = C
Given:
The general solution of the differential equation
$\frac{dx}{x}+\frac{dy}{y}=0$
is _____
Hint:
The above sum will be solved by the formula of derivatives.
Solution:
\begin{aligned} &\frac{dx}{x}+\frac{dy}{y}=0 \\ &\Rightarrow \frac{dx}{x}=-\frac{dy}{y} \\ &\Rightarrow log\, x=-log\, y+log\, C \\ &\Rightarrow log\, x+log\, y= C \\ &\Rightarrow log\, xy=log\, C \\ &\therefore xy=C \end{aligned}
So, the answer is xy = C

Differential Equation exercise Fill in the blank question 30

Order=3 and Degree=1
Given:
The order and degree of the differential equation
$\frac{\mathrm{d} ^{3}y}{\mathrm{d} x^{3}}-3\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+2\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{4}=y^{4}$
are ____ and ____ respectively.
Hint:
Order = Highest Derivative
Degree = Highest power of highest order Derivative
Solution:
Order = Highest Derivative = 3
Degree = Highest power of highest Derivative = 1
So, the order is 3 and degree is 1.

Differential Equation exercise Fill in the blank question 31

$\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+y=0$
Hint:
Use simple differentiation w.r.t x
Given:
The differential equation for which y=a cos x + b sin x is a solution, is ____
Solution:
$y=a\, cos\, x+b\, sin\, x \qquad \qquad \dots(i)$
differentiating both side w.r.t x, we get
\begin{aligned} &\frac{\mathrm{d} y}{\mathrm{d} x}=a\left ( \frac{dcos\, x}{dx} \right )+b\left ( \frac{dsin\, x}{dx} \right ) \\ &=a(-sin\, x)+b(cos\, x) \\ &=-asin\, x+bcos\, x \end{aligned}
Again differentiating both side w.r.t x, we get
Now, we have to verify
\begin{aligned} &\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+y=0 \end{aligned}
Taking L.H.S
\begin{aligned} &\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=(-acos\, x-bsin\, x) \\ &\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=-y \qquad \qquad [using(i)] \\ &\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+y=0 \end{aligned}
∴So, the differential equation is
$\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+y=0$

Differential Equation exercise Fill in the blank question 32

Rectangular Hyperbola
Hint:
Ratio of abscissa and ordinate = rectangular hyperbola
Given:
The curve for which the slope of the tangent at any point is equal to the ratio of abscissa and ordinate of the point is ______
Solution:
The curve for which the slope of the tangent at any point is equal to the ratio of abscissa and ordinate of the point is ‘Rectangular Hyperbola’
So the answer is Rectangular Hyperbola.

Differential Equation exercise Fill in the blank question 33

Order=1 and Degree=3
Given:
Family y = Ax + A3 of curves will correspond to a differential equation of order and degree _____
Hint:
Order=Highest Derivative
Solution:
Given, family of curves is
$y=Ax+A^3 \qquad \qquad \dots(i)$
$\frac{\mathrm{d} y}{\mathrm{d} x}=A$
Putting the value of A in equation (i)
$y=x\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{3}$
∴ Order = 1
∴ Degree= Highest power of Derivative=3
∴So, the order is 1 and degree is 3

Differential Equation exercise Fill in the blank question 34

Hyperbola
Given:
The differential equation xdy + ydx = 0 represents the family of ____
Hint:
By the help of integral sums will get solve.
Solution:
\begin{aligned} &xdy + ydx=0 \\ &\Rightarrow xdy = -ydx \\ &\Rightarrow \frac{dy}{y}=-\frac{dx}{x} \end{aligned}
Integrating both sides
\begin{aligned} &\int \frac{dy}{y}=-\int \frac{dx}{x} \\ &\Rightarrow log\, y=-log\, x+log \, C \\ &\Rightarrow log\, y+log\, x=log \, C \\ &\therefore xy=C \end{aligned}
So, the above differential equation represents the Hyperbola.

Differential Equation exercise Fill in the blank question 35

$(x^{2}-y^{2})dy=2xydx$
Hint:
To remove the arbitrary constant we have differentiate the given equation.
Given:
The differential equation of the family of curves x2 + y2 - 2ay = 0, where a is arbitrary constant is _____
Solution:
\begin{aligned} &x^{2}+y^{2}=2ay \\ &\Rightarrow \frac{x^{2}+y^{2}}{y}=2a \\ &\Rightarrow \frac{y\frac{\mathrm{d} y}{\mathrm{d} x}(x^{2}+y^{2})-(x^{2}+y^{2})\frac{\mathrm{d} y}{\mathrm{d} x}}{y^{2}}=0 \\ &\Rightarrow y\left [ 2x+2y\frac{\mathrm{d} y}{\mathrm{d} x} \right ]-(x^{2}+y^{2})\frac{\mathrm{d} y}{\mathrm{d} x}=0 \\ &\Rightarrow 2xy+2y^{2}\frac{\mathrm{d} y}{\mathrm{d} x}-(x^{2}+y^{2})\frac{\mathrm{d} y}{\mathrm{d} x}=0 \end{aligned}
\begin{aligned} &\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}(2y^{2}-x^{2}-y^{2})+2xy=0 \\ &\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}(y^{2}-x^{2})+2xy=0 \\ &\Rightarrow (x^{2}-y^{2})\frac{\mathrm{d} y}{\mathrm{d} x}=2xy \\ &\Rightarrow (x^{2}-y^{2})dy=2xydx \end{aligned}
\begin{aligned} & (x^{2}-y^{2})dy=2xydx \end{aligned}

Differential Equation exercise Fill in the blank question 36

Order = 2, Degree = 1
Hint:
Order=Highest Derivative
Degree=Highest power of highest Derivative
Given:
The order and degree of the differential equation
$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{4}$
are ____respectively.
Solution:
By using chain we will evaluate the derivative
$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{4}$
$\Rightarrow 4\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{3}\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}$
∴Order = Highest Derivative i.e. 2
∴ Degree = The highest power the highest derivative is 1
So the order and degree are 2,1 respectively.

Differential Equation exercise Fill in the blank question 37

$\frac{1}{x}$
Given:
The integrating factor of differential equation
$x\frac{\mathrm{d} y}{\mathrm{d} x}-y=log\, x$
is ______
Hint:
By the help of integrating factor the sum will solve.
Solution:
$x\frac{\mathrm{d} y}{\mathrm{d} x}-y=log\, x$
$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}-\frac{y}{x}=\frac{log\, x}{x}$
Comparing the above equation with the standard linear differential equation, we get
$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q$
Here,
$P=\frac{-1}{x}\: and\: Q=\frac{log\, x}{x}$
So, integrating factor
$e^{\int P\, dx}=e^{\int-\frac{1}{x} dx}=e^{-log\, x}=\frac{1}{x}$
$\frac{1}{x}$

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