NCERT Solutions for Exercise 9.6 Class 12 Maths Chapter 9- Differential Equations

Question:2 Solve for general solution:

$\frac{dy}{dx}+3y=e^{-2x}$

Answer:

Given equation is
$\frac{dy}{dx}+3y=e^{-2x}$
This is $\frac{dy}{dx}+py=Q$ type where p = 3 and $Q=e^{-2x}$
Now,
$I.F.=e^{\int pdx}=e^{\int 3dx}=e^{3x}$
Now, the solution of given differential equation is given by the relation
$Y(I.F.)=\int (Q\times I.F.)dx+C$
$Y(e^{3x})=\int (e^{-2x}\times e^{3x})dx+C$
$$\begin{gathered} Y(e^{3x})=\int (e^x)dx+C \\ Y(e^{3x})=e^x+C \\ Y=e^{-2x}+C{e}^{-3x} \end{gathered}$$
Therefore, the general solution is $Y=e^{-2x}+C{e}^{-3x}$

Question:3 Find the general solution

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

Answer:

Given equation is
$\frac{dy}{dx}+\frac{y}{x}=x^2$
This is $\frac{dy}{dx}+py=Q$ type where $p=\frac{1}{x}$ and $Q=x^2$
Now,
$I.F.=e^{\int pdx}=e^{\int \frac{1}{x}dx}=e^{\log x}=x$
Now, the solution of given differential equation is given by relation
$y(I.F.)=\int (Q\times I.F.)dx+C$
$y(x)=\int (x^2\times x)dx+C$
$$\begin{gathered} y(x)=\int (x^3)dx+C \\ y.x=\frac{x^4}{4}+C \end{gathered}$$
Therefore, the general solution is $yx=\frac{x^4}{4}+C$

Question:4 Solve for General Solution.

$\frac{dy}{dx}+(\sec x)y=\tan x(0\le x<\frac{\pi }{2})$

Answer:

Given equation is
$\frac{dy}{dx}+(\sec x)y=\tan x(0\le x<\frac{\pi }{2})$
This is $\frac{dy}{dx}+py=Q$ type where $p=\sec x$ and $Q=\tan x$
Now,
$I.F.=e^{\int pdx}=e^{\int \sec xdx}=e^{\log |\sec x+\tan x|}=\sec x+\tan x$ $(\because 0\le x\le \frac{\pi }{2}\sec x>0,\tan x>0)$
Now, the solution of given differential equation is given by relation
$y(I.F.)=\int (Q\times I.F.)dx+C$
$y(\sec x+\tan x)=\int ((\sec x+\tan x)\times \tan x)dx+C$
$$\begin{gathered} y(\sec x+\tan x)=\int (\sec x\tan x+{\tan }^{2}x)dx+C \\ y(\sec x+\tan x)=\sec x+\int ({\sec }^{2}x-1)dx+C \\ y(\sec x+\tan x)=\sec x+\tan x-x+C \end{gathered}$$
Therefore, the general solution is $y(\sec x+\tan x)=\sec x+\tan x-x+C$

Question:5 Find the general solution.

${\cos }^{2}x\frac{dy}{dx}+y=\tan x(0\le x<\frac{\pi }{2})$

Answer:

Given equation is
${\cos }^{2}x\frac{dy}{dx}+y=\tan x(0\le x<\frac{\pi }{2})$
we can rewrite it as
$\frac{dy}{dx}+{\sec }^{2}xy={\sec }^{2}x\tan x$
This is $\frac{dy}{dx}+py=Q$ where $p={\sec }^{2}x$ and $Q={\sec }^{2}x\tan x$
Now,
$I.F.=e^{\int pdx}=e^{\int {\sec }^{2}xdx}=e^{\tan x}$
Now, the solution of given differential equation is given by relation
$y(I.F.)=\int (Q\times I.F.)dx+C$
$y(e^{\tan x})=\int (({\sec }^{2}x\tan x)\times e^{\tan x})dx+C$
$y{e}^{\tan x}=\int {\sec }^{2}x\tan x{e}^{\tan x}dx+C$
take
$$\begin{gathered} e^{\tan x}=t \\ \Rightarrow {\sec }^{2}x.e^{\tan x}dx=dt \end{gathered}$$
$$\begin{gathered} \int t.\log tdt=\log t.\int tdt-\int (\frac{d(\log t)}{dt}.\int tdt)dt \\ \int t.\log tdt=\log t.\frac{t^2}{2}-\int (\frac{1}{t}.\frac{t^2}{2})dt \\ \int t.\log tdt=\log t.\frac{t^2}{2}-\int \frac{t}{2}dt \\ \int t.\log tdt=\log t.\frac{t^2}{2}-\frac{t^2}{4} \\ \int t.\log tdt=\frac{t^2}{4}(2\log t-1) \end{gathered}$$
Now put again $t=e^{\tan x}$
$\int {\sec }^{2}x\tan x{e}^{\tan x}dx=\frac{e^{2\tan x}}{4}(2\tan x-1)$
Put this value in our equation

$y{e}^{\tan x}=\frac{e^{2\tan x}}{4}(2\tan x-1)+C$
Therefore, the general solution is $y=\frac{e^{\tan x}}{4}(2\tan x-1)+C{e}^{-\tan x}$

Question:6 Solve for General Solution.

$x\frac{dy}{dx}+2y=x^2\log x$

Answer:

Given equation is
$x\frac{dy}{dx}+2y=x^2\log x$
Wr can rewrite it as
$\frac{dy}{dx}+2.\frac{y}{x}=x\log x$
This is $\frac{dy}{dx}+py=Q$ type where $p=\frac{2}{x}$ and $Q=x\log x$
Now,
$I.F.=e^{\int pdx}=e^{\int \frac{2}{x}dx}=e^{2\log x}=e^{\log x^2}=x^2$ $(\because 0\le x\le \frac{\pi }{2}\sec x>0,\tan x>0)$
Now, the solution of given differential equation is given by relation
$y(I.F.)=\int (Q\times I.F.)dx+C$
$y(x^2)=\int (x\log x\times x^2)dx+C$
$x^{2}y=\int x^3\log x+C$
Let
$$\begin{gathered} I=\int x^3\log x \\ I=\log x\int x^{3}dx-\int (\frac{d(\log x)}{dx}.\int x^{3}dx)dx \\ I=\log x.\frac{x^4}{4}-\int (\frac{1}{x}.\frac{x^4}{4})dx \\ I=\log x.\frac{x^4}{4}-\int \left(\frac{x^3}{4}\right)dx \\ I=\log x.\frac{x^4}{4}-\frac{x^4}{16} \end{gathered}$$
Put this value in our equation
$$\begin{gathered} x^{2}y=\log x.\frac{x^4}{4}-\frac{x^4}{16}+C \\ y=\frac{x^2}{16}(4\log x-1)+C.x^{-2} \end{gathered}$$
Therefore, the general solution is $y=\frac{x^2}{16}(4\log x-1)+C.x^{-2}$

Question:7 Solve for general solutions.

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

Answer:

Given equation is
$x\log x\frac{dy}{dx}+y=\frac{2}{x}\log x$
we can rewrite it as
$\frac{dy}{dx}+\frac{y}{x\log x}=\frac{2}{x^2}$
This is $\frac{dy}{dx}+py=Q$ type where $p=\frac{1}{x\log x}$ and $Q=\frac{2}{x^2}$
Now,
$I.F.=e^{\int pdx}=e^{\int \frac{1}{x\log x}dx}=e^{\log (\log x)}=\log x$
Now, the solution of given differential equation is given by relation
$y(I.F.)=\int (Q\times I.F.)dx+C$
$y(\log x)=\int ((\frac{2}{x^2})\times \log x)dx+C$

take
$I=\int ((\frac{2}{x^2})\times \log x)dx$
$$\begin{gathered} I=\log x.\int \frac{2}{x^2}dx-\int (\frac{d(\log x)}{dt}.\int \frac{x^2}{2}dx)dx \\ I=-\log x.\frac{2}{x}+\int (\frac{1}{x}.\frac{2}{x})dx \\ I=-\log x.\frac{2}{x}+\int \frac{2}{x^2}dx \\ I=-\log x.\frac{2}{x}-\frac{2}{x} \end{gathered}$$
Put this value in our equation

$y\log x=-\frac{2}{x}(\log x+1)+C$
Therefore, the general solution is $y\log x=-\frac{2}{x}(\log x+1)+C$

Question:8 Find the general solution.

$(1+x^2)dy+2xydx=\cot xdx(x\ne 0)$

Answer:

Given equation is
$(1+x^2)dy+2xydx=\cot xdx(x\ne 0)$
we can rewrite it as
$\frac{dy}{dx}+\frac{2xy}{(1+x^2)}=\frac{\cot x}{1+x^2}$
This is $\frac{dy}{dx}+py=Q$ type where $p=\frac{2x}{1+x^2}$ and $Q=\frac{\cot x}{1+x^2}$
Now,
$I.F.=e^{\int pdx}=e^{\int \frac{2x}{1+x^2}dx}=e^{\log (1+x^2)}=1+x^2$
Now, the solution of the given differential equation is given by the relation
$y(I.F.)=\int (Q\times I.F.)dx+C$
$y(1+x^2)=\int ((\frac{\cot x}{1+x^2})\times (1+x^2))dx+C$
$$\begin{gathered} y(1+x^2)=\int \cot xdx+C \\ y(1+x^2)=\log |\sin x|+C \\ y=(1+x^2{)}^{-1}\log |\sin x|+C(1+x^2{)}^{-1} \end{gathered}$$
Therefore, the general solution is $y=(1+x^2{)}^{-1}\log |\sin x|+C(1+x^2{)}^{-1}$

Question:9 Solve for general solution.

$x\frac{dy}{dx}+y-x+xy\cot x=0(x\ne 0)$

Answer:

Given equation is
$x\frac{dy}{dx}+y-x+xy\cot x=0(x\ne 0)$
we can rewrite it as
$\frac{dy}{dx}+y.(\frac{1}{x}+\cot x)=1$
This is $\frac{dy}{dx}+py=Q$ type where $p=(\frac{1}{x}+\cot x)$ and $Q=1$
Now,
$I.F.=e^{\int pdx}=e^{\int (\frac{1}{x}+\cot x)dx}=e^{\log x+\log |\sin x|}=x.\sin x$
Now, the solution of the given differential equation is given by the relation
$y(I.F.)=\int (Q\times I.F.)dx+C$
$y(x.\sin x)=\int 1\times x\sin xdx+C$
$y(x.\sin x)=\int x\sin xdx+C$
Lets take
$$\begin{gathered} I=\int x\sin xdx \\ I=x.\int \sin xdx-\int (\frac{d(x)}{dx}.\int \sin xdx)dx \\ I=-x.\cos x+\int (\cos x)dx \\ I=-x\cos x+\sin x \end{gathered}$$
Put this value in our equation
$$\begin{gathered} y(x.\sin x)=-x\cos x+\sin x+C \\ y=-\cot x+\frac{1}{x}+\frac{C}{x\sin x} \end{gathered}$$
Therefore, the general solution is $y=-\cot x+\frac{1}{x}+\frac{C}{x\sin x}$

Question:10 Find the general solution.

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

Answer:

Given equation is
$(x+y)\frac{dy}{dx}=1$
we can rewrite it as
$$\begin{gathered} \frac{dy}{dx}=\frac{1}{x+y} \\ x+y=\frac{dx}{dy} \\ \frac{dx}{dy}-x=y \end{gathered}$$
This is $\frac{dx}{dy}+px=Q$ type where $p=-1$ and $Q=y$
Now,
$I.F.=e^{\int pdy}=e^{\int -1dy}=e^{-y}$
Now, the solution of given differential equation is given by relation
$x(I.F.)=\int (Q\times I.F.)dy+C$
$x(e^{-y})=\int y\times e^{-y}dy+C$
$x{e}^{-y}=\int y.e^{-y}dy+C$
Lets take
$$\begin{gathered} I=\int y{e}^{-y}dy \\ I=y.\int e^{-y}dy-\int (\frac{d(y)}{dy}.\int e^{-y}dy)dy \\ I=-y.e^{-y}+\int e^{-y}dy \\ I=-y{e}^{-y}-e^{-y} \end{gathered}$$
Put this value in our equation
$$\begin{gathered} x.e^{-y}=-e^{-y}(y+1)+C \\ x=-(y+1)+C{e}^y \\ x+y+1=C{e}^y \end{gathered}$$
Therefore, the general solution is $x+y+1=C{e}^y$

Question:11 Solve for general solution.

$ydx+(x-y^2)dy=0$

Answer:

Given equation is
$ydx+(x-y^2)dy=0$
we can rewrite it as
$\frac{dx}{dy}+\frac{x}{y}=y$
This is $\frac{dx}{dy}+px=Q$ type where $p=\frac{1}{y}$ and $Q=y$
Now,
$I.F.=e^{\int pdy}=e^{\int \frac{1}{y}dy}=e^{\log y}=y$
Now, the solution of given differential equation is given by relation
$x(I.F.)=\int (Q\times I.F.)dy+C$
$x(y)=\int y\times ydy+C$
$xy=\int y^{2}dy+C$
$xy=\frac{y^3}{3}+C$
$x=\frac{y^2}{3}+\frac{C}{y}$
Therefore, the general solution is $x=\frac{y^2}{3}+\frac{C}{y}$

Question:12 Find the general solution.

$(x+3y^2)\frac{dy}{dx}=y(y>0)$

Answer:

Given equation is
$(x+3y^2)\frac{dy}{dx}=y(y>0)$
we can rewrite it as
$\frac{dx}{dy}-\frac{x}{y}=3y$
This is $\frac{dx}{dy}+px=Q$ type where $p=\frac{-1}{y}$ and $Q=3y$
Now,
$I.F.=e^{\int pdy}=e^{\int \frac{-1}{y}dy}=e^{-\log y}=y^{-1}=\frac{1}{y}$
Now, the solution of given differential equation is given by relation
$x(I.F.)=\int (Q\times I.F.)dy+C$
$x(\frac{1}{y})=\int 3y\times \frac{1}{y}dy+C$
$\frac{x}{y}=\int 3dy+C$
$\frac{x}{y}=3y+C$
$x=3y^2+Cy$
Therefore, the general solution is $x=3y^2+Cy$

Question:13 Solve for particular solution.

$\frac{dy}{dx}+2y\tan x=\sin x;y=0whenx=\frac{\pi }{3}$

Answer:

Given equation is
$\frac{dy}{dx}+2y\tan x=\sin x;y=0whenx=\frac{\pi }{3}$
This is $\frac{dy}{dx}+py=Q$ type where $p=2\tan x$ and $Q=\sin x$
Now,
$I.F.=e^{\int pdx}=e^{\int 2\tan xdx}=e^{2\log |\sec x|}={\sec }^{2}x$
Now, the solution of given differential equation is given by relation
$y(I.F.)=\int (Q\times I.F.)dx+C$
$y({\sec }^{2}x)=\int ((\sin x)\times {\sec }^{2}x)dx+C$
$$\begin{gathered} y({\sec }^{2}x)=\int (\sin \times \frac{1}{\cos x}\times \sec x)dx+C \\ y({\sec }^{2}x)=\int \tan x\sec xdx+C \\ y.{\sec }^{2}x=\sec x+C \end{gathered}$$
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when $x=\frac{\pi }{3}$
at $x=\frac{\pi }{3}$
$$\begin{gathered} 0.\sec \frac{\pi }{3}=\sec \frac{\pi }{3}+C \\ C=-2 \end{gathered}$$
Now,

$$\begin{gathered} y.{\sec }^{2}x=\sec x-2 \\ \frac{y}{{\cos }^{2}x}=\frac{1}{\cos x}-2 \\ y=\cos x-2{\cos }^{2}x \end{gathered}$$
Therefore, the particular solution is $y=\cos x-2{\cos }^{2}x$

Question:14 Solve for particular solution.

$(1+x^2)\frac{dy}{dx}+2xy=\frac{1}{1+x^2};y=0whenx=1$

Answer:

Given equation is
$(1+x^2)\frac{dy}{dx}+2xy=\frac{1}{1+x^2};y=0whenx=1$
we can rewrite it as
$\frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{1}{(1+x^2{)}^2}$
This is $\frac{dy}{dx}+py=Q$ type where $p=\frac{2x}{1+x^2}$ and $Q=\frac{1}{(1+x^2{)}^2}$
Now,
$I.F.=e^{\int pdx}=e^{\int \frac{2x}{1+x^2}dx}=e^{\log |1+x^2|}=1+x^2$
Now, the solution of given differential equation is given by relation
$y(I.F.)=\int (Q\times I.F.)dx+C$
$y(1+x^2)=\int (\frac{1}{(1+x^2{)}^2}\times (1+x^2))dx+C$
$$\begin{gathered} y(1+x^2)=\int \frac{1}{(1+x^2)}dx+C \\ y(1+x^2)={\tan }^{-1}x+C \end{gathered}$$
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when x = 1
at x = 1
$$\begin{gathered} 0.(1+1^2)={\tan }^{-1}1+C \\ C=-\frac{\pi }{4} \end{gathered}$$
Now,
$y(1+x^2)={\tan }^{-1}x-\frac{\pi }{4}$
Therefore, the particular solution is $y(1+x^2)={\tan }^{-1}x-\frac{\pi }{4}$

Question:15 Find the particular solution.

$\frac{dy}{dx}-3y\cot x=\sin 2x;y=2whenx=\frac{\pi }{2}$

Answer:

Given equation is
$\frac{dy}{dx}-3y\cot x=\sin 2x;y=2whenx=\frac{\pi }{2}$
This is $\frac{dy}{dx}+py=Q$ type where $p=-3\cot x$ and $Q=\sin 2x$
Now,
$I.F.=e^{\int pdx}=e^{-3\cot xdx}=e^{-3\log |\sin x|}={\sin }^{-3}x=\frac{1}{{\sin }^{3}x}$
Now, the solution of given differential equation is given by relation
$y(I.F.)=\int (Q\times I.F.)dx+C$
$y(\frac{1}{{\sin }^{3}x})=\int (\sin 2x\times \frac{1}{{\sin }^{3}x})dx+C$
$\frac{y}{{\sin }^{3}x}=\int (2\sin x\cos x\times \frac{1}{{\sin }^{3}x})dx+C$
$\frac{y}{{\sin }^{3}x}=\int (2\times \frac{\cos x}{\sin x}\times \frac{1}{\sin x})dx+C$
$\frac{y}{{\sin }^{3}x}=\int (2\times \cot x\times cosecx)dx+C$
$\frac{y}{{\sin }^{3}x}=-2cosecx+C$
Now, by using boundary conditions we will find the value of C
It is given that y = 2 when $x=\frac{\pi }{2}$
at $x=\frac{\pi }{2}$
$$\begin{gathered} \frac{2}{{\sin }^3\frac{\pi }{2}}=-2cosec\frac{\pi }{2}+C \\ 2=-2+C \\ C=4 \end{gathered}$$
Now,
$y=4{\sin }^{3}x-2{\sin }^{2}x$
Therefore, the particular solution is $y=4{\sin }^{3}x-2{\sin }^{2}x$

Question:16 Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

Answer:

Let f(x , y) is the curve passing through origin
Then, the slope of tangent to the curve at point (x , y) is given by $\frac{dy}{dx}$
Now, it is given that
$$\begin{gathered} \frac{dy}{dx}=y+x \\ \frac{dy}{dx}-y=x \end{gathered}$$
It is $\frac{dy}{dx}+py=Q$ type of equation where $p=-1andQ=x$
Now,
$I.F.=e^{\int pdx}=e^{\int -1dx}=e^{-x}$
Now,
$y(I.F.)=\int (Q\times I.F.)dx+C$
$y(e^{-x})=\int (x\times e^{-x})dx+C$
Now, Let
$$\begin{gathered} I=\int (x\times e^{-x})dx \\ I=x.\int e^{-x}dx-\int (\frac{d(x)}{dx}.\int e^{-x}dx)dx \\ I=-x{e}^{-x}+\int e^{-x}dx \\ I=-x{e}^{-x}-e^{-x} \\ I=-e^{-x}(x+1) \end{gathered}$$
Put this value in our equation
$y{e}^{-x}=-e^{-x}(x+1)+C$
Now, by using boundary conditions we will find the value of C
It is given that curve passing through origin i.e. (x , y) = (0 , 0)
$$\begin{gathered} 0.e^{-0}=-e^{-0}(0+1)+C \\ C=1 \end{gathered}$$
Our final equation becomes
$$\begin{gathered} y{e}^{-x}=-e^{-x}(x+1)+1 \\ y+x+1=e^x \end{gathered}$$
Therefore, the required equation of the curve is $y+x+1=e^x$

Question:17 Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Answer:

Let f(x , y) is the curve passing through point (0 , 2)
Then, the slope of tangent to the curve at point (x , y) is given by $\frac{dy}{dx}$
Now, it is given that
$$\begin{gathered} \frac{dy}{dx}+5=y+x \\ \frac{dy}{dx}-y=x-5 \end{gathered}$$
It is $\frac{dy}{dx}+py=Q$ type of equation where $p=-1andQ=x-5$
Now,
$I.F.=e^{\int pdx}=e^{\int -1dx}=e^{-x}$
Now,
$y(I.F.)=\int (Q\times I.F.)dx+C$
$y(e^{-x})=\int ((x-5)\times e^{-x})dx+C$
Now, Let
$$\begin{gathered} I=\int ((x-5)\times e^{-x})dx \\ I=(x-5).\int e^{-x}dx-\int (\frac{d(x-5)}{dx}.\int e^{-x}dx)dx \\ I=-(x-5)e^{-x}+\int e^{-x}dx \\ I=-x{e}^{-x}-e^{-x}+5e^{-x} \\ I=-e^{-x}(x-4) \end{gathered}$$
Put this value in our equation
$y{e}^{-x}=-e^{-x}(x-4)+C$
Now, by using boundary conditions we will find the value of C
It is given that curve passing through point (0 , 2)
$$\begin{gathered} 2.e^{-0}=-e^{-0}(0-4)+C \\ C=-2 \end{gathered}$$
Our final equation becomes
$$\begin{gathered} y{e}^{-x}=-e^{-x}(x-4)-2 \\ y=4-x-2e^x \end{gathered}$$
Therefore, the required equation of curve is $y=4-x-2e^x$