Ex 9.6 Class 12 Maths Ncert Solutions

It is given that \( \frac{d y}{d x}+2 y=\sin x \)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( p=2 \) and \( Q=\sin x \))
Now, I.F. \( ={e}^{\int {pdx}}={e}^{\int 2 {dx}}={e}^{2 x} \)
Thus, the solution of the given differential equation is given by the relation:
\({y}(\text {I. F.})=\int({Q} \times \text {I. F.}) {dx}+{C}\)
\(\Rightarrow y e^{2 x}=\int \sin x . e^{2 x} d x+C\quad \ldots\ldots\text{(1)}\)
Let \( {I}=\int \sin x \cdot e^{2 x} d x \)
\(\Rightarrow {I}=\sin x \cdot \int e^{2 x} d x-\int\left(\frac{d}{d x}(\sin x) \cdot {e}^{\int {d} {d}}\right) d x\)
\(=\sin x \cdot \frac{e^{2 x}}{2}-\int\left(\cos x \cdot \frac{e^{2 x}}{2}\right) d x\)
\(\left.=\frac{e^{2 x} \sin x}{2}-\frac{1}{2}\left[\cos \int e^{2 x}-\int\left(\frac{d}{d x}(\cos x) \cdot \iint e^{2 x} d x\right)\right)\right]\)
\(=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{2}-\frac{1}{4} \int\left(\sin x \cdot e^{2 x}\right) d x\)
\(\Rightarrow \frac{e^{2 x}}{2}(2 \sin x-\cos x)-\frac{1}{4} {I}\)
\(\Rightarrow \frac{5}{4} {I}=\frac{e^{2 x}}{4}(2 \sin x-\cos x)\)
\(\Rightarrow {I}=\frac{e^{2 x}}{5}(2 \sin x-\cos x)\)
Now, putting the value of I in (1), we get,
\(\Rightarrow y e^{2 x}=\frac{e^{2 x}}{5}(2 \sin x-\cos x)+C\)
\(\Rightarrow y=\frac{1}{5}(2 \sin x-\cos x)+C e^{-2 x}\)
Therefore, the required general solution of the given differential equation is
\(y=\frac{1}{5}(2 \sin x-\cos x)+C e^{-2 x}\)

It is given that \( \frac{d y}{d x}+3 y=e^{-2 x} \)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( {p}=3 \) and \( {Q}=e^{-2 x} \))
Now, I.F. \( =e^{\int p d x}=e^{\int 3 d x}=e^{3 x} \)
Thus, the solution of the given differential equation is given by the relation:
\(y(\text {I.F.})=\int(Q \times \text {I. F.}) d x+C\)
\(\Rightarrow y^{3 x}=\int\left(e^{-2 x} \times e^{2 x}\right) d x+C\)
\(\Rightarrow y^{3 x}=e^{x}+C\)
\(\Rightarrow y={e}^{-2 x}+{Ce}^{-3 x}\)
Therefore, the required general solution of the given differential equation is \( y=e^{-2 x}+C^{-3 x} \)

It is given that \( \frac{d y}{d x}+\frac{y}{x}=x^{2} \)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( {p}=\frac{1}{x} \) and \( {Q}=x^{2} \))
Now, I.F. \( =\int(Q \times \) I.F.\( ) {dx}+{C} \)
\(\Rightarrow y(x)=\int\left(x^{2} \cdot x\right) d x+C\)
\(\Rightarrow x y=\int\left(x^{3}\right) d x+C\)
\(\Rightarrow x y=\frac{x^{4}}{4}+C\)
Therefore, the required general solution of the given differential equation is \( x y=\frac{x^{4}}{4}+C \).

It is given that \( \frac{d y}{d x}+(\sec x) y=\tan x \)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( {p}=\sec x \) and \( \left.{Q}=\tan x\right) \)
Now, I.F. \( =\int(Q \times \) I.F.\( ) d x+C \)
\(\Rightarrow {y}(\sec x+\tan x)=\int \tan x(\sec x+\tan x) {dx}+{C}\)
\(\Rightarrow {y}(\sec x+\tan x)=\int \sec x \tan x d x+\int \tan ^{2} x d x+{C}\)
\(\Rightarrow {y}(\sec x+\tan x)=\sec x+\int\left(\sec ^{2} x-1\right) d x+C\)
\(\Rightarrow {y}(\sec x+\tan x)=\sec x+\tan x-x+{C}\)
Therefore, the required general solution of the given differential equation is
\(y(\sec x+\tan x)=\sec x+\tan x-x+C\)

It is given that \( \cos ^{2} x \frac{d y}{d x}+y=\tan x \)
\(\Rightarrow \frac{d y}{d x}+\sec ^{2} x \cdot y=\sec ^{2} x \tan x\)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( {p}=\sec ^{2} x \) and \(Q =\sec ^{2} x \tan x\))
Now, I.F. \( =e^{\int p d x}=e^{\int \sec ^{2} x d x}=e^{\tan x} \)
Thus, the solution of the given differential equation is given by the relation:
\(y(\text {I.F.})=\int(Q \times \text { I.F.}) d x+C\)
\(\Rightarrow y \cdot e^{\tan x}=\int e^{\tan x} d x+C\quad \ldots\ldots\text{(1)}\)
Now, Let \( {t}=\tan x \)
\(\Rightarrow \frac{d}{d x}(\tan x)=\frac{d t}{d x}\)
\(\Rightarrow \sec ^{2} x=\frac{d t}{d x}\)
\(\Rightarrow \sec ^{2} {xdx}={dt}\)
Thus, the equation (1) becomes,
\(\Rightarrow y \cdot e^{\tan x}=\int\left(e^{t} \cdot t\right) d t+C\)
\(\Rightarrow y \cdot e^{\tan x}=\int\left(t \cdot e^{t}\right) d t+C\)
\(\Rightarrow y \cdot e^{\tan x}=t \int e^{t} d t-\int\left(\frac{d}{d t}(t) \cdot \int e^{t} d t\right)+C\)
\(\Rightarrow y \cdot e^{\tan x}=t \cdot e^{t}-\int e^{t} d t+C\)
\(\Rightarrow {t} {e}^{\tan x}=({t}-1) {e}^{{t}}+{C}\)
\(\Rightarrow {t} {e}^{\tan x}=(\tan x-1) {e}^{\tan x}+{C}\)
\(\Rightarrow {y}=(\tan x-1)+{C} {e}^{\tan x}\)
Therefore, the required general solution of the given differential equation is
\(y=(\tan x-1)+C e^{-\tan x} .\)

It is given that \( x \frac{d y}{d x}+2 y=x^{2} \log x \)
\(\Rightarrow \frac{d y}{d x}+\frac{2}{x} y=x^{2} \log x\)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( {p}=\frac{2}{x} \) and \(Q =x \log x \))
Now, I.F. \( =e^{\int p d x}=e^{\int \frac{2}{x} d x}=e^{2(\log x)}=e^{\log x^{2}}=x^{2} \)
Thus, the solution of the given differential equation is given by the relation:
\(\text {y(I.F.})=\int({Q} \times {I} . {F} .) {dx}+{C}\)
\(\Rightarrow y \cdot x^{2}=\int\left(x \log x \cdot x^{2}\right) d x+C\)
\(\Rightarrow x^{2} y=\int\left(x^{3} \log x\right) d x+C\)
\(\Rightarrow x^{2} y=\log x \cdot \int x^{3} d x-\int\left[\frac{d}{d x}(\log x) \cdot \int x^{3} d x\right] d x+C\)
\(\Rightarrow x^{2} y=\log x \cdot \frac{x^{4}}{4}-\int\left(\frac{1}{x} \cdot \frac{x^{4}}{4}\right) d x+C\)
\(\Rightarrow x^{2} y=\frac{x^{4} \log x}{4}-\frac{1}{x} \cdot \frac{x^{4}}{4}+C\)
\(\Rightarrow x^{2} y=\frac{1}{16} x^{4}-(4 \log x-1)+C\)
\(\Rightarrow y=\frac{1}{16} x^{2}-(4 \log x-1)+C x^{-2}\)
Therefore, the required general solution of the given differential equation
\(y=\frac{1}{16} x^{2}-(4 \log x-1)+C x^{-2}\)

It is given that \( x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x \)
\(\Rightarrow \frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x^{2}}\)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( {p}=\frac{1}{x \log x} \) and \(Q =\frac{2}{x^{2}} \))
Now, I.F. \( =e^{\int p d x}=e^{\int \frac{1}{x \log x} d x}=e^{\log (\log x)}=\log x \)
Thus, the solution of the given differential equation is given by the relation:
\(y(\text {I.F.})=\int({Q} \times \text {I. F.}) {d} x+{C}\)
\(\Rightarrow y \cdot \log x=\int\left[\frac{2}{x^{2}} \cdot \log x\right] d x+C\quad \ldots\ldots\text{(1)}\)
Now, \( \int\left[\frac{2}{x^{2}} \cdot \log x\right] d x=2 \int\left(\log \frac{1}{x^{2}}\right) d x \)
\(=2\left[\log x \cdot \int \frac{1}{x^{2}} d x-\int\left\{\frac{d}{d x}(\log x) \cdot \int \frac{1}{x^{2}} d x\right\} d x\right]\)
\(=2\left[\log x\left(-\frac{1}{x}\right)-\int\left(\frac{1}{x} \cdot\left(-\frac{1}{x}\right)\right)\right]\)
\(=2\left[-\frac{\log x}{x}+\int \frac{1}{x^{2}} d x\right]\)
\(=-\frac{2}{x}(1+\log x)\)
Now, substituting the value in (1), we get,
\(\Rightarrow y \cdot \log x=-\frac{2}{x}(1+\log x)+C\)
Therefore, the required general solution of the given differential equation is
\(y \cdot \log x=-\frac{2}{x}(1+\log x)+C\)

It is given that \( \left(1+x^{2}\right) d y+2 {xy} {dx}=\cot {xdx} \)
\(\Rightarrow \frac{d y}{d x}+\frac{2 x y}{\left(1+x^{2}\right)}=\frac{\cot x}{1+x^{2}}\)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( {p}=\frac{2 x y}{\left(1+x^{2}\right)} \) and \(Q =\frac{\cot x}{1+x^{2}}\))
Now, I.F. \( =e^{\int p d x}=e^{\int \frac{2 x y}{\left(1+x^{2}\right)} d x}=e^{\log \left(1+x^{2}\right)}=1+x^{2} \)
Thus, the solution of the given differential equation is given by the relation:
\(y(\text {I.F.})=\int(Q \times \text { I.F.}) {dx}+{C}\)
\(\Rightarrow {y} \cdot\left(1+x^{2}\right)=\int\left[\frac{\cot x}{1+x^{2}} \cdot\left(1+x^{2}\right)\right] {dx}+{C}\)
\(\Rightarrow {y} \cdot\left(1+x^{2}\right)=\int \cot x {dx}+{C}\)
\(\Rightarrow {y}\left(1+x^{2}\right)=\log |\sin x|+{C}\)
Therefore, the required general solution of the given differential equation is
\(y\left(1+x^{2}\right)=\log |\sin x|+C\)

It is given that \( x \frac{d y}{d x}+y-x+x y \cot x=0 \)
\(\Rightarrow x \frac{d y}{d x}+y(1+x \cot x)=x\)
\(\Rightarrow x \frac{d y}{d x}+\left(\frac{1}{x}+\cot x\right) y=1\)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( {p}=\frac{1}{x} \) and \( {Q}=1 \))
\(\text{Now, I.F. } =e^{\int p d x}=e^{\int\left(\frac{1}{x}+\cot x\right) d y}=e^{\log x+\log (\sin x)}=e^{\log (x \sin x)}=x \sin x \)
Thus, the solution of the given differential equation is given by the relation:
\(\text { y(I.F.})=\int({Q} \times \text {I. F.}) {dx}+{C}\)
\(\Rightarrow {y} \cdot(x \sin x)=\int[1 \times x \sin x] {dx}+{C}\)
\(\Rightarrow {y} \cdot(x \sin x)=\int[x \sin x] {dx}+{C}\)
\(\Rightarrow {y}(x \sin x)=x \int \sin x d x-\int\left[\frac{d}{d x}(x) \cdot \int \sin x d x\right]+C\)
\(\Rightarrow {y}(x \sin x)=x(-\cos x)-\int 1 \cdot(-\cos x) d x+C\)
\(\Rightarrow {y}(x \sin x)=-x \cos x+\sin x+{C}\)
\(\Rightarrow y=\frac{-x \cos x}{x \sin x}+\frac{\sin x}{x \sin x}+\frac{C}{x \sin x}\)
\(\Rightarrow y=\cot x+\frac{1}{x}+\frac{C}{x \sin x}\)
Therefore, the required general solution of the given differential equation is
\(y=\cot x+\frac{1}{x}+\frac{C}{x \sin x}\)

It is given that \( (x+y) \frac{d y}{d x}=1 \)
\(\Rightarrow \frac{d y}{d x}=\frac{1}{x+y}\)
\(\Rightarrow \frac{d x}{d y}=x+y\)
\(\Rightarrow \frac{d x}{d y}-x=y\)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( {p}=-1 \) and \( {Q}=y \))
Now, I.F. \( =e^{\int p d y}=e^{\int-d y}=e^{-y} \)
Thus, the solution of the given differential equation is given by the relation:
\(x(\text {I.F.})=\int({Q} \times \text {I. F.}) {dy}+{C}\)
\(\Rightarrow x e^{-y}=\int\left[y . e^{-y}\right] {dy}+{C}\)
\(\Rightarrow x e^{-y}=y \int e^{-d y}-\int\left[\frac{d}{d x}(y) \cdot \int e^{-y} d y\right]+C\)
\(\Rightarrow x e^{-y}=y\left(-e^{-y}\right)-\int\left(-e^{-y}\right) d y+C\)
\(\Rightarrow x e^{-y}=-y e^{-y}-\int e^{-y} d y+C\)
\(\Rightarrow x e^{-y}=-y e^{-y}-e^{-y}+C\)
\(\Rightarrow x=-{y}-1+{Ce}^{{y}}\)
\(\Rightarrow x+{y}+1={Ce}^{{y}}\)
Therefore, the required general solution of the given differential equation is
\(x+{y}+1={Ce}^{{y}}\).

It is given that \( y d x+\left(x-y^{2}\right) d y=0 \)
\(\Rightarrow y d x=\left(y^{2}-x\right) d y\)
\(\Rightarrow \frac{d x}{d y}=\frac{\left(y^{2}-x\right)}{y}=y-\frac{x}{y}\)
\(\Rightarrow \frac{d x}{d y}+\frac{x}{y}=y\)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( {p}=\frac{1}{y} \) and \( {Q}=y \))
Now, I.F. \( =e^{\int p d y}=e^{\int \frac{d y}{y}}=e^{\log y}=y \)
Thus, the solution of the given differential equation is given by the relation:
\(x( \text{I.F.})=\int({Q} \times \text{I.F.}) {dy}+{C}\)
\(\Rightarrow x \cdot y=\int[y \cdot y] {dy}+{C}\)
\(\Rightarrow x \cdot y=\int y^{2} {dy}+{C}\)
\(\Rightarrow x \cdot y=\frac{y^{3}}{3}+C\)
\(\Rightarrow x y=\frac{y^{3}}{3}+\frac{c}{y}\)
Therefore, the required general solution of the given differential equation is \( x y=\frac{y^{3}}{3}+\frac{C}{y} \)

It is given that \( \left(x+3 y^{2}\right) \frac{d y}{d x}=y \)
\(\Rightarrow \frac{d y}{d x}=\frac{y}{x+3 y^{2}}\)
\(\Rightarrow \frac{d x}{d y}=\frac{x+3 y^{2}}{y}=\frac{x}{y}+3 y\)
\(\Rightarrow \frac{d x}{d y}-\frac{x}{y}=3 y\)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( {p}=-\frac{1}{y} \) and \( {Q}=3 y \))
Now, I.F. \( =e^{\int p d y}=e^{\int \frac{d y}{y}}=e^{-\log y}=e^{\log \left(\frac{1}{y}\right)}=\frac{1}{y} \)
Thus, the solution of the given differential equation is given by the relation:
\(x( \text{I.F.})=\int({Q} \times \text{I.F.}) {dy}+{C}\)
\(\Rightarrow x \cdot \frac{1}{y}=\int\left[3 y \cdot \frac{1}{y}\right] {dy}+{C}\)
\(\Rightarrow \frac{x}{y}=3 y+C\)
\(\Rightarrow x=3 y^{2}+C y\)
Therefore, the required general solution of the given differential equation is \( x=3 y^{2}+C y \).

It is given that \( \frac{d y}{d x}+2 y \tan x=\sin x \)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( {p}=2 \tan x \) and \( Q =\sin x) \)
Now, I.F. \( =e^{\int p d x}=e^{\int 2 \tan x d x}=e^{2 \log (\sec x)}=e^{\log \left(\sec ^{2} x\right)}=\sec ^{2} x \)
Thus, the solution of the given differential equation is given by the relation:
\(y(\text {I.F.})=\int(Q \times \text { I.F.}) d x+C\)
\(\Rightarrow y .\left(\sec ^{2} x\right)=\int\left[\sin x \cdot \sec ^{2} x\right] d x+C\)
\(\Rightarrow y .\left(\sec ^{2} x\right)=\int[\sec x \cdot \tan x] d x+C\)
\(\Rightarrow y \cdot\left(\sec ^{2} x\right)=\sec x+C\quad \ldots\ldots\text{(1)}\)
Now, it is given that \( {y}=0 \) at \( x=\frac{\pi}{3} \)
\(0 \times \sec ^{2} \frac{\pi}{3}=\sec \frac{\pi}{3}+C\)
\(\Rightarrow 0=2+C\)
\(\Rightarrow C=-2\)
Now, Substituting the value of \( {C}=-2 \) in (1), we get,
\(\Rightarrow y \cdot\left(\sec ^{2} x\right)=\sec x-2\)
\(\Rightarrow {y}=\cos x-2 \cos ^{2} x\)
Therefore, the required general solution of the given differential equation is
\(y=\cos x-2 \cos ^{2} x\)

It is given that \( \left(1+x^{2}\right) \frac{d y}{d x}+2 x y=\frac{1}{1+x^{2}} \)
\(\Rightarrow \frac{d y}{d x}+\frac{2 x y}{\left(1+x^{2}\right)}=\frac{1}{\left(1+x^{2}\right)^{2}}\)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( {p}=\frac{2 x}{\left(1+x^{2}\right)} \) and \( {Q}= \frac{1}{\left(1+x^{2}\right)^{2}}\))
Now, I.F. \( =e^{\int p d x}=e^{\int \frac{2 x}{\left(1+x^{2}\right)} d x}=e^{\log \left(1+x^{2}\right)}=1+x^{2} \)
Thus, the solution of the given differential equation is given by the relation:
\(y(\text {I.F.})=\int(Q \times \text {I. F.}) d x+C\)
\(\Rightarrow y \cdot\left(1+x^{2}\right)=\int\left[\frac{1}{\left(1+x^{2}\right)^{2}} \cdot\left(1+x^{2}\right)\right] d x+C\)
\(\Rightarrow y \cdot\left(1+x^{2}\right)=\int \frac{1}{\left(1+x^{2}\right)} d x+C\)
\(\Rightarrow y \cdot\left(1+x^{2}\right)=\tan ^{-1} x+C\quad \ldots\ldots\text{(1)}\)
Now, it is given that \( {y}=0 \) at \( x=1 \)
\(0 \times \sec ^{2} \frac{\pi}{3}=\sec \frac{\pi}{3}+C\)
\(\Rightarrow 0=\tan ^{-1} 1+{C}\)
\(\Rightarrow {C}=-\frac{\pi}{4}\)
Now, Substituting the value of \( {C}=-\frac{\pi}{4} \) in (1), we get,
\(\Rightarrow y \cdot\left(\sec ^{2} x\right)=\sec x-2\)
\(\Rightarrow y \cdot\left(1+x^{2}\right)=\tan ^{-1} x-\frac{\pi}{4}\)
Therefore, the required general solution of the given differential equation is
\( y .\left(1+x^{2}\right)=\tan ^{-1} x-\frac{\pi}{4} \)

It is given that \( \frac{d y}{d x}-3 y \cot x=\sin 2 x \)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( {p}=-3 \cot x \) and \( Q =\sin 2 x) \)
Now, I.F. \( =e^{\int p d x}=e^{-3 \int \cot x d x}=e^{-3 \log |\sin x|}=e^{\log \left|\frac{1}{\sin ^{3} x}\right|}=\frac{1}{\sin ^{3} x} \)
Thus, the solution of the given differential equation is given by the relation:
\({y}(\text {I.F.})=\int({Q} \times \text {I. F.}) {dx}+{C}\)
\(\Rightarrow {y} \cdot \frac{1}{\sin ^{3} x}=\int\left[\sin 2 x \cdot \frac{1}{\sin ^{3} x}\right] {dx}+{C}\)
\(\Rightarrow {y} \cdot \operatorname{cosec}^{3} x=2=\int(\cot x \operatorname{cosec} x) {dx}+{C}\)
\(\Rightarrow {y} \operatorname{cosec}^{3} x=2 \operatorname{cosec} x+{C}\)
\(\Rightarrow y=-\frac{2}{\operatorname{cosec}^{2} x}+\frac{3}{\operatorname{cosec}^{3} x}\)
\(\Rightarrow {y}=-2 \sin ^{2} x+{C} \sin ^{3} x\quad \ldots\ldots\text{(1)}\)
Now, it is given that \( {y}=2 \) when \( x=\frac{\pi}{2} \)
Thus, we get,
\(2=-2+C\)
\(\Rightarrow C=4\)
Now, Substituting the value of \( {C}=4 \) in (1), we get,
\(y=-2 \sin ^{2} x+4 \sin ^{3} x\)
\(\Rightarrow y=4 \sin ^{3} x-2 \sin ^{2} x\)
Therefore, the required general solution of the given differential equation is
\({y}=4 \sin ^{3} x-2 \sin ^{2} x\)

Let \( {F}(x, {y}) \) be the curve passing through origin and let \( (x, {y}) \) be a point on the curve.
We know the slope of the tangent to the curve at \( (x, {y}) \) is \( \frac{d y}{d x} \).
According to the given conditions, we get,
\(\frac{d y}{d x}=x+y\)
\(\Rightarrow \frac{d y}{d x}-y=x\)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( {p}=-1 \) and \( {Q}=x \))
Now, I.F. \( =e^{\int p d x}=e^{\int(-1) d x}=e^{-x} \)
Thus, the solution of the given differential equation is given by the relation:
\({y}( \text{I.F.})=\int({Q} \times \text {I. F.}) {dy}+{C}\)
\(\Rightarrow {y} e^{-x}=x e^{-x} {dx}+{C}\quad \ldots\ldots\text{(1)}\)
Now, \( \int x e^{-x} d x=x \int e^{-x} d x-\int\left[\frac{d}{d x}(x) . \int e^{-x} d x\right] d x \)
\(=x\left(e^{-x}\right)-\int\left(-e^{-x}\right) d x\)
\(=x\left(e^{-x}\right)-\left(-e^{-x}\right)\)
\(=-e^{-x}(x+1)\)
Thus, from equation (1), we get,
\(\Rightarrow {y} e^{-x}=-e^{-x}(x+1)+{C}\)
\(\Rightarrow {y}=-(x+1)+{Ce}^{x}\)
\(\Rightarrow x+{y}+1={Ce}^{x}\quad \ldots\ldots\text{(2)}\)
Now, it is given that curve passes through origin.
Thus, equation (2) becomes:
\(1={C}\)
\(\Rightarrow {C}=1\)
Substituting \( {C}=1 \) in equation (2), we get,
\(x+y-1=e^{x}\)
Therefore, the required general solution of the given differential equation is
\(x+y-1=e^{x}\)

Let \( {F}(x, {y} ) \) be the curve and let (\( x, {y} \)) be a point on the curve.
We know the slope of the tangent to the curve at \( (x, {y}) \) is \( \frac{d y}{d x} \).
According to the given conditions, we get,
\(\frac{d y}{d x}+5=x+y\)
\(\Rightarrow \frac{d y}{d x}-y=x-5\)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( {p}=-1 \) and \( {Q}=-5 \))
Now, I.F. \( =e^{\int p d x}=e^{\int(-1) d x}=e^{-x} \)
Thus, the solution of the given differential equation is given by the relation:
\(\text { y(I.F.})=\int({Q} \times \text { I. F.}) {dy}+\text { C }\)
\(\Rightarrow {y} e^{-x}=(x-5) e^{-x} {dx}+{C}\quad \ldots\ldots\text{(1)}\)
\(\text {Now, } \int(x-5) e^{-x} d x=(x-5) \int e^{-x} d x-\int\left[\frac{d}{d x}(x-5) \cdot \int e^{-x} d x\right] d x\)
\(=(x-5)\left(e^{-x}\right)-\int\left(-e^{-x}\right) d x\)
\(=(x-5)\left(e^{-x}\right)-\left(-e^{-x}\right)\)
\(=(4-x)-e^{-x}\)
Thus, from equation (1), we get,
\(\Rightarrow {y} e^{-x}=(4-x) e^{-x}+{C}\)
\(\Rightarrow {y}=4-x+C {e}^{x}\)
\(\Rightarrow x+{y}-4=C {e}^{x}\)
Now, it is given that curve passes through \( (0,2) \).
Thus, equation (2) becomes:
\(0+2-4={Ce}^{0}\)
\(\Rightarrow-2={C}\)
\(\Rightarrow {C}=-2\)
Substituting \( C=-2 \) in equation (2), we get,
\(x+{y}-4=-2 {e}^{x}\)
\(\Rightarrow {y}=4-x-2 {e}^{x}\)
Therefore, the required general solution of the given differential equation is
\(y=4-x-2 e^{x}\)

It is given that \( \frac{d y}{d x}-y=2 x^{2} \)
\(\Rightarrow \frac{d y}{d x}-\frac{y}{x}=2 x\)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( {p}=-\frac{1}{x} \) and \( {Q}=2 x \))
Now, I.F. \( =e^{\int p d x}=e^{\int-\frac{1}{x} d x}=e^{\log \left(x^{-1}\right)}=x^{-1}=\frac{1}{x} \)

It is given that \( \left(1-y^{2}\right) \frac{d y}{d x}+y x=a y \)
\(\Rightarrow \frac{d y}{d x}-\frac{y x}{1-y^{2}}=\frac{a y}{1-y^{2}}\)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( {p}=\frac{y}{1-y^{2}} \) and \( {Q}= \frac{a}{1-y^{2}}\))
Now, I.F. \( =e^{\int p d y}=e^{\int \frac{y}{1-y^{2}} d y}=e^{\frac{1}{2} \log \left(1-y^{2}\right)}=e^{\log \left[\frac{1}{\sqrt{1-y^{2}}}\right]} \)
\(=\frac{1}{\sqrt{1-y^{2}}}\)