Class 12 Maths Chapter 9 Miscellaneous Exercise Solutions

It is given that equation is \( \frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}-6 y=\log x \)
\(\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}-6 y-\log x=0\)
We can see that the highest order derivative present in the differential is \( \frac{d^{2} y}{d x^{2}} \).
Thus, its order is two. It is polynomial equation in \( \frac{d^{2} y}{d x^{2}} \). The highest power raised to \( \frac{d^{2} y}{d x^{2}} \) is \(1 \).
Therefore, its degree is one.

It is given that equation is \( \left(\frac{d y}{d x}\right)^{3}-4\left(\frac{d y}{d x}\right)^{2}+7 y=\sin x \)
\(\left(\frac{d y}{d x}\right)^{3}-4\left(\frac{d y}{d x}\right)^{2}+7 y-\sin x=0\)
We can see that the highest order derivative present in the differential is \( \frac{d y}{d x} \).
Thus, its order is one. It is polynomial equation in \( \frac{d y}{d x} \) The highest power raised to \( \frac{d y}{d x} \) is \(3 \).
Therefore, its degree is three.

It is given that equation is \( \frac{d^{4} y}{d x^{4}}-\sin \left(\frac{d^{3} y}{d x^{3}}\right)=0 \)
\(\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}-6 y-\log x=0\)
We can see that the highest order derivative present in the differential is \( \frac{d^{4} y}{d x^{4}} \).
Thus, its order is four. The given differential equation is not a polynomial equation.
Therefore, its degree is not defined.

It is given that \( x y=a e^{x}+b e^{-x}+x^{2} \)
Now, differentiating both sides w.r.t. \(x\), we get,
\(\frac{d y}{d x}=a \frac{d}{d x}\left(e^{x}\right)+b \frac{d}{d x}\left(e^{-x}\right)+\frac{d}{d x}\left(x^{2}\right)\)
\(\Rightarrow \frac{d y}{d x}=a e^{x}+b e^{-x}+2 x\)
Now, Again differentiating both sides w.r.t. \( x \), we get,
\(\frac{d}{d x}\left(y^{\prime}\right)=\frac{d}{d x}\left(a e^{x}+b e^{-x}+2 x\right)\)
\(\Rightarrow \frac{d^{2} y}{d x^{2}}=a e^{x}+b e^{-x}+2\)
Now, Substituting the values of \( \frac{d y}{d x} \), and \( \frac{d^{2} y}{d x^{2}} \) in the given differential equations, we get,
\( \text {LHS }=x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-x y+x^{2}-2\)
\(={x}\left({ae}^{{x}}+{be}^{-{x}}+2\right)+2\left({ae}^{{x}}-{be}^{-{x}}+2\right)-{x}\left({ae}^{{x}}+{be}^{-{x}}+{x}^{2}\right)+{x}^{2}-2\)
\(=\left({axe}^{{x}}+{bxe^{- \textrm {x } } + 2 { x }}\right)+2\left({ae}^{{x}}-{be}^{-{x}}+2\right)-{x}\left({ae}^{{x}}+{be}^{-{x}}+{x}^{2}\right)+{x}^{2}-2\)
\(=2 {ae}^{{x}}-2 {be}^{-{x}}+{x}^{2}+6 {x}-2\)
\(\neq 0\)
\(\Rightarrow \text { LHS } \neq \text { RHS }\)
Therefore, the given function is not the solution of the corresponding differential equation.

It is given that \( =e^{x}(a \cos x+b \sin x)=a e^{x} \cos x+b e^{x} \sin x \)
Now, differentiating both sides w.r.t. \( x \), we get,
\(\frac{d y}{d x}=a \frac{d}{d x}\left(e^{x} \cos x\right)+b \frac{d}{d x}\left(e^{x} \sin x\right)\)
\(\Rightarrow \frac{d y}{d x}=a\left(e^{x} \cos x-e^{x} \sin x\right)+b\left(e^{x} \sin x-e^{x} \cos x\right)\)
\(\Rightarrow \frac{d y}{d x}=(a+b) e^{x} \cos x+(b-a) e^{x} \sin x\)
Now, again differentiating both sides w.r.t. \(x \), we get,
\( \frac{d^{2} y}{d x^{2}}=(a+b) \cdot \frac{d}{d x}\left(e^{x} \cos x\right)+(b-a) \frac{d}{d x}\left(e^{x} \sin x\right)\)
\(=(a+b) \cdot\left[e^{x} \cos x-e^{x} \sin x\right]+(b-a)\left[e^{x} \sin x-e^{x} \cos x\right]\)
\(=e^{x}[a \cos x-a \sin x+b \cos x-b \sin x+b \sin x+b \cos x-a \sin x-a \cos x] \)
\(=\left[2 e^{x}(b \cos x-a \sin x)\right] \)
Now, Substituting the values of \( \frac{d y}{d x} \), and \( \frac{d^{2} y}{d x^{2}} \) in the given differential equations, we get,
\(\text {LHS }=\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+2 y\)
\(=2 {e}^{x}({b} \cos x-{a} \sin x)-2 {e}^{x}[({a}+{b}) \cos x+({b}-{a}) \sin x]\) \(+2 {e}^{x}({a} \cos x+{b} \sin x)\)
\(={e}^{x}[(2 {b} \cos x-2 {a} \sin x)-(2 {a} \cos x+2 {~b} \cos x)\) \(-(2 {b} \sin x-2 {a} \sin x)+(2 {a \cos} x+2 {b \sin} x)]\)
\(={e}^{x}[(2 {b}-2 {a}-2 {~b}+2 {a}) \cos x]+{e}^{x}[(-2 {a}-2 {b}+2 {a}+2 {~b} \sin x]\)
\(=0=\text { RHS }\)
Therefore, the given function is the solution of the corresponding differential equation.

It is given that \( =y=x \sin 3 x \)
Now, differentiating both sides w.r.t. \(x \), we get,
\(\frac{d y}{d x}=\frac{d}{d x}(x \sin 3 x)=\sin 3 x+x \cdot \cos 3 x \cdot 3\)
\(\Rightarrow \frac{d y}{d x}=\sin 3 x+3 x \cos 3 x\)
Now, again differentiating both sides w.r.t. \( x \), we get,
\(\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(x \sin 3 x)+3 \frac{d}{d x}(x \cos 3 x)\)
\(\Rightarrow \frac{d^{2} y}{d x^{2}}=3 x \cos 3 x+3[\cos 3 x+x(-\sin 3 x) .3]\)
\(\Rightarrow \frac{d^{2} y}{d x^{2}}=6 \cos 3 x-9 x \sin 3 x\)
Now, substituting the value of \( \frac{d^{2} y}{d x^{2}} \) in the LHS of the given differential equation, we get,
\(\frac{d^{2} y}{d x^{2}}+9 y-6 \cos 3 x\)
\(=(6 \cdot \cos 3 x-9 x \sin 3 x)+9 x \sin 3 x-6 \cos 3 x\)
\(=0=\text{RHS}\)
Therefore, the given function is the solution of the corresponding differential equation.

It is given that \( x^{2}=2 {y}^{2} \log {y} \)
Now, differentiating both sides w.r.t. \(x \), we get,
\(2 x=2 \cdot \frac{d}{d x}\left(y^{2} \log y\right)\)
\(\Rightarrow x=\left[2 y \cdot \log y \cdot \frac{d y}{d x}+y^{2} \cdot \frac{1}{y} \cdot \frac{d y}{d x}\right]\)
\(\Rightarrow x=\frac{d y}{d x}(2 y \log y+y)\)
\(\Rightarrow \frac{d y}{d x}=\frac{x}{y(1+2 \log y)}\)
Now, substituting the value of \( \frac{d y}{d x} \) in the LHS of the given differential equation, we get,
\(\left(x^{2}+y^{2}\right) \frac{d y}{d x}-x y=\left(2 y^{2} \log +y^{2}\right) \cdot \frac{x}{y(1+2 \log y)}-x y\)
\(=y^{2}(1+2 \log y) \cdot \frac{x}{y(1+2 \log y)}-x y\)
\(={xy}-{xy}\)
\(=0\)
Therefore, the given function is the solution of the corresponding differential equation.

It is given that \( (x-a)^{2}+2 y^{2}=a^{2} \)
\(\Rightarrow x^{2}+a^{2}-2 a x+2 y^{2}=a^{2}\)
\(\Rightarrow 2 y^{2}=2 a x-x^{2}\quad \ldots\ldots\text{(1)}\)
Now, differentiating both sides w.r.t. \( x \), we get,
\( 2 y \frac{d y}{d x}=\frac{2 a-2 x}{2} \)
\( \Rightarrow \frac{d y}{d x}=\frac{a-x}{2 y} \)
\( \Rightarrow \frac{d y}{d x}=\frac{2 a x-2 x^{2}}{4 x y} \quad \ldots\ldots\text{(2)}\)
So, equation (1), we get,
\(2 {ax}=2 {y}^{2}+x^{2}\)
On substituting this value in equation (3), we get,
\(\frac{d y}{d x}=\frac{2 y^{2}+x^{2}-2 x^{2}}{4 x y}\)
\(\Rightarrow \frac{d y}{d x}=\frac{2 y^{2}-x^{2}}{4 x y}\)
Therefore, the differential equation of the family of curves is given as \( \frac{d y}{d x}=\frac{2 y^{2}-x^{2}}{4 x y} \).

It is given that \( \left(x^{3}-3 x^{2}\right) d x=\left(y^{3}-3 x^{2} y\right) d y \)
\(\Rightarrow \frac{d y}{d x}=\frac{x^{3}-3 x y^{2}}{y^{3}-3 x^{2} y}\quad \ldots\ldots\text{(1)}\)
Now, let us take \( {y}={vx} \)
\(\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)\)
\(\Rightarrow \frac{{dy}}{{dx}}={v}+x \frac{{dv}}{{dx}}\)
Now, substituting the values of y and \( \frac{{dv}}{{dx}} \) in equation (1), we get,
\(v+x \frac{d v}{d x}=\frac{x^{3}-3 x(v x)^{2}}{(v x)^{3}-3 x^{2}(v x)}\)
\(\Rightarrow v+x \frac{d v}{d x}=\frac{1-3 v^{2}}{v^{3}-3 v}\)
\(\Rightarrow x \frac{d v}{d x}=\frac{1-3 v^{2}}{v^{3}-3 v}-v\)
\(\Rightarrow x \frac{d v}{d x}=\frac{1-3 v^{2}-v\left(v^{3}-3 v\right)}{v^{3}-3 v}\)
\(\Rightarrow \frac{v^{3}-3 v}{1-3 v^{4}} d v=\frac{d x}{x}\)
On integrating both sides we get,
\( \int \frac{v^{3}-3 v}{1-3 v^{4}} {~d} v=\log x+\log C^{\prime} \quad \ldots\ldots\text{(2)}\)
Now, \( \int \frac{v^{3}-3 v}{1-3 v^{4}} d v=\int \frac{v^{3}}{1-v^{4}} d v-3 \int \frac{v d v}{1-v^{4}} \)
\( \Rightarrow \int \frac{v^{3}-3 v}{1-3 {v}^{4}} {dv}={I}_{1}-3 {I}_{2} \) where \( {I}_{1} \int \frac{{v}^{3}}{1-{v}^{4}} {dv} \) and \( {I}_{2}=\int \frac{{vdv}}{1-{v}^{4}} \quad \ldots\ldots\text{(3)}\)
Let \( 1-{v}^{4}=t \)
\(\Rightarrow \frac{d}{d v}\left(1-v^{4}\right)=\frac{d t}{d v}\)
\(\Rightarrow-4 v^{3}=\frac{d t}{d v}\)
\(\Rightarrow v^{3} d v=-\frac{d t}{4}\)
Now, \( {I}_{1}=\int-\frac{{dt}}{4}=-\frac{1}{4} \log t=-\frac{1}{4} \log \left(1-{v}^{4}\right) \)
and \( {I}_{2}=\int \frac{{vdv}}{1-{v}^{4}}=\int \frac{{vdv}}{1-\left({v}^{2}\right)^{2}} \)
Let \( {v}^{2}={p} \)
\(\frac{{d}}{{dv}}\left({v}^{2}\right)=\frac{{dp}}{{dv}}\)
\(\Rightarrow 2 {v}=\frac{{dp}}{{dv}}\)
\(\Rightarrow {vdv}=\frac{d p}{2}\)
\(\therefore {I}_{2}=\frac{1}{2} \int \frac{{dt}}{1-{p}^{2}}=\frac{1}{2 \times 2} \log \left|\frac{1+p}{1-p}\right|=\frac{1}{4}\left|\frac{1+{v}^{2}}{1-{v}^{2}}\right|\)
Now, substituting the values of \( {I}_{1} \) and \( {I}_{2} \) in equation (3), we get,
\(\int\left(\frac{{v}^{3}-3 y}{1-{v}^{4}}\right) {dv}-\frac{3}{4} \log \left|\frac{1+{v}^{2}}{1-{v}^{2}}\right|\)
Thus, equation (2), becomes,
\(-\frac{1}{4} \log \left(1-{v}^{4}\right)-\frac{3}{4} \log \left|\frac{1+{v}^{2}}{1-{v}^{2}}\right|=\log x+\log C^{\prime}\)
\(\Rightarrow-\frac{1}{4} \log \left[\left(1-{v}^{4}\right)\left(\frac{1+{v}^{2}}{1-{v}^{2}}\right)^{3}\right]=\log C^{\prime} x\)
\(\Rightarrow \frac{\left(1+{v}^{2}\right)^{4}}{\left(1-{v}^{2}\right)^{2}}=\left(C^{\prime} x\right)^{-4}\)
\(\Rightarrow \frac{\left(1+\frac{y^{2}}{x^{2}}\right)^{4}}{\left(1-\frac{y^{2}}{x^{2}}\right)^{2}}=\frac{1}{C^{\prime 4} x^{4}}\)
\(\Rightarrow\left(x^{2}-y^{2}\right)^{2}=C^{\prime} 4\left(x^{2}+y^{2}\right)^{4}\)
\(\Rightarrow\left(x^{2}-y^{2}\right)=C^{\prime 2}\left(x^{2}+y^{2}\right)\)
\(\Rightarrow\left(x^{2}-y^{2}\right)=C\left(x^{2}+y^{2}\right), \text { where } {C}=C^{\prime 2}\)
Therefore, the result is proved.

We know that the equation of a circle in the first quadrant with centre (a, a) and radius a which touches the coordinate axes is:
\((x-a)^{2}+(y-a)^{2}=a^{2}\quad \ldots\ldots\text{(1)}\)

Now differentiating above equation w.r.t. \(x \), we get,
\(2(x-{a})+2({y}-{a}) \frac{d y}{d x}=0\)
\(\Rightarrow(x-{a})+({y}-{a}) {y}^{\prime}=0\)
\(\Rightarrow x-{a}+{yy}^{\prime}-{ay}^{\prime}=0\)
\(\Rightarrow x+{yy}^{\prime}-{a}\left(1+{y}^{\prime}\right)=0\)
\(\Rightarrow {a}=\frac{x+y y^{\prime}}{1+y^{\prime}}\)
Now, substituting the value of a in equation (1), we get,
\({\left[x-\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right)\right]^{2}+\left[y-\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right)\right]^{2}=\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right)^{2}}\)
\(\Rightarrow\left[\frac{(x-y) y^{\prime}}{1+y^{\prime}}\right]^{2}+\left[\frac{y-x}{1+y^{\prime}}\right]^{2}=\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right)^{2}\)
\(\Rightarrow(x-{y})^{2} \cdot {y}^{\prime 2}+(x-{y})^{2}=\left(x+yy^{\prime}\right)^{2}\)
\(\Rightarrow(x-{y})^{2}\left[1+\left({y}^{\prime}\right)^{2}\right]=\left(x+y y^{\prime}\right)^{2}\)
Therefore, the required differential equation of the family of circles is
\((x-y)^{2}\left[1+\left(y^{\prime}\right)^{2}\right]=\left(x+y y^{\prime}\right)^{2}\)

It is given that \( \frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0 \)
\(\Rightarrow \frac{d y}{d x}=-\sqrt{\frac{1-y^{2}}{1-x^{2}}}\)
\(\Rightarrow \frac{d y}{\sqrt{1-y^{2}}}=-\frac{d x}{\sqrt{1-x^{2}}}\)
On integrating, we get,
\(\sin ^{-1} y=\sin ^{-1} x+C\)
\(\Rightarrow \sin ^{-1} x=\sin ^{-1} y+C\)

It is given that \( \frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}=0 \)
\(\Rightarrow \frac{d y}{d x}=-\left(\frac{y^{2}+y+1}{x^{2}+x+1}\right)\)
\(\Rightarrow \frac{d y}{y^{2}+y+1}=\frac{-d x}{x^{2}+x+1}\)
\(\Rightarrow \frac{d y}{y^{2}+y+1}+\frac{d x}{x^{2}+x+1}=0\)
On integrating both sides, we get,
\(\int \frac{d y}{y^{2}+y+1}+\int \frac{d x}{x^{2}+x+1}={C}\)
\(\Rightarrow \int \frac{d y}{\left(y+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}+\int \frac{d x}{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}={C}\)
\(\Rightarrow \frac{2}{\sqrt{3}} \tan ^{-1}\left[\frac{y+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right]+\frac{2}{\sqrt{3}} \tan ^{-1}\left[\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right]={C}\)
\(\Rightarrow \tan ^{-1}\left[\frac{2 y+1}{\sqrt{3}}\right]+\tan ^{-1}\left[\frac{2 x+1}{\sqrt{3}}\right]=\frac{\sqrt{3}}{2} {C}\)
\(\Rightarrow \tan ^{-1}\left[\frac{\frac{2 y+1}{\sqrt{3}}+\frac{2 x+1}{\sqrt{3}}}{1-\frac{2 y+1}{\sqrt{3}} \cdot \frac{12+1}{\sqrt{3}}}\right]=\frac{\sqrt{3}}{2} {C}\)
\(\Rightarrow \tan ^{-1}\left[\frac{2 \sqrt{3}(x+y+1)}{3-4 x y-2 x-2 y-1}\right]=\frac{\sqrt{3}}{2} {C}\)
\(\Rightarrow \tan ^{-1}\left[\frac{2 \sqrt{3}(x+y+1)}{2(1-x-y-2 x y)}\right]=\frac{\sqrt{3}}{2} {C}\)
\(\Rightarrow \frac{\sqrt{3}(x+y+1)}{(1-x-y-2 x y)}=\tan \left(\frac{\sqrt{3}}{2} {C}\right)\)
Let \( \tan \left(\frac{\sqrt{3}}{2} {C}\right)={B} \)
Then,
\(x+y+1=\frac{2 {B}}{\sqrt{3}}(1-x-y-2 x y)\)
Now, let \( A=\frac{2 B}{\sqrt{3}} \), then, we have,
\(x+y+1=A(1-x-y-2 x y)\)

It is given that \( \sin x \cos {ydx}+\cos x \sin {ydy}=0 \)
\(\Rightarrow \frac{\sin x \cos y d x+\cos x \sin y d y}{\cos x \cos y}=0\)
\(\Rightarrow \tan x d x+\tan y d y=0\)
So, on integrating both sides, we get,
\(\log (\sec x)+\log (\sec {y})=\log {C}\)
\(\Rightarrow \log (\sec x \cdot \sec {y})=\log {C}\)
\(\Rightarrow \sec x \cdot \sec {y}=C\)
The curve passes through point \( \left(0, \frac{\pi}{4}\right) \)
Thus, \( 1 \times \sqrt{2}={C} \)
\(\Rightarrow {C}=\sqrt{2}\)
On substituting \( {C}=\sqrt{2} \) in equation (1), we get,
\(\sec x \cdot \sec {y}=\sqrt{2}\)
\(\Rightarrow \sec x \cdot \frac{1}{\cos y}=\sqrt{2}\)
\(\Rightarrow \cos y=\frac{\sec x}{\sqrt{2}}\)
Therefore, the required equation of the curve is \( \cos y=\frac{\sec x}{\sqrt{2}} \)

It is given that \( \left(1+e^{2 x}\right) d y+\left(1+y^{2}\right) e^{x} d x=0 \)
\(\Rightarrow \frac{d y}{1+y^{2}}+\frac{e^{x} d x}{1+e^{2 x}}=0\)
On integrating both sides, we get,
\(\tan ^{-1} y+\int \frac{e^{x} d x}{1+e^{2 x}}=C\quad \ldots\ldots\text{(1)}\)
Let \( {e}^{x}={t} \)
\(\Rightarrow {e}^{2 x}={t}^{2}\)
\(\Rightarrow \frac{d}{d x}\left(e^{x}\right)=\frac{d t}{d x}\)
\(\Rightarrow e^{x}=\frac{d t}{d x}\)
\(\Rightarrow {e}^{x} {dx}={dt}\)
Substituting the value in equation (1), we get,
\(\tan ^{-1} y+\int \frac{d t}{1+t^{2}}=C\)
\(\Rightarrow \tan ^{-1} {y}+\tan ^{-1} {t}={C}\)
\(\Rightarrow \tan ^{-1} {y}+\tan ^{-1}\left({e}^{x}\right)={C}\quad \ldots\ldots\text{(2)}\)
Now, \( {y}=1 \) at \( x=0 \)
Therefore, equation (2) becomes:
\(\tan ^{-1} 1+\tan ^{-1} 1={C}\)
\(\Rightarrow \frac{\pi}{4}+\frac{\pi}{4}={C}\)
\(\Rightarrow C=\frac{\pi}{4}\)
Substituting \( C=\frac{\pi}{4} \) in (2), we get,
\(\tan ^{-1} {y}+\tan ^{-1}\left({e}^{x}\right)=\frac{\pi}{4}\)

It is given that \( y e^{\frac{x}{y}}=\left(x e^{\frac{x}{y}}+y^{2}\right) d y \)
\(
\Rightarrow y e^{\frac{x}{y} }\frac{d x}{d y}=x e^{\frac{x}{y}}+y^{2}\)
\(
\Rightarrow e^{\frac{x}{y}}\left[y \cdot \frac{d x}{d y}-x\right]=y^{2}\)
\(
\Rightarrow e^{\frac{x}{y}} \frac{\left[y \cdot \frac{d x}{d y}-x\right]}{y^{2}}=1\quad \ldots\ldots\text{(1)}
\)
Let \( e^{\frac{x}{y}}=z \)
Differentiating it w.r.t. \(y\), we get,
\(\frac{d}{d y}\left(e^{\frac{x}{y}}\right)=\frac{d z}{d y}\)
\(\Rightarrow e^{\frac{x}{y}} \cdot \frac{d}{d y}\left(\frac{x}{y}\right)=\frac{d z}{d y}\)
\(\Rightarrow e^{\frac{x}{y}}\left[\frac{y \cdot \frac{d x}{d y}-x}{y^{2}}\right]=\frac{d z}{d y}\quad \ldots\ldots\text{(2)}\)
From equation (1) and equation (2), we get,
\(\frac{d z}{d y}=1\)
\(\Rightarrow {dz}={dy}\)
On integrating both sides, we get,
\({z}={y}+{C}\)
\(\Rightarrow e^{\frac{x}{y}}=y+C\)

It is given that \( (x-y)(d x+d y)=d x-d y \)
\(\Rightarrow(x-{y}+1) {dy}=(1-x+{y}) {dx}\)
\(\Rightarrow \frac{d y}{d x}=\frac{1-x+y}{x-y+1}\)
\(\Rightarrow \frac{d y}{d x}=\frac{1-(x-y)}{x-y+1}\quad \ldots\ldots\text{(1)}\)
Let \( x-{y}={t} \)
\(\Rightarrow \frac{d(x-y)}{d x}=\frac{d t}{d x}\)
\(\Rightarrow 1-\frac{d y}{d x}=\frac{d t}{d x}\)
\(\Rightarrow 1-\frac{d t}{d x}=\frac{d y}{d x}\)
Now, let us substitute the value of \( x-y \) and \( \frac{d y}{d x} \) in equation (1), we get,
\(1-\frac{d t}{d x}=\frac{1-t}{1+t}\)
\(\Rightarrow \frac{d t}{d x}=1-\left(\frac{1-t}{1+t}\right)\)
\(\Rightarrow \frac{d t}{d x}=\frac{(1+t)-(1-t)}{1+t}\)
\(\Rightarrow \frac{d t}{d x}=\frac{2 t}{1+t}\)
\(\Rightarrow\left(\frac{1+t}{t}\right) d t=2 d x\)
\(\Rightarrow\left(1+\frac{1}{t}\right) d t=2 d x\quad \ldots\ldots\text{(2)}\)
On integrating both side, we get,
\({t}+\log |{t}|=2 x+{C}\)
\(\Rightarrow(x-{y})+\log |x-{y}|=2 x+{C}\)
\(\Rightarrow \log |x-{y}|=x+{y}+{C}\quad \ldots\ldots\text{(3)}\)
Now, \( {y}=-1 \) at \( x=0 \)
Then, equation (3), we get,
\(\log 1=0-1+C\)
\(\Rightarrow C=1\)
Substituting \( {C}=1 \) in equation (3), we get,
\(\log |x-y|=x+y+1\)
Therefore, a particular solution of the given differential equation is \( \log |x -{y} |=x+{y}+1 \).

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

It is given that \( \frac{d y}{d x}+y \cot x=4 x \operatorname{cosec} x \)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( {p}=\cot x \) and \( {Q}= \) \( 4 x \operatorname{cosec}x) \)
Now, I.F. \( =e^{\int p d x}=e^{\int \cot x d x}=e^{\log |\sin x|}=\sin x \)
Thus, the solution of the given differential equation is given by the relation:
\(\text {y(I.F.)}=\int(Q \times \text { I.F.}) d x+C\)
\(\Rightarrow y \sin x=\int 2 x \operatorname{cosec} x d x+C\)
\(=4 \int x d x+C\)
\(=4 . \frac{x^{2}}{2}+C\)
\(\Rightarrow y \sin x=2 x^{2}+C\quad \ldots\ldots\text{(1)}\)
Now \( {y}=0 \) at \( x=\frac{\pi}{2} \)
Therefore, equation (1), we get,
\( 0=2 \times \frac{\pi^{2}}{4}+C \)
\( \Rightarrow C=\frac{\pi^{2}}{4} \)
Now, substituting \( {C}=\frac{\pi^{2}}{4} \) in equation (1), we get,
\({y} \sin x=2 x^{2}-\frac{\pi^{2}}{4}\)
Therefore, the required particular solution of the given differential equation is
\({y} \sin x=2 x^{2}-\frac{\pi^{2}}{4}\)

It is given that \( (x+1) \frac{d y}{d x}=2 e^{-y}-1 \)
\(\Rightarrow \frac{d y}{2 e^{-y}-1}=\frac{d x}{x+1}\)
\(\Rightarrow \frac{e^{y} d y}{2-e^{y}}=\frac{d x}{x+1}\)
On integrating both sides, we get,
\(\int \frac{{e}^{{y}} {dy}}{2-{e}^{{y}}}=\log |x+1|+\log {C}\quad \ldots\ldots\text{(1)}\)
Let \( 2-{e}^{{y}}=t \)
\(\therefore \frac{d}{d t}\left(2-{e}^{{y}}\right)=\frac{d t}{d y}\)
\(\Rightarrow-{e}^{{y}}=\frac{d t}{d y}\)
\(\Rightarrow {e}^{{y}} {dt}=-{dt}\)
Substituting value in equation (1), we get,
\(\int \frac{-{dt}}{{t}}=\log |x+1|+\log {C}\)
\(\Rightarrow-\log |{t}|=\log |{C}(x+1)|\)
\(\Rightarrow-\log \left|2-{e}^{{y}}\right|=\log |{C}(x+1)|\)
\(\Rightarrow \frac{1}{2-{e}^{{y}}}={C}(x+1)\)
\( \Rightarrow 2-{e}^{{y}}=\frac{1}{c(x+1)}\quad \ldots\ldots\text{(2)} \)
Now, at \( x=0 \) and \( y=0 \), equation (2) becomes,
\( \Rightarrow 2-1=\frac{1}{C} \)
\( \Rightarrow {C}=1 \)
Now, substituting the value of C I equation (2), we get,
\( \Rightarrow 2-{e}^{{y}}=\frac{1}{(x+1)} \)
\( \Rightarrow {e}^{{y}}=2-\frac{1}{(x+1)} \)
\( \Rightarrow {e}^{{y}}=\frac{2 x+2-1}{(x+1)} \)
\( \Rightarrow {e}^{{y}}=\frac{2 x+1}{(x+1)} \)
\( \Rightarrow y=\log \left|\frac{2 x+1}{x+1}\right| \cdot(x \neq-1) \)
Therefore, the required particular solution of the given differential equation is
\( y=\log \left|\frac{2 x+1}{x+1}\right| \cdot(x \neq-1) \)

Let the population at any instant (t) be \( y \).
Now it is given that the rate of increase of population is proportional to the number of inhabitants at any instant.
\(\therefore \frac{d y}{d t} \alpha y\)
\(\Rightarrow \frac{d y}{d t}=k y(k \text { is a constant})\)
\(\Rightarrow \frac{d y}{y}=k d t\)
Now, integrating both sides, we get,
\(\log {y}={kt}+{C}\quad \ldots\ldots\text{(1)}\)
According to given conditions,
In the year \( 1999, {t}=0 \) and \( {y}=20000 \)
\(\Rightarrow \log 20000=C\quad \ldots\ldots\text{(2)}\)
Also, in the year \(2004, t=5 \) and \( {y}=25000 \)
\(\Rightarrow \log 25000={k} .5+{C}\)
\(\Rightarrow \log 25000=5 {k}+\log 20000\)
\(\Rightarrow 5 k=\log \left(\frac{25000}{20000}\right)=\log \left(\frac{5}{4}\right)\)
\(\Rightarrow k=\frac{1}{5} \log \left(\frac{5}{4}\right)\quad \ldots\ldots\text{(3)}\)
Also, in the year \(2009, t=10 \)
Now, substituting the values of \( {t}, {k} \) and c in equation (1), we get
\(\log y=10 \times \frac{1}{5} \log \left(\frac{5}{4}\right)+\log (20000)\)
\(\Rightarrow \log y=\log \left[20000 \times\left(\frac{5}{4}\right)^{2}\right]\)
\(\Rightarrow y=20000 \times \frac{5}{4} \times \frac{5}{4}\)
\(\Rightarrow y=31250\)
Therefore, the population of the village in 2009 will be 31250.

It is given that \( \frac{y d x-x d x}{y}=0 \)
\(\Rightarrow \frac{y d x-x d x}{y}=0\)
\(\Rightarrow \frac{1}{x} d x-\frac{1}{y} d y=0\)
Integrating both sides, we get,
\(\log |x|-\log |{y}|=\log {k}\)
\(\Rightarrow \log \left|\frac{x}{y}\right|=\log k\)
\(\Rightarrow \frac{x}{y}=k\)
\(\Rightarrow y=\frac{1}{k} x\)
\( \Rightarrow {y}={Cx} \) where \( {C}=\frac{1}{k} \).

The integrating factor of the given differential equation \( \frac{d x}{d y}+{P}_{1} x={Q}_{1} \) is \( e^{\int {P}_{1} d y} \).
Thus, the general solution of the differential equation is given by,
\(x({I} . {F} .)=\int({Q} \times {I} . {F} .) {dy}+{C}\)
\(\Rightarrow x . e^{\int {P}_{1} d y}=\int\left({Q}_{1} e^{\int {P}_{1} d y}\right) {dy}+{C}\)

It is given that \( e^{x} d y+\left(y e^{x}+2 x\right) d x=0 \)
\(\Rightarrow e^{x} \frac{d y}{d x}+y e^{x}+2 x=0\)
\(\Rightarrow \frac{d y}{d x}+y=-2 x e^{-x}\)
This is equation in the form of \( \frac{d y}{d x}+p y=Q \) (where, \( {p}=1 \) and \( {Q}=-2 {xe}^{-x} \))
Now, I.F. \( =e^{\int p d x}=e^{\int d x}=e^{x} \)
Thus, the solution of the given differential equation is given by the relation:
\(\text {y(I.F.})=\int(Q \times \text {I. F.}) d x+C\)
\(\Rightarrow y e^{x}=\int\left(-2 x e^{-x} \times e^{x}\right) d x+C\)
\(\Rightarrow y e^{2 \sqrt{x}}=\int 2 x d x+C\)
\(\Rightarrow {ye}^{x}=-x^{2}+{C}\)
\(\Rightarrow {ye}^{x}+x^{2}={C}\)