Ex 9.5 Class 12 Maths Ncert Solutions

\(\frac{d y}{d x}=\frac{x^{2}+y^{2}}{x^{2}+x y}\)
Let \( {f}(x, {y})=\frac{x^{2}+y^{2}}{x^{2}+x y} \)
Here, putting \( x={kx} \) and \( {y}={ky} \)
\({F}({kx}, {ky})=\frac{(k x)^{2}+(k y)^{2}}{(k x)^{2}+k x \cdot k y}\)
\(=\frac{k^{2}}{k^{2}} \cdot \frac{x^{2}+y^{2}}{x^{2}+x y}\)
\(={k}^{0} \cdot {f}(x, {y})\)
Therefore, the given differential equation is homogeneous.
\(\left(x^{2}+{xy}\right) {dy}=\left(x^{2}+{y}^{2}\right) {dx}\)
\(\frac{d y}{d x}=\frac{x^{2}+y^{2}}{x^{2}+x y}\)
To solve it we make the substitution.
\(y=v x\)
Differentiating eq. with respect to \( x \), we get
\(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(v+x \frac{d v}{d x}=\frac{x^{2}+(v x)^{2}}{x^{2}+x \cdot v x}\)
\(v+x \frac{d v}{d x}=\frac{x^{2}\left(1+v^{2}\right)}{x^{2}(1+v)}\)
\(v+x \frac{d v}{d x}=\frac{1+v^{2}}{1+v}\)
\(x \frac{d v}{d x}=\frac{1+v^{2}}{1+v}-v=\frac{1+v^{2}-v-v^{2}}{1+v}\)
\(x \frac{d v}{d x}=\frac{1-v}{1+v}\)
\(\frac{1+v}{1-v} d v=\frac{1}{x} d x\)
Integrating on both side,
\(\int \frac{1+{v}}{1-{v}} {dv}=\int \frac{1}{x} {dx}\)
\(\int\left(-1+\frac{2}{1-{v}}\right) {dv}=\int \frac{1}{x} {dx}\)
\(-{v}-2 \log |1-{v}|=\log |x|+\log C\)
\(-\frac{y}{x}-2 \log \left|1-\frac{y}{x}\right|=\log |x|+\log C\)
\(-\frac{y}{x}=2 \log \left|1-\frac{y}{x}\right|+\log |x|+\log C\)
\(-\frac{y}{x}=\log \frac{(x-y)^{2}}{x^{2}}+\log |x|+\log C\)
\(-\frac{y}{x}=\log \frac{(x-y)^{2}}{x^{2}} \cdot C x\)
\(-\frac{y}{x}=\log \frac{(x-y)^{2}}{x} C\)
\(\frac{C(x-y)^{2}}{x}=e^{-\frac{y }{ x}}\)
\(C(x-y)^{2}=x e^{-\frac{y }{ x}}\)
\((x-y)^{2}=k x e^{-\frac{y }{ x}}\)
The required solution of the differential equation.

\(y^{\prime}=\frac{x+y}{x}\)
\(\frac{d y}{d x}=\frac{x+y}{x}\)
Let \( {f}(x, {y})=\frac{x+y}{x} \)
Here, putting \( x={kx} \) and \( {y}={ky} \)
\({f}({kx}, {ky})=\frac{k x+k y}{k x}\)
\(=\frac{k}{k} \cdot \frac{x+y}{x}\)
\(={k}^{0} \cdot {f}(x, {y})\)
Therefore, the given differential equation is homogeneous.
\(y^{\prime}=\frac{x+y}{x}\)
\(\frac{d y}{d x}=\frac{x+y}{x}\)
To solve it we make the substitution.
\(y=v x\)
Differentiating eq. with respect to \( x \), we get
\(\frac{d y}{d x}={v}+x \frac{{dv}}{{dx}}\)
\({v}+x \frac{{dv}}{{dx}}=\frac{x+{vx}}{x}\)
\({v}+x \frac{{dv}}{{dx}}=1+{v}\)
\(x \frac{{dv}}{{dx}}=1\)
\({dv}=\frac{1}{x} {dx}\)
\(\int d v=\int \frac{1}{x} {dx}\)
\({v}=\log x+{C}\)
\(\frac{y}{x}=\log x+C\)
\({y}=x \log x+{Cx}
\)
The required solution of the differential equation.

\((x-{y}) {dy}=(x+{y}) {dx}\)
\(\frac{d y}{d x}=\frac{x+y}{x-y}\)
Let \( {f}(x, {y})=\frac{x+y}{x-y} \)
Here, putting \( x={kx} \) and \( {y}={ky} \)
\({f}({kx}, {ky})=\frac{k x+k y}{k x-k y}\)
\({f}({kx}, {ky})=\frac{x+y}{x}\)
\(={k}^{0} . {f}(x, {y})\)
Therefore, the given differential equation is homogeneous.
\((x-y) d y-(x+y) d x=0\)
\(\frac{d y}{d x}=\frac{x+y}{x-y}\)
To make it we make the substitution.
\(y=v x\)
Differentiating eq. with respect to \(x \), we get
\(\frac{d y}{d x}={v}+x \frac{{dv}}{{dx}}\)
\({v}+x \frac{{dv}}{{dx}}=\frac{x+{vx}}{x-{vx}}\)
\({v}+x \frac{{dv}}{{dx}}=\frac{1+{v}}{1-{v}}\)
\(x \frac{{dv}}{{dx}}=\frac{1+{v}}{1-{v}}-{v}\)
\(x \frac{{dv}}{{dx}}=\frac{1+{v}-{v}+{v}^{2}}{1-{v}}\)
\(x \frac{{dv}}{{dx}}=\frac{1+{v}^{2}}{1-{v}}\)
\(\frac{1-{v}}{1+{v}^{2}} {dv}=\frac{1}{x} {dx}\)
Integrating both sides we get,
\(\int \frac{1-v}{1+v^{2}} d v=\int \frac{1}{x} d x\)
\(\int \frac{1}{1+v^{2}} d v-\int \frac{v}{1+v^{2}} d v=\int \frac{1}{x} d x\quad \ldots\ldots\text{(i)}\)
Let \( {I}_{1}=\int \frac{{v}}{1+{v}^{2}} {dv} \)
Put \( 1+{v}^{2}={t} \)
\(2 {vdv}={dt}\)
\({vdv}=\frac{1}{2} d t\)
\(\frac{1}{2} \int \frac{1}{t} d t\)
\(\frac{1}{2} \log t\)
\(\frac{1}{2} \log \left(1+v^{2}\right)\)
\(\therefore \tan ^{-1} {v}-\frac{1}{2} \log \left(1+{v}^{2}\right)=\log x+C \quad(\therefore \text {From }({i}))\)
\(\tan ^{-1} \frac{y}{x}=\log x+\frac{1}{2} \log \left(\frac{x^{2}+y^{2}}{x^{2}}\right)+C\)
\(\tan ^{-1} \frac{y}{x}=\frac{1}{2}\left(\log \left(\frac{x^{2}+y^{2}}{x^{2}} \times x^{2}\right)\right)+C\)
\(\tan ^{-1} \frac{y}{x}=\frac{1}{2}\left(\log x^{2}+y^{2}\right)+C\)
The required solution of the differential equation.

\(2 {xy} {dy}=-\left(x^{2}-{y}^{2}\right) {dx}\)
\(\frac{d y}{d x}=-\frac{x^{2}-y^{2}}{2 x y}\)
Let \( {f}(x, {y})=-\frac{x^{2}-y^{2}}{2 x y} \)
Here, putting \( x={kx} \) and \( {y}={ky} \)
\({f}({kx}, {ky})=\frac{k^{2} x^{2}-k^{2} y^{2}}{2 k^{2} x y}\)
\({f}({kx}, {ky})=\frac{k^{2}}{k^{2}} \cdot \frac{x^{2}-y^{2}}{2 x y}\)
\(={k}^{0} \cdot {f}(x, {y})\)
Therefore, the given differential equation is homogeneous.
\(\left(x^{2}-y^{2}\right) d x+2 x y d y=0\)
\(2 x y d y=-\left(x^{2}-y^{2}\right) d x\)
\(\frac{d y}{d x}=-\frac{x^{2}-y^{2}}{2 x y}\)
To solve it we make the substitution.
\(y=v x\)
Differentiating eq. with respect to \( x \), we get
\(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(v+x \frac{d v}{d x}=-\frac{x^{2}-v^{2} x^{2}}{2 x \cdot {vx}}\)
\({v}+x \frac{{dv}}{{dx}}=-\frac{x^{2}\left(1-{v}^{2}\right)}{2 {vx}^{2}}\)
\(x \frac{{dv}}{{dx}}=-\frac{1-{v}^{2}}{2 {v}}-{v}\)
\(x \frac{{dv}}{{dx}}=\frac{-1+{v}^{2}-2 {v}^{2}}{2 {v}}\)
\(x \frac{{dv}}{{dx}}=\frac{-1+{v}^{2}}{2 {v}}\)
\(-\frac{2 {v}}{1+{v}^{2}} {dv}=\frac{1}{x} {dx}\)
\(\frac{2 {v}}{1+{v}^{2}} {dv}=-\frac{1}{x} {dx}\)
Integrating both sides, we get
\( \int \frac{2 {v}}{1+{v}^{2}} {dv}=\int \frac{1}{x} {dx}\quad \ldots\ldots\text{(i)} \)
Let \( {I}_{1}=\int \frac{2 {v}}{1+{v}^{2}} {d} v \)
Put \( 1+{v}^{2}={t} \)
\(2 {vdv}={dt}\)
\({vdv}=\frac{1}{2} {dt}\)
\(\int \frac{1}{t} {dt}\)
\(\log ({t})\)
\(\therefore \log \left(1+{v}^{2}\right)=-\log x+\log C\quad\) (\(\therefore\) From (i) eq.)
\(\log \left(1+\left(\frac{y}{x}\right)^{2}\right)=-\log x+\log C\)
\(\log \left(\frac{x^{2}+y^{2}}{x^{2}}\right)=\log \frac{C}{x}\)
\(x^{2}+y^{2}=C x\)
The required solution of the differential equation.

\(\frac{d y}{d x}=\frac{x^{2}-2 y^{2}+x y}{x^{2}}\)
Let \( {f}(x, {y})=\frac{x^{2}-2 y^{2}+x y}{x^{2}} \)
Here, putting \( x={kx} \) and \( {y}={ky} \)
\({f}({kx}, {ky})=\frac{k^{2} x^{2}-2 k^{2} y^{2}+k x k y}{k^{2} x^{2}}\)
\({f}({kx}, {ky})=\frac{k^{2}}{k^{2}} \cdot \frac{x^{2}-2 y^{2}+x y}{x^{2}}\)
\(={k}^{0} \cdot {f}(x, {y})\)
Therefore, the given differential equation is homogeneous.
\(x^{2} \frac{d y}{d x}=x^{2}-2 y^{2}+x y\)
\(\frac{d y}{d x}=\frac{x^{2}-2 y^{2}+x y}{x^{2}}\)
To solve it we make the substitution.
\(y=v x\)
Differentiating eq. with respect to \( x \), we get
\(\frac{d y}{d x}={v}+x \frac{{dv}}{{dx}}\)
\({v}+x \frac{{dv}}{{dx}}=\frac{x^{2}-2 {v}^{2} x^{2}+x \cdot {vx}}{x^{2}}\)
\({v}+x \frac{{dv}}{{dx}}=\frac{1-2 {v}^{2}+{v}}{1}\)
\(x \frac{{dv}}{{dx}}=1-2 {v}^{2}+{v}\)
\(x \frac{{dv}}{{dx}}=1-2 {v}^{2}\)
\(x \frac{{dv}}{{dx}}=\frac{-1+{v}^{2}}{2 {v}}\)
\(\frac{1}{1-2 {v}^{2}} {dv}=\frac{1}{x} {dx}\)
Integrating both sides, we get
\(\int \frac{1}{1-2 v^{2}} d v=\int \frac{1}{x} d x\)
\(\int \frac{1}{1-(\sqrt{2} v)^{2}} d v=\int \frac{1}{x} d x\)
\(\int \frac{1}{1^{2}-(\sqrt{2} v)^{2}} d v=\int \frac{1}{x} d x\)
\(\frac{1}{\sqrt{2}} \cdot \frac{1}{2.1} \log \left|\frac{1+\sqrt{2} v}{1-\sqrt{2} v}\right|=\log |x|+C\)
\(\frac{1}{2 \sqrt{2}} \log \left|\frac{1+\sqrt{2} \frac{y}{x}}{1-\sqrt{2} \frac{y}{x}}\right|=\log |x|+C\)
\(\frac{1}{2 \sqrt{2}} \log \left|\frac{x+\sqrt{2} \frac{y}{x}}{x-\sqrt{2} \frac{y}{x}}\right|=\log |x|+C\)
The required solution of the differential equation.

\(x d y=\left(\sqrt{x^{2}+y^{2}}+y\right) d x\)
\(\frac{d y}{d x}=\frac{\left(\sqrt{x^{2}+y^{2}}+y\right)}{x}\)
Let \( {f}(x, {y})=\frac{\left(\sqrt{x^{2}+y^{2}}+y\right)}{x} \)
Here, putting \( x={kx} \) and \( {y}={ky} \)
\({f}({kx}, {ky})=\frac{\sqrt{k^{2} x^{2}+k^{2} y^{2}}+k y}{k x}\)
\({f}({kx}, {ky})=\frac{k}{k} \cdot \frac{\sqrt{x^{2}+y^{2}}+y}{x}\)
\(={k}^{0} . {f}(x, {y})\)
Therefore, the given differential equation is homogeneous.
\(x d y-y d x=\sqrt{x^{2}+y^{2}} d x\)
\(x d y=\left(\sqrt{x^{2}+y^{2}}+y\right) d x\)
\(\frac{d y}{d x}=\frac{\left(\sqrt{x^{2}+y^{2}}+y\right)}{x}\)
To solve it we make the substitution.
\(y=v x\)
Differentiating eq. with respect to \( x \), we get
\(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(v+x \frac{d v}{d x}=\frac{\sqrt{x^{2}+x^{2} v^{2}}+v x}{x}\)
\(v+x \frac{d v}{d x}=\frac{x \sqrt{1+v^{2}}+v x}{x}\)
\(v+x \frac{d v}{d x}=\sqrt{1+v^{2}}+v\)
\(x \frac{d v}{d x}=\sqrt{1+v^{2}}\)
\(x \frac{d v}{d x}=\frac{-1+v^{2}}{2 v}\)
\(\frac{1}{\sqrt{1+v^{2}}} d v=\frac{1}{x} d x\)
Integrating both sides, we get
\(\int \frac{1}{\sqrt{1+{v}^{2}}} {dv}=\int \frac{1}{x} {dx}\)
\(\log \left({v}+\sqrt{1+{v}^{2}}\right)=\log x+\log C \quad\left(\therefore \int \frac{1}{\sqrt{x^{2}+{a}^{2}}}=\log \left(x+\sqrt{x^{2}+{a}^{2}}\right)\right)\)
\(\log \left(\frac{y}{x}+\sqrt{1+\frac{y^{2}}{x^{2}}}\right)=\log C x\)
\(\frac{y}{x}+\sqrt{1+\frac{y^{2}}{x^{2}}}=C x\)
\(\frac{y}{x}+\sqrt{\frac{x^{2}+y^{2}}{x^{2}}}=C x\)
\(\frac{y}{x}+\sqrt{\frac{x^{2}+y^{2}}{x}}=C x\)
\(\frac{y}{x}+\sqrt{x^{2}+y^{2}}=C x^{2}\)
The required solution of the differential equation.

\(\frac{d y}{d x}=\frac{\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y}{\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x}\)
Let \( {f}(x, {y})=\frac{\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y}{\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x} \)
Here, putting \( x={kx} \) and \( {y}={ky} \)
\({f}({kx}, {ky})=\frac{\left\{k x \cos \left(\frac{k y}{k x}\right)+k y \sin \left(\frac{k y}{k x}\right)\right\} k y}{\left\{k y \sin \left(\frac{k y}{k x}\right)-k x \cos \left(\frac{k y}{k x}\right)\right\} k x}\)
\({f}({kx}, {ky})=\frac{k^{2}}{k^{2}} \cdot \frac{\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y}{\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x}\)
\(={k}^{0} \cdot {f}(x, {y})\)
Therefore, the given differential equation is homogeneous.
\(x d y-y d x=\sqrt{x^{2}+y^{2}} d x\)
\(x d y=\left(\sqrt{x^{2}+y^{2}}+y\right) d x\)
\(\frac{d y}{d x}=\frac{\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y}{\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x}\)
To solve it we make the substitution.
\(y=v x\)
Differentiating eq. with respect to \( x \), we get
\(\frac{d y}{d x}={v}+x \frac{{dv}}{{dx}}\)
\({v}+x \frac{{dv}}{{dx}}=\frac{d y}{d x}=\frac{\{x \cos ({v})+y \sin ({v})\} v x}{\{{vx} \sin ({v})-x \cos ({v})\} x}\)
\({v}+x \frac{{dv}}{{dx}}=\frac{\{\cos ({v})+y \sin ({v})\} {v}}{\{{v} \sin ({v})-\cos ({v})\}}\)
\(x \frac{{dv}}{{dx}}=\frac{\{\cos ({v})+y \sin ({v})\} {v}}{\{{v} \sin ({v})-\cos ({v})\}}-{v}\)
\(x \frac{{dv}}{{dx}}=\frac{{v} \cos ({v})+{v}^{2} \sin ({v})-{v}^{2} \sin ({v})+{v} \cos ({v})}{{v} \sin ({v})-\cos ({v})}\)
\(x \frac{{dv}}{{dx}}=\frac{2 {v} \cos ({v})}{{v} \sin ({v})-\cos ({v})}\)
\(\frac{{v} \sin ({v})-\cos ({v})}{2 {v} \cos ({v})} {dv}=\frac{1}{x} {dx}\)
\(\frac{{v} \sin ({v})}{2 {v} \cos ({v})} {dv}-\frac{-\cos ({v})}{2 {v} \cos ({v})} {dv}=\frac{1}{x} {dx}\)
\(\frac{1}{2} \tan {v} {dv}-\frac{1}{2} \cdot \frac{1}{{v}} {dv}=\frac{1}{x} d x\)
Integrating both sides, we get
\(\frac{1}{2} \int \tan {vdv}-\frac{1}{2} \cdot \int \frac{1}{{v}} {dv}=\int \frac{1}{x} d x\)
\(\frac{1}{2} \log \sec {v}-\frac{1}{2} \log {v}=\log x+\log {k}\)
\(\log \sec {v}-\log {v}=2 \log {kx}\)
\(\log \sec \left(\frac{y}{x}\right)-\log \left(\frac{y}{x}\right)=2 \log k x\)
\(\log \left(\frac{x}{y} \sec \left(\frac{y}{x}\right)\right)-\log (k x)^{2}\)
\(\frac{x}{y} \sec \left(\frac{y}{x}\right)=k^{2} x^{2}\)
\(\frac{1}{x y \cos \left(\frac{y}{x}\right)}=k^{2}\)
\(x y \cos \left(\frac{y}{x}\right)=\frac{1}{k^{2}}\)
\(C=\frac{1}{k^{2}}\)
\(x y \cos \left(\frac{y}{x}\right)=C\)
The required solution of the differential equation.

\(x \frac{d y}{d x}=y-x \sin \left(\frac{y}{x}\right)\)
\(\frac{d y}{d x}=\frac{y-x \sin \left(\frac{y}{x}\right)}{x}\)
Let \( {f}(x, {y})=\frac{y-x \sin \left(\frac{y}{x}\right)}{x} \)
Here, putting \( x={kx} \) and \( {y}={ky} \)
\({f}({kx}, {ky})=\frac{k y-k x \sin \left(\frac{k y}{k x}\right)}{k x}\)
\({f}({kx}, {ky})=\frac{k^{2}}{k^{2}} \cdot \frac{y-x \sin \left(\frac{y}{x}\right)}{x}\)
\(={k}^{0} . {f}(x, {y})\)
Therefore, the given differential equation is homogeneous.
\(x \frac{d y}{d x}=y-x \sin \left(\frac{y}{x}\right)\)
\(\frac{d y}{d x}=\frac{y-x \sin \left(\frac{y}{x}\right)}{x}\)
To solve it we make the substitution.
\(y=v x\)
Differentiating eq. with respect to \( x \), we get
\(\frac{d y}{d x}={v}+x \frac{{dv}}{{dx}}\)
\({v}+x \frac{{dv}}{{dx}}=\frac{{vx}-x \sin \left(\frac{{vx}}{x}\right)}{x}\)
\({v}+x \frac{{dv}}{{dx}}={v}-\sin {v}\)
\(x \frac{{dv}}{{dx}}=-\sin {v}\)
\(x \frac{{dv}}{{dx}}=-\frac{1}{x} {dx}\)
\(\operatorname{cosec} {dv}=-\frac{1}{x} {dx}\)
Integrating both side, we get
\(\int \operatorname{cosec} d v=-\int \frac{1}{x} d x\)
\(\log (\operatorname{cosec} v-\cot v)=-\log x+\log C\)
\(\log \left(\operatorname{cosec} \frac{y}{x}-\cot \frac{y}{x}\right)=\log \frac{c}{x}\)
\(\operatorname{cosec} \frac{y}{x}-\cot \frac{y}{x}=\frac{C}{x}\)
\(\frac{1}{\sin \frac{y}{x}}-\frac{\cos \frac{y}{x}}{\sin \frac{y}{x}}=\frac{C}{x}\)
\(1-\cos \frac{y}{x}=\frac{c}{x} \cdot \sin \frac{y}{x}\)
\(x\left(1-\cos \frac{y}{x}\right)=C \sin \frac{y}{x}\)
The required solution of the differential equation.

\(y d x+x \log \left(\frac{y}{x}\right) d y-2 x d y=0\)
\(x \log \left(\frac{y}{x}\right) d y-2 x d y=-y d x\)
\(\left(x \log \left(\frac{y}{x}\right) d y-2 x\right) d y=-y d x\)
\(\frac{d y}{d x}=\frac{-y}{x \log \left(\frac{y}{x}\right) d y-2 x}\)
\(\frac{d y}{d x}=\frac{-y}{2 x-x \log \left(\frac{y}{x}\right)}\)
Let \( {f}(x, {y})=\frac{y-x \sin \left(\frac{y}{x}\right)}{x} \)
Here, putting \( x={kx} \) and \( {y}={ky} \)
\({f}({kx}, {ky})=\frac{k y}{2 k x-k x \log \left(\frac{k y}{k x}\right)}\)
\({f}({kx}, {ky})=\frac{k^{2}}{k^{2}} \cdot \frac{y}{2 x-x \log \left(\frac{y}{x}\right)}\)
\(={k}^{0} \cdot {f}(x, {y})\)
Therefore, the given differential equation is homogeneous.
\(y d x+x \log \left(\frac{y}{x}\right) d y-2 x d y=0\)
\(x \log \left(\frac{y}{x}\right) d y-2 x d y=-y d x\)
\(\left(x \log \left(\frac{y}{x}\right) d y-2 x\right) d y=-y d x\)
\(\frac{d y}{d x}=\frac{-y}{x \log \left(\frac{y}{x}\right) d y-2 x}\)
\(\frac{d y}{d x}=\frac{-y}{2 x-x \log \left(\frac{y}{x}\right)}\)
To solve it we make the substitution.
\(y=v x\)
Differentiating eq. with respect to \( x \), we get
\(\frac{d y}{d x}={v}+x \frac{{dv}}{{dx}}\)
\({v}+x \frac{{dv}}{{dx}}=\frac{{vx}}{2 x-x \log \left(\frac{y}{x}\right)}\)
\({v}+x \frac{{dv}}{{dx}}=\frac{{v}}{2-\log {v}}\)
\(x \frac{{dv}}{{dx}}=\frac{{v}}{2-\log {v}}-{v}\)
\(x \frac{{dv}}{{dx}}=\frac{{v}-2 {v}+{v} \log {v}}{2-\log {v}}\)
\(x \frac{{dv}}{{dx}}=\frac{-{v}+{v} \log {v}}{2-\log {v}}\)
\(\frac{2-\log v}{v(\log v-1)} d v=\frac{1}{x} d x\)
\(\frac{1-(\log v-1)}{v(\log v-1)} d v=\frac{1}{x} d x\)
\(\frac{1}{v(\log v-1)} d v-\frac{1}{v} d v=\frac{1}{x} d x\)
Integrating both sides, we get
\(\int \frac{1}{v(\log v-1)} d v-\int \frac{1}{v} d v=\int \frac{1}{x} d x\quad \ldots\ldots\text{(i)}\)
Let \( {I}_{1}=\int \frac{1}{{v}(\log {v}-1)} {dv} \)
Put, \( \log {v}-1={t} \)
\(\frac{1}{{v}} {dv}={dt}\)
\(\int \frac{1}{t} {dt}\)
\(\log {t}\)
\(\log (\log {v}-1)\)
\(\therefore \log (\log {v}-1)-\log ({v})=\log (x)+\log ({c})\quad(\text {From (i) eq.})\)
\(\log \left(\frac{\log {v}-1}{{v}}\right)=\log (C x)\)
\(\frac{\log {v}-1}{{v}}=C x\)
\(\frac{\log \left(\frac{{y}}{x}\right)-1}{\frac{{y}}{x}}=C x\)
\(\log \left(\frac{{y}}{x}\right)-1=C y\)
The required solution of the differential equation.

\(\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y=0\)
\(\left(1+e^{\frac{x}{y}}\right) d x=-e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y\)
\(\frac{d y}{d x}=\frac{-e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)}{\left(1+e^{\frac{x}{y}}\right)}\)
Let \( {f}(x, {y})=\frac{-e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)}{\left(1+e^{\frac{x}{y}}\right)} \)
Here, putting \( x={kx} \) and \( {y}={ky} \)
\({f}({kx}, {ky})=\frac{-e^{\frac{k x}{k y}}\left(1-\frac{k x}{k y}\right)}{\left(1+e^{\frac{k x}{k y}}\right)}\)
\(=\frac{-e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)}{\left(1+e^{\frac{x}{y}}\right)}\)
\(={k}^{0} . {f}(x, {y})\)
Therefore, the given differential equation is homogeneous.
\(\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y=0\)
\(\left(1+e^{\frac{x}{y}}\right) d x=-e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y\)
\(\frac{d x}{d y}=\frac{-e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)}{\left(1+e^{\frac{x}{y}}\right)}\)
To solve it we make the substitution.
\(y=v x\)
Differentiating eq. with respect to \(x \), we get
\(\frac{d x}{d y}={v}+{y} \frac{{dv}}{{dy}}\)
\({v}+x \frac{{dv}}{{dy}}=\frac{-e^{\frac{-{vy}}{y}}\left(1-\frac{{vy}}{y}\right)}{\left(1+e^{\frac{{vy}}{y}}\right)}\)
\(x \frac{{dv}}{{dy}}=\frac{-e^{{v}}(1-{v})}{1+e^{{v}}}\)
\(\frac{-e^{{v}}(1-{v})}{1+e^{{v}}} {dv}=\frac{1}{y} {dy}\)
Integrating both sides, we get
\(\int \frac{-e^{{v}}(1-{v})}{1+e^{{v}}} {dv}=\int \frac{1}{y} {dy}\quad \ldots\ldots\text{(i)}\)
Let \( {I}_{1}=\int \frac{-e^{{v}}(1-{v})}{1+e^{{v}}} {dv} \)
Put \( e^{v}+v=t \)
\(\left(e^{v}+1\right) d v=d t\)
\(e^{v}+1=\frac{d t}{d v}\)
\(d v=\frac{d t}{e^{v}+1}\)
\(\int \frac{1}{t} {dt}\)
\(\log {t}\)
\(\log \left({e}^{{v}}+{v}\right)\)
\(\therefore \log \left({e}^{{v}}+{v}\right)=-\log {y}+\log {C}\quad(\therefore \text {From (i) eq.})\)
\(\log \left({e}^{\frac{x }{ {y}}}+\frac{x}{y}\right)=-\log {y}+\log {C}\)
\(\log \left({e}^{\frac{x }{ {y}}}+\frac{x}{y}\right)=\log \frac{C}{y}\)
\({e}^{\frac{x }{ {y}}}+\frac{x}{y}=\frac{C}{y}\)
Multiply by \(y\) on both side, we get
\(y e^{\frac{x }{ y}}+x=C\)
\(x+y e^{\frac{x }{ y}}=C\)
The required solution of the differential equation.

\((x+y) d y+(x-y) d x=0\)
\(\frac{d y}{d x}=-\frac{(x-y)}{(x+y)}\)
Let \( {f}(x, {y})=-\frac{(x-y)}{(x+y)} \)
Here, putting \( x={kx} \) and \( {y}={ky} \)
\({f}({kx}, {ky})=-\frac{(k x-k y)}{(k x+k y)}\)
\({f}({kx}, {ky})=\frac{k}{k} \cdot-\frac{(x-y)}{(x+y)}\)
\(={k}^{0} \cdot {f}(x, {y})\)
Therefore, the given differential equation is homogeneous.
\((x+y) d y+(x-y) d x=0\)
\(\frac{d y}{d x}=-\frac{(x-y)}{(x+y)}\)
To solve it we make the substitution.
\(y=v x\)
Differentiating eq. with respect to \( x \), we get
\(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(v+x \frac{d v}{d x}=-\frac{(x-v x)}{(x+v x)}\)
\(v+x \frac{d v}{d x}=-\frac{(1-v)}{(1+v)}\)
\(x \frac{d v}{d x}=-\frac{(1-v)}{(1+v)}-v\)
\(x \frac{d v}{d x}=\frac{-1+v-v-v^{2}}{(1+v)}\)
\(x \frac{d v}{d x}=\frac{-\left(1+v^{2}\right)}{(1+v)}\)
\(\frac{1+v}{1+v^{2}} d v=-\frac{1}{x} d x\)
Integrating both sides, we get
\( \int \frac{1+v}{1+v^{2}} d v=-\int \frac{1}{x} d x \)
\( \int \frac{1+v}{1+{v}^{2}} {dv}+\int \frac{{v}}{1+{v}^{2}} {dv}=-\int \frac{1}{x} {dx} \)
\(\tan ^{-1} {v}+\frac{1}{2} \log \left(1+{v}^{2}\right)=-\log x+C\)
\(\tan ^{-1} \frac{{y}}{x}+\frac{1}{2} \log \left(1+\left(\frac{{y}}{x}\right)^{2}\right)=-\log x+C\)
\({y}=1 \text { when } x=1\)
\(\tan ^{-1} \frac{1}{1}+\frac{1}{2} \log \left(1+\left(\frac{1}{1}\right)^{2}\right)=-\log 1+C\)
\(\frac{\pi}{4}+\frac{1}{2} \log 2=0+C\)
\({C}=\frac{\pi}{4}+\frac{1}{2} \log 2\)
\(\therefore \tan ^{-1} \frac{{y}}{x}+\frac{1}{2} \log \left(1+\left(\frac{{y}}{x}\right)^{2}\right)=-\log x+C\)
Where, \( C=\frac{\pi}{4}+\frac{1}{2} \log 2 \)
\(\therefore \tan ^{-1} \frac{{y}}{x}+\frac{1}{2} \log \left(1+\left(\frac{{y}}{x}\right)^{2}\right)\)
\(=-\log x+\frac{\pi}{4}+\frac{1}{2} \log 2\)
\(2 \tan ^{-1} \frac{{y}}{x}+\log \left(\frac{x^{2}+y^{2}}{x^{2}}\right)\)
\(=-2 \log x+\frac{\pi}{2}+\log 2\)
\(2 \tan ^{-1} \frac{{y}}{x}+\log \left(\frac{x^{2}+y^{2}}{x^{2}}\right)+\log x^{2}+\frac{\pi}{2}+\log 2\)
\(2 \tan ^{-1} \frac{{y}}{x}+\log \left(x^{2}+y^{2}\right)=\frac{\pi}{2}+\log 2\)
The required solution of the differential equation.

\(x^{2} d y+\left(x y+y^{2}\right) d x=0\)
\(\frac{d y}{d x}=-\frac{\left(x y+y^{2}\right)}{x^{2}}\)
Let \( {f}(x, {y})=-\frac{\left(x y+y^{2}\right)}{x^{2}} \)
Here, putting \( x={kx} \) and \( {y}={ky} \)
\({f}({kx}, {ky})=-\frac{\left(k x k y+k^{2} y^{2}\right)}{k^{2} x^{2}}\)
\({f}({kx}, {ky})=\frac{k^{2}}{k^{2}} \cdot-\frac{\left(x y+y^{2}\right)}{x^{2}}\)
\(={k}^{0} . {f}(x, {y})\)
Therefore, the given differential equation is homogeneous.
\(x^{2} d y+\left(x y+y^{2}\right) d x=0\)
\(\frac{d y}{d x}=-\frac{\left(x y+y^{2}\right)}{x^{2}}\)
To solve it we make the substitution.
\(y=v x\)
Differentiating eq. with respect to \( x \), we get
\(\frac{d y}{d x}={v}+x \frac{{dv}}{{dx}}\)
\({v}+x \frac{{dv}}{{dx}}=-\frac{\left(x . {vx}+{v}^{2} x^{2}\right)}{x^{2}}\)
\({v}+x \frac{{dv}}{{dx}}=-{v}-{v}^{2}\)
\({v}+x \frac{{dv}}{{dx}}=-{v}-{v}^{2}-{v}\)
\(x \frac{d v}{d x}=-v(v+2)\)
\(\frac{1}{v(v+2)} d v=-\frac{1}{x} d x\)
Integrating both sides, we get
\(\int \frac{1}{v(v+2)} d v=-\int \frac{1}{x} d x\)
\(\frac{1}{2} \int \frac{2}{v(v+2)} d v=-\int \frac{1}{x} d x\)
\(\frac{1}{2} \int \frac{2+v-v}{v(v+2)} d v=-\int \frac{1}{x} d x\)
\(\frac{1}{2} \int \frac{2}{v(v+2)} d v=-\int \frac{1}{x} d x\)
\(\frac{1}{2} \int\left(\frac{2}{v(v+2)}-\frac{v}{v(v+2)}\right) d v=-\int \frac{1}{x} d x\)
\(\frac{1}{2} \int\left(\frac{1}{v}-\frac{1}{v+2}\right) d v=-\int \frac{1}{x} d x\)
\(\frac{1}{2}(\log -\log ({v}+2))=-\log x+\log C\)
\(\frac{1}{2}\left(\log \frac{{v}}{{v}+2}\right)=\log \frac{C}{x}\)
\(\log \left(\frac{\frac{{y}}{x}}{\frac{{y}}{x}+2}\right)=2 \log \frac{C}{x}\)
\(\log \left(\frac{y}{y+2 x}\right)=2 \log \left(\frac{C}{x}\right)^{2}\)
\(\frac{y}{{y}+2 x}=\left(\frac{C}{x}\right)^{2}\)
\(\frac{x^{2} y}{y+2 x}={C}^{2}\)
\(y=1 \text { where } x=1\)
\({C}^{2}=\frac{1}{1+2}=\frac{1}{3}\)
\(\therefore \frac{x^{2} y}{y+2 x}=\frac{1}{3}\)
\(3 x^{2} y=y+2 x\)
\(y+2 x=3 x^{2} y\)
The required solution of the differential equation.

\({\left[x \sin ^{2}\left(\frac{y}{x}\right)-y\right] d x=-x d y}\)
\({\left[x \sin ^{2}\left(\frac{y}{x}\right)-y\right]=-x \frac{d y}{d x}}\)
\(\frac{d y}{d x}=-\left[\frac{x \sin ^{2}\left(\frac{y}{x}\right)-y}{x}\right]\)
\({f}({x}, {y})=-\left[\frac{x \sin ^{2}\left(\frac{y}{x}\right)-y}{x}\right]\)
Here, putting \( {x}={kx} \) and \( {y}={ky} \)
\( {f}({kx}, {ky})=-\left[\frac{k x \sin ^{2}\left(\frac{k y}{k x}\right)-k y}{k x}\right]\)
\({f}({kx}, {ky})=\frac{k}{k} \cdot-\left[\frac{x \sin ^{2}\left(\frac{y}{x}\right)-y}{x}\right]\)
\(={k}^{0} \cdot {f}({x}, {y})\)
Therefore, the given differential equation is homogeneous.
\( {\left[x \sin ^{2}\left(\frac{y}{x}\right)-y\right] d x=-x d y}\)
\({\left[x \sin ^{2}\left(\frac{y}{x}\right)-y\right]=-x \frac{d y}{d x}}\)
\(\frac{d y}{d x}=-\left[\frac{x \sin ^{2}\left(\frac{y}{x}\right)-y}{x}\right]\)
To solve it we make the substitution.
\(y=v x\)
Differentiating eq. with respect to \( x \), we get
\( \frac{d y}{d x}={v}+{x} \frac{{dv}}{{dx}}\)
\({v}+{x} \frac{{dv}}{{dx}}=-\frac{\left[x \sin ^{2}\left(\frac{{vx}}{{x}}\right)-{vx}\right]}{x}\)
\({v}+{x} \frac{{dv}}{{dx}}=-\left[\frac{x \sin ^{2} {v}-{vx}}{x}\right]\)
\({v}+{x} \frac{{dv}}{{dx}}=-\sin ^{2} {v}-{v}\)
\({x} \frac{{dv}}{{dx}}=\left[\sin ^{2} {v}-{v}\right]-{v}\)
\({x} \frac{{dv}}{{dx}}=\sin ^{2} {v}+{v}-{v}\)
\({x} \frac{{dv}}{{dx}}=-\sin ^{2} {v}\)
\(\frac{1}{\sin ^{2} v} {dv}=-\frac{1}{x} {dx}\)
Integrating both sides, we get
\( \int \frac{1}{\sin ^{2} v} {dv}=-\int \frac{1}{x} {dx}\)
\(\int \operatorname{cosec}^{2} {vdv}=-\log {x}-\log {C}\)
\(-\cot {v}=-\log {x}-\log {C}\)
\( \cot {v}=\log {x}+\log {C}\)
\(\cot \frac{y}{x}=\log (C x)\)
\(y=\frac{\pi}{4} \text { when } {x}=1\)
\(\cot \frac{\frac{\pi}{4}}{x}=\log (C \cdot 1)\)
\(\cot \frac{\pi}{4}=\log {C}\)
\(1={C}\)
\({e}^{1}={C}\)
\(\therefore \cot \frac{y}{x}=\log (e x)\)
The required solution of the differential equation.

\(\frac{d y}{d x}-\frac{y}{x}+\operatorname{cosec}\left(\frac{y}{x}\right)=0\)
\(\frac{d y}{d x}=\frac{y}{x}-\operatorname{cosec}\left(\frac{y}{x}\right)\)
Let \( {f}(x, {y})=\frac{y}{x}-\operatorname{cosec}\left(\frac{y}{x}\right) \)
Here, putting \( x={kx} \) and \( {y}={ky} \)
\({f}({kx}, {ky})=\frac{k y}{k x}-\operatorname{cosec}\left(\frac{k y}{k x}\right)\)
\(=\frac{y}{x}-\operatorname{cosec}\left(\frac{y}{x}\right)\)
\(={k}^{0} . {f}(x, {y})\)
Therefore, the given differential equation is homogeneous.
\(\frac{d y}{d x}-\frac{y}{x}+\operatorname{cosec}\left(\frac{y}{x}\right)=0\)
\(\frac{d y}{d x}=\frac{y}{x}-\operatorname{cosec}\left(\frac{y}{x}\right)\)
To solve it we make the substitution.
\(y=v x\)
Differentiating eq. with respect to \( x \), we get
\(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(v+x \frac{d v}{d x}=\frac{v x}{x}+\operatorname{cosec}\left(\frac{v x}{x}\right)\)
\(v+x \frac{d v}{d x}=v+\operatorname{cosec} v\)
\(x \frac{{dv}}{{dx}}=-\operatorname{cosec} {v}\)
\(\frac{1}{\operatorname{cosec} {v}} {dv}=-\frac{1}{x} {dx}\)
Integrating both sides, we get
\(\int \sin {v} d v=-\frac{1}{x} {dx}\)
\(-\cos {v}=-\log x+{C}\)
\(-\cos \frac{y}{x}=-\log x+{C}\)
\({y}=0 \text { when } x=1\)
\(-\cos \frac{0}{1}=-\log 1+C\)
\(-1=C\)
\(\therefore-\cos \frac{y}{x}=-\log x+1\)
\(\cos \frac{y}{x}=-\log x+\log {e}\)
The required solution of the differential equation.

\(2 x y+y^{2}-2 x^{2} \frac{d y}{d x}=0\)
\(\frac{d y}{d x}=\frac{2 x y+y^{2}}{2 x^{2}}\)
Let \( {f}(x, {y})=\frac{2 x y+y^{2}}{2 x^{2}} \)
Here, putting \( x={kx} \) and \( {y}={ky} \)
\({f}({kx}, {ky})=\frac{2 k x k y+(k y)^{2}}{2(k x)^{2}}\)
\(=\frac{k^{2}}{k^{2}} \cdot \frac{2 x y+y^{2}}{2 x^{2}}\)
\(={k}^{0} \cdot {f}(x, {y})\)
Therefore, the given differential equation is homogeneous.
\(2 x y+y^{2}-2 x^{2} \frac{d y}{d x}=0\)
\(\frac{d y}{d x}=\frac{2 x y+y^{2}}{2 x^{2}}\)
To solve it we make the substitution.
\(y=v x\)
Differentiating eq. with respect to \( x \), we get
\(\frac{d y}{d x}={v}+x \frac{{dv}}{{dx}}\)
\({v}+x \frac{{dv}}{{dx}}=\frac{2 x \cdot {vx}+({vx})^{2}}{2 x^{2}}\)
\({v}+x \frac{{dv}}{{dx}}=\frac{2 {v}+{v}^{2}}{2}\)
\({v}+x \frac{{dv}}{{dx}}={v}+\frac{1}{2} {v}^{2}\)
\(x \frac{{dv}}{{dx}}=\frac{1}{2} {v}^{2}\)
\(2 \frac{1}{{v}^{2}} {dv}=\frac{1}{x} {dx}\)
Integrating both sides, we get
\(\int 2 \frac{1}{{v}^{2}} {dv}=\int \frac{1}{x} {dx}\)
\(-\frac{2}{{v}}=\log x+C\)
\(-\frac{2}{\frac{y}{x}}=\log x+C\)
\(-\frac{2 x}{y}=\log x+C\)
\({y}=2 \text { when } x=1\)
\(-\frac{2.1}{2}=\log 1+C\)
\(-1={C}\)
\(\therefore-\frac{2 x}{y}=\log x-1\)
\(\frac{2 x}{y}=1-\log x\)
\(y=\frac{2 x}{1-\log |x|} ; x \neq e, x \neq 0\)

Since \( \frac{d x}{d y} \) is given equal to \( h\left(\frac{x}{y}\right) \)
\( h\left(\frac{x}{y}\right) \) is a function of \( \frac{x}{y} \).
Therefore, we shall substitute, \( x=v y \)

\( (4 x+6 y+5) d y-(3 y+2 x+4) d x=0 \)
It cannot be homogeneous as we can see that \( (4 x+6 y+5) \) is not homogeneous.

\( (x y) d x-\left(x^{3}+y^{3}\right) d y=0 \)
It cannot be homogeneous as the \(xy\) which multiplies with \(dx \) and \( x^{3}+ {y}^{3} \) which multiplies with \(dy\) are not of same degree.

\( \left(x^{3}+2 y^{2}\right) d x+2 x y d y=0 \)
Similarly, it cannot be homogeneous as the \( x^{3}+2 y^{2} \) which multiplies with \(dx\) and \(2 xy \) which multiplies with \(dy\) are not of same degree.

\( y^{2} d x+\left(x^{2}-x y-y^{2}\right) d y=0 \)
\(\frac{d y}{d x}=-\frac{x^{2}-x y-y^{2}}{y^{2}}\)
Let \( {f}(x, {y})=-\frac{x^{2}-x y-y^{2}}{y^{2}} \)
Here, putting \( x={kx} \) and \( {y}={ky} \)
\({f}({kx}, {ky})=-\frac{(k x)^{2}-k x k y-(k y)^{2}}{(k y)^{2}}\)
\(=\frac{k^{2}}{k^{2}} \cdot-\frac{x^{2}-x y-y^{2}}{y^{2}}\)
\(={k}^{0} \cdot {f}(x, {y})\)
Therefore, the given differential equations are homogeneous.