Solution

The given differential equation i.e., $(x^{2}+x y) d y=(x^{2}+y^{2}) d x$ can be written as:

$\frac{d y}{d x}=\frac{x^{2}+y^{2}}{x^{2}+x y}$

Let $F(x, y)=\frac{x^{2}+y^{2}}{x^{2}+x y}$.

Now, $F(\lambda x, \lambda y)=\frac{(\lambda x)^{2}+(\lambda y)^{2}}{(\lambda x)^{2}+(\lambda x)(\lambda y)}=\frac{x^{2}+y^{2}}{x^{2}+x y}=\lambda^{0} \cdot F(x, y)$

This shows that equation (1) is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

Differentiating both sides with respect to $x$, we get:

$\frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $v$ and $\frac{d y}{d x}$ in equation (1), we get:

$ \begin{aligned} & v+x \frac{d v}{d x}=\frac{x^{2}+(v x)^{2}}{x^{2}+x(v x)} \\ & \Rightarrow v+x \frac{d v}{d x}=\frac{1+v^{2}}{1+v} \\ & \Rightarrow x \frac{d v}{d x}=\frac{1+v^{2}}{1+v}-v=\frac{(1+v^{2})-v(1+v)}{1+v} \\ & \Rightarrow x \frac{d v}{d x}=\frac{1-v}{1+v} \\ & \Rightarrow(\frac{1+v}{1-v})=d v=\frac{d x}{x} \\ & \Rightarrow(\frac{2-1+v}{1-v}) d v=\frac{d x}{x} \\ & \Rightarrow(\frac{2}{1-v}-1) d v=\frac{d x}{x} \end{aligned} $

Integrating both sides, we get:

$ \begin{aligned} & -2 \log (1-v)-v=\log x-\log k \\ & \Rightarrow v=-2 \log (1-v)-\log x+\log k \\ & \Rightarrow v=\log [\frac{k}{x(1-v)^{2}}] \\ & \Rightarrow \frac{y}{x}=\log [\frac{k}{x(1-\frac{y}{x})^{2}}] \\ & \Rightarrow \frac{y}{x}=\log [\frac{k x}{(x-y)^{2}}] \\ & \Rightarrow \frac{k x}{(x-y)^{2}}=e^{\frac{y}{x}} \\ & \Rightarrow(x-y)^{2}=k x e^{-\frac{y}{x}} \end{aligned} $

This is the required solution of the given differential equation.

Solution

The given differential equation is:

$y^{\prime}=\frac{x+y}{x}$

$\Rightarrow \frac{d y}{d x}=\frac{x+y}{x}$

Let $F(x, y)=\frac{x+y}{x}$.

Now, $F(\lambda x, \lambda y)=\frac{\lambda x+\lambda y}{\lambda x}=\frac{x+y}{x}=\lambda^{0} F(x, y)$

Thus, the given equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

Differentiating both sides with respect to $x$, we get:

$\frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{x+v x}{x}$

$\Rightarrow v+x \frac{d v}{d x}=1+v$

$x \frac{d v}{d x}=1$

$\Rightarrow d v=\frac{d x}{x}$

Integrating both sides, we get:

$v=\log x+C$

$\Rightarrow \frac{y}{x}=\log x+C$

$\Rightarrow y=x \log x+C x$

This is the required solution of the given differential equation.

Solution

The given differential equation is:

$(x-y) d y-(x+y) d x=0$

$\Rightarrow \frac{d y}{d x}=\frac{x+y}{x-y}$

Let $F(x, y)=\frac{x+y}{x-y}$.

$\therefore F(\lambda x, \lambda y)=\frac{\lambda x+\lambda y}{\lambda x-\lambda y}=\frac{x+y}{x-y}=\lambda^{0} \cdot F(x, y)$

Thus, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{x+v x}{x-v x}=\frac{1+v}{1-v}$

$x \frac{d v}{d x}=\frac{1+v}{1-v}-v=\frac{1+v-v(1-v)}{1-v}$

$\Rightarrow x \frac{d v}{d x}=\frac{1+v^{2}}{1-v}$

$\Rightarrow \frac{1-v}{(1+v^{2})} d v=\frac{d x}{x}$

$\Rightarrow(\frac{1}{1+v^{2}}-\frac{v}{1-v^{2}}) d v=\frac{d x}{x}$

Integrating both sides, we get:

$ \begin{aligned} & \tan ^{-1} v-\frac{1}{2} \log (1+v^{2})=\log x+C \\ & \Rightarrow \tan ^{-1}(\frac{y}{x})-\frac{1}{2} \log [1+(\frac{y}{x})^{2}]=\log x+C \\ & \Rightarrow \tan ^{-1}(\frac{y}{x})-\frac{1}{2} \log (\frac{x^{2}+y^{2}}{x^{2}})=\log x+C \\ & \Rightarrow \tan ^{-1}(\frac{y}{x})-\frac{1}{2}[\log (x^{2}+y^{2})-\log x^{2}]=\log x+C \\ & \Rightarrow \tan ^{-1}(\frac{y}{x})=\frac{1}{2} \log (x^{2}+y^{2})+C \end{aligned} $

This is the required solution of the given differential equation.

Solution

The given differential equation is:

$(x^{2}-y^{2}) d x+2 x y d y=0$

$\Rightarrow \frac{d y}{d x}=\frac{-(x^{2}-y^{2})}{2 x y}$

Let $F(x, y)=\frac{-(x^{2}-y^{2})}{2 x y}$.

$\therefore F(\lambda x, \lambda y)=[\frac{(\lambda x)^{2}-(\lambda y)^{2}}{2(\lambda x)(\lambda y)}]=\frac{-(x^{2}-y^{2})}{2 x y}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$ $\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=-[\frac{x^{2}-(v x)^{2}}{2 x \cdot(v x)}]$

$v+x \frac{d v}{d x}=\frac{v^{2}-1}{2 v}$

$\Rightarrow x \frac{d v}{d x}=\frac{v^{2}-1}{2 v}-v=\frac{v^{2}-1-2 v^{2}}{2 v}$

$\Rightarrow x \frac{d v}{d x}=-\frac{(1+v^{2})}{2 v}$

$\Rightarrow \frac{2 v}{1+v^{2}} d v=-\frac{d x}{x}$

Integrating both sides, we get:

$\log (1+v^{2})=-\log x+\log C=\log \frac{C}{x}$

$\Rightarrow 1+v^{2}=\frac{C}{x}$

$\Rightarrow[1+\frac{y^{2}}{x^{2}}]=\frac{C}{x}$

$\Rightarrow x^{2}+y^{2}=C x$

This is the required solution of the given differential equation.

Solution

The given differential equation is:

$x^{2} \frac{d y}{d x}=x^{2}-2 y^{2}+x y$ $\frac{d y}{d x}=\frac{x^{2}-2 y^{2}+x y}{x^{2}}$

Let $F(x, y)=\frac{x^{2}-2 y^{2}+x y}{x^{2}}$.

$\therefore F(\lambda x, \lambda y)=\frac{(\lambda x)^{2}-2(\lambda y)^{2}+(\lambda x)(\lambda y)}{(\lambda x)^{2}}=\frac{x^{2}-2 y^{2}+x y}{x^{2}}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{x^{2}-2(v x)^{2}+x \cdot(v x)}{x^{2}}$

$\Rightarrow v+x \frac{d v}{d x}=1-2 v^{2}+v$

$\Rightarrow x \frac{d v}{d x}=1-2 v^{2}$

$\Rightarrow \frac{d v}{1-2 v^{2}}=\frac{d x}{x}$

$\Rightarrow \frac{1}{2} \cdot \frac{d v}{\frac{1}{2}-v^{2}}=\frac{d x}{x}$

$\Rightarrow \frac{1}{2} \cdot[\frac{d v}{(\frac{1}{\sqrt{2}})^{2}-v^{2}}]=\frac{d x}{x}$

Integrating both sides, we get:

$ \begin{aligned} & \frac{1}{2} \cdot \frac{1}{2 \times \frac{1}{\sqrt{2}}} \log |\frac{\frac{1}{\sqrt{2}}+v}{\frac{1}{\sqrt{2}}-v}|=\log |x|+C \\ & \Rightarrow \frac{1}{2 \sqrt{2}} \log |\frac{\frac{1}{\sqrt{2}}+\frac{y}{x}}{\frac{1}{\sqrt{2}}-\frac{y}{x}}|=\log |x|+C \\ & \Rightarrow \frac{1}{2 \sqrt{2}} \log |\frac{x+\sqrt{2} y}{x-\sqrt{2} y}|=\log |x|+C \end{aligned} $

This is the required solution for the given differential equation.

Solution

$x d y-y d x=\sqrt{x^{2}+y^{2}} d x$

$\Rightarrow x d y=[y+\sqrt{x^{2}+y^{2}}] d x$

$\frac{d y}{d x}=\frac{y+\sqrt{x^{2}+y^{2}}}{x^{2}}$

Let $F(x, y)=\frac{y+\sqrt{x^{2}+y^{2}}}{x^{2}}$.

$\therefore F(\lambda x, \lambda y)=\frac{\lambda x+\sqrt{(\lambda x)^{2}+(\lambda y)^{2}}}{\lambda x}=\frac{y+\sqrt{x^{2}+y^{2}}}{x}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $v$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{v x+\sqrt{x^{2}+(v x)^{2}}}{x}$

$\Rightarrow v+x \frac{d v}{d x}=v+\sqrt{1+v^{2}}$

$\Rightarrow \frac{d v}{\sqrt{1+v^{2}}}=\frac{d x}{x}$

Integrating both sides, we get:

$\log |v+\sqrt{1+v^{2}}|=\log |x|+\log C$

$\Rightarrow \log |\frac{y}{x}+\sqrt{1+\frac{y^{2}}{x^{2}}}|=\log |C x|$

$\Rightarrow \log |\frac{y+\sqrt{x^{2}+y^{2}}}{x}|=\log |C x|$

$\Rightarrow y+\sqrt{x^{2}+y^{2}}=C x^{2}$

This is the required solution of the given differential equation.

Solution

The given differential equation is:

$$x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right) y d x=y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right) x d y$$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d y}{d x}=v+x=\frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$ \begin{aligned} & v+x \frac{d v}{d x}=\frac{(x \cos v+v x \sin v) \cdot v x}{(v x \sin v-x \cos v) \cdot x} \\ & \Rightarrow v+x \frac{d v}{d x}=\frac{v \cos v+v^{2} \sin v}{v \sin v-\cos v} \\ & \Rightarrow x \frac{d v}{d x}=\frac{v \cos v+v^{2} \sin v}{v \sin v-\cos v}-v \\ & \Rightarrow x \frac{d v}{d x}=\frac{v \cos v+v^{2} \sin v-v^{2} \sin v+v \cos v}{v \sin v-\cos v} \\ & \Rightarrow x \frac{d v}{d x}=\frac{2 v \cos v}{v \sin v-\cos v} \\ & \Rightarrow[\frac{v \sin v-\cos v}{v \cos v}] d v=\frac{2 d x}{x} \\ & \Rightarrow(\tan v-\frac{1}{v}) d v=\frac{2 d x}{x} \end{aligned} $

Integrating both sides, we get:

$\log (\sec v)-\log v=2 \log x+\log C$

$\Rightarrow \log (\frac{\sec v}{v})=\log (C x^{2})$

$\Rightarrow(\frac{\sec v}{v})=C x^{2}$

$\Rightarrow \sec v=C x^{2} v$

$\Rightarrow \sec (\frac{y}{x})=C \cdot x^{2} \cdot \frac{y}{x}$

$\Rightarrow \sec (\frac{y}{x})=C x y$

$\Rightarrow \cos (\frac{y}{x})=\frac{1}{C x y}=\frac{1}{C} \cdot \frac{1}{x y}$

$\Rightarrow x y \cos (\frac{y}{x})=k \quad(k=\frac{1}{C})$

This is the required solution of the given differential equation.

Solution

$x \frac{d y}{d x}-y+x \sin (\frac{y}{x})=0$

$\Rightarrow x \frac{d y}{d x}=y-x \sin (\frac{y}{x})$

$\Rightarrow \frac{d y}{d x}=\frac{y-x \sin (\frac{y}{x})}{x}$

Let $F(x, y)=\frac{y-x \sin (\frac{y}{x})}{x}$.

$\therefore F(\lambda x, \lambda y)=\frac{\lambda y-\lambda x \sin (\frac{\lambda y}{\lambda x})}{\lambda x}=\frac{y-x \sin (\frac{y}{x})}{x}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{v x-x \sin v}{x}$

$\Rightarrow v+x \frac{d v}{d x}=v-\sin v$

$\Rightarrow-\frac{d v}{\sin v}=\frac{d x}{x}$

$\Rightarrow cosec v d v=-\frac{d x}{x}$

Integrating both sides, we get:

$\log |cosec v-\cot v|=-\log x+\log C=\log \frac{C}{x}$

$\Rightarrow cosec(\frac{y}{x})-\cot (\frac{y}{x})=\frac{C}{x}$

$\Rightarrow \frac{1}{\sin (\frac{y}{x})}-\frac{\cos (\frac{y}{x})}{\sin (\frac{y}{x})}=\frac{C}{x}$

$\Rightarrow x[1-\cos (\frac{y}{x})]=C \sin (\frac{y}{x})$

This is the required solution of the given differential equation.

Solution

$y d x+x \log (\frac{y}{x}) d y-2 x d y=0$

$\Rightarrow y d x=[2 x-x \log (\frac{y}{x})] d y$

$\Rightarrow \frac{d y}{d x}=\frac{y}{2 x-x \log (\frac{y}{x})}$

Let $F(x, y)=\frac{y}{2 x-x \log (\frac{y}{x})}$.

$\therefore F(\lambda x, \lambda y)=\frac{\lambda y}{2(\lambda x)-(\lambda x) \log (\frac{\lambda y}{\lambda x})}=\frac{y}{2 x-\log (\frac{y}{x})}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$ $\Rightarrow \frac{d y}{d x}=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{v x}{2 x-x \log v}$

$\Rightarrow v+x \frac{d v}{d x}=\frac{v}{2-\log v}$

$\Rightarrow x \frac{d v}{d x}=\frac{v}{2-\log v}-v$

$\Rightarrow x \frac{d v}{d x}=\frac{v-2 v+v \log v}{2-\log v}$

$\Rightarrow x \frac{d v}{d x}=\frac{v \log v-v}{2-\log v}$

$\Rightarrow \frac{2-\log v}{v(\log v-1)} d v=\frac{d x}{x}$

$\Rightarrow[\frac{1+(1-\log v)}{v(\log v-1)}] d v=\frac{d x}{x}$

$\Rightarrow[\frac{1}{v(\log v-1)}-\frac{1}{v}] d v=\frac{d x}{x}$

Integrating both sides, we get:

$$ \begin{align*} & \int \frac{1}{v(\log v-1)} d v-\int \frac{1}{v} d v=\int \frac{1}{x} d x \\ & \Rightarrow \int \frac{d v}{v(\log v-1)}-\log v=\log x+\log C \tag{2} \end{align*} $$

$\Rightarrow$ Let $\log v-1=t$

$\Rightarrow \frac{d}{d v}(\log v-1)=\frac{d t}{d v}$

$\Rightarrow \frac{1}{v}=\frac{d t}{d v}$

$\Rightarrow \frac{d v}{v}=d t$

Therefore, equation (1) becomes:

$\Rightarrow \int \frac{d t}{t}-\log v=\log x+\log C$

$\Rightarrow \log t-\log (\frac{y}{x})=\log (C x)$

$\Rightarrow \log [\log (\frac{y}{x})-1]-\log (\frac{y}{x})=\log (C x)$

$\Rightarrow \log [\frac{\log (\frac{y}{x})-1}{\frac{y}{x}}]=\log (C x)$

$\Rightarrow \frac{x}{y}[\log (\frac{y}{x})-1]=C x$

$\Rightarrow \log (\frac{y}{x})-1=C y$

This is the required solution of the given differential equation.

Solution

$(1+e^{\frac{x}{y}}) d x+e^{\frac{x}{y}}(1-\frac{x}{y}) d y=0$

$\Rightarrow(1+e^{\frac{x}{y}}) d x=-e^{\frac{x}{y}}(1-\frac{x}{y}) d y$ $\Rightarrow \frac{d x}{d y}=\frac{-e^{\frac{x}{y}}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}$

Let $F(x, y)=\frac{-e^{\frac{x}{y}}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}$.

$\therefore F(\lambda x, \lambda y)=\frac{-e^{\frac{\lambda x}{\lambda y}}(1-\frac{\lambda x}{\lambda y})}{1+e^{\frac{\lambda x}{\lambda y}}}=\frac{-e^{\frac{x}{y}}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$x=v y$

$\Rightarrow \frac{d}{d y}(x)=\frac{d}{d y}(v y)$

$\Rightarrow \frac{d x}{d y}=v+y \frac{d v}{d y}$

Substituting the values of $x$ and $\frac{d x}{d y}$ in equation (1), we get:

$v+y \frac{d v}{d y}=\frac{-e^{v}(1-v)}{1+e^{v}}$

$\Rightarrow y \frac{d v}{d y}=\frac{-e^{v}+v e^{v}}{1+e^{v}}-v$

$\Rightarrow y \frac{d v}{d y}=\frac{-e^{v}+v e^{v}-v-v e^{v}}{1+e^{v}}$

$\Rightarrow y \frac{d v}{d y}=-[\frac{v+e^{v}}{1+e^{v}}]$

$\Rightarrow[\frac{1+e^{v}}{v+e^{v}}] d v=-\frac{d y}{y}$

Integrating both sides, we get: $\Rightarrow \log (v+e^{v})=-\log y+\log C=\log (\frac{C}{y})$

$\Rightarrow[\frac{x}{y}+e^{\frac{x}{y}}]=\frac{C}{y}$

$\Rightarrow x+y e^{\frac{x}{y}}=C$

This is the required solution of the given differential equation.

Solution

$(x+y) d y+(x-y) d x=0$

$\Rightarrow(x+y) d y=-(x-y) d x$

$\Rightarrow \frac{d y}{d x}=\frac{-(x-y)}{x+y}$

Let $F(x, y)=\frac{-(x-y)}{x+y}$.

$\therefore F(\lambda x, \lambda y)=\frac{-(\lambda x-\lambda y)}{\lambda x-\lambda y}=\frac{-(x-y)}{x+y}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get: $v+x \frac{d v}{d x}=\frac{-(x-v x)}{x+v x}$

$\Rightarrow v+x \frac{d v}{d x}=\frac{v-1}{v+1}$

$\Rightarrow x \frac{d v}{d x}=\frac{v-1}{v+1}-v=\frac{v-1-v(v+1)}{v+1}$

$\Rightarrow x \frac{d v}{d x}=\frac{v-1-v^{2}-v}{v+1}=\frac{-(1+v^{2})}{v+1}$

$\Rightarrow \frac{(v+1)}{1+v^{2}} d v=-\frac{d x}{x}$

$\Rightarrow[\frac{v}{1+v^{2}}+\frac{1}{1+v^{2}}] d v=-\frac{d x}{x}$

Integrating both sides, we get:

$\frac{1}{2} \log (1+v^{2})+\tan ^{-1} v=-\log x+k$

$\Rightarrow \log (1+v^{2})+2 \tan ^{-1} v=-2 \log x+2 k$

$\Rightarrow \log [(1+v^{2}) \cdot x^{2}]+2 \tan ^{-1} v=2 k$

$\Rightarrow \log [(1+\frac{y^{2}}{x^{2}}) \cdot x^{2}]+2 \tan ^{-1} \frac{y}{x}=2 k$

$\Rightarrow \log (x^{2}+y^{2})+2 \tan ^{-1} \frac{y}{x}=2 k$

Now, $y=1$ at $x=1$.

$\Rightarrow \log 2+2 \tan ^{-1} 1=2 k$

$\Rightarrow \log 2+2 \times \frac{\pi}{4}=2 k$

$\Rightarrow \frac{\pi}{2}+\log 2=2 k$

Substituting the value of $2 k$ in equation (2), we get:

$\log (x^{2}+y^{2})+2 \tan ^{-1}(\frac{y}{x})=\frac{\pi}{2}+\log 2$

This is the required solution of the given differential equation.

Solution

$x^{2} d y+(x y+y^{2}) d x=0$

$\Rightarrow x^{2} d y=-(x y+y^{2}) d x$

$\Rightarrow \frac{d y}{d x}=\frac{-(x y+y^{2})}{x^{2}}$

Let $F(x, y)=\frac{-(x y+y^{2})}{x^{2}}$.

$\therefore F(\lambda x, \lambda y)=\frac{[\lambda x \cdot \lambda y+(\lambda y)^{2}]}{(\lambda x)^{2}}=\frac{-(x y+y^{2})}{x^{2}}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$ \begin{aligned} & v+x \frac{d v}{d x}=\frac{-[x \cdot v x+(v x)^{2}]}{x^{2}}=-v-v^{2} \\ & \Rightarrow x \frac{d v}{d x}=-v^{2}-2 v=-v(v+2) \\ & \Rightarrow \frac{d v}{v(v+2)}=-\frac{d x}{x} \\ & \Rightarrow \frac{1}{2}[\frac{(v+2)-v}{v(v+2)}] d v=-\frac{d x}{x} \\ & \Rightarrow \frac{1}{2}[\frac{1}{v}-\frac{1}{v+2}] d v=-\frac{d x}{x} \end{aligned} $

Integrating both sides, we get:

$$ \begin{align*} & \frac{1}{2}[\log v-\log (v+2)]=-\log x+\log C \\ & \Rightarrow \frac{1}{2} \log (\frac{v}{v+2})=\log \frac{C}{x} \\ & \Rightarrow \frac{v}{v+2}=(\frac{C}{x})^{2} \\ & \Rightarrow \frac{\frac{y}{y}}{\frac{y}{x}+2}=(\frac{C}{x})^{2} \\ & \Rightarrow \frac{y}{y+2 x}=\frac{C^{2}}{x^{2}} \\ & \Rightarrow \frac{x^{2} y}{y+2 x}=C^{2} \tag{2} \end{align*} $$

Now, $y=1$ at $x=1$.

$\Rightarrow \frac{1}{1+2}=C^{2}$

$\Rightarrow C^{2}=\frac{1}{3}$

Substituting $C^{2}=\frac{1}{3}$ in equation (2), we get:

$ \begin{aligned} & \frac{x^{2} y}{y+2 x}=\frac{1}{3} \\ & \Rightarrow y+2 x=3 x^{2} y \end{aligned} $

This is the required solution of the given differential equation.

Solution

$[x \sin ^{2}(\frac{y}{x})-y] d x+x d y=0$

$\Rightarrow \frac{d y}{d x}=\frac{-[x \sin ^{2}(\frac{y}{x})-y]}{x}$

Let $F(x, y)=\frac{-[x \sin ^{2}(\frac{y}{x})-y]}{x}$.

$\therefore F(\lambda x, \lambda y)=\frac{-[\lambda x \cdot \sin ^{2}(\frac{\lambda x}{\lambda y})-\lambda y]}{\lambda x}=\frac{-[x \sin ^{2}(\frac{y}{x})-y]}{x}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve this differential equation, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x=\frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{-[x \sin ^{2} v-v x]}{x}$

$\Rightarrow v+x \frac{d v}{d x}=-[\sin ^{2} v-v]=v-\sin ^{2} v$

$\Rightarrow x \frac{d v}{d x}=-\sin ^{2} v$

$\Rightarrow \frac{d v}{\sin ^{2} v}=-\frac{d x}{d x}$

$\Rightarrow cosec^{2} v d v=-\frac{d x}{x}$

Integrating both sides, we get: $-\cot v=-\log |x|-C$

$\Rightarrow \cot v=\log |x|+C$

$\Rightarrow \cot (\frac{y}{x})=\log |x|+\log C$

$\Rightarrow \cot (\frac{y}{x})=\log |Cx|$

Now, $y=\frac{\pi}{4}$ at $x=1$

$\Rightarrow \cot (\frac{\pi}{4})=\log |C|$

$\Rightarrow 1=\log C$

$\Rightarrow C=e^{1}=e$

Substituting $C=e$ in equation (2), we get:

$\cot (\frac{y}{x})=\log |e x|$

This is the required solution of the given differential equation.

Solution

$\frac{d y}{d x}-\frac{y}{x}+cosec(\frac{y}{x})=0$

$\Rightarrow \frac{d y}{d x}=\frac{y}{x}-cosec(\frac{y}{x})$

Let $F(x, y)=\frac{y}{x}-cosec(\frac{y}{x})$.

$\therefore F(\lambda x, \lambda y)=\frac{\lambda y}{\lambda x}-cosec(\frac{\lambda y}{\lambda x})$

$\Rightarrow F(\lambda x, \lambda y)=\frac{y}{x}-cosec(\frac{y}{x})=F(x, y)=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=v-cosec v$

$\Rightarrow-\frac{d v}{cosec v}=-\frac{d x}{x}$

$\Rightarrow-\sin v d v=\frac{d x}{x}$

Integrating both sides, we get:

$\cos v=\log x+\log C=\log |C x|$

$\Rightarrow \cos (\frac{y}{x})=\log |C x|$

This is the required solution of the given differential equation.

Now, $y=0$ at $x=1$.

$\Rightarrow \cos (0)=\log C$

$\Rightarrow 1=\log C$

$\Rightarrow C=e^{1}=e$

Substituting $C=e$ in equation (2), we get:

$\cos (\frac{y}{x})=\log |(e x)|$

This is the required solution of the given differential equation.

Solution

$2 x y+y^{2}-2 x^{2} \frac{d y}{d x}=0$

$\Rightarrow 2 x^{2} \frac{d y}{d x}=2 x y+y^{2}$

$\Rightarrow \frac{d y}{d x}=\frac{2 x y+y^{2}}{2 x^{2}}$

Let $F(x, y)=\frac{2 x y+y^{2}}{2 x^{2}}$.

$\therefore F(\lambda x, \lambda y)=\frac{2(\lambda x)(\lambda y)+(\lambda y)^{2}}{2(\lambda x)^{2}}=\frac{2 x y+y^{2}}{2 x^{2}}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y=v x$

$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$

$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the value of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{2 x(v x)+(v x)^{2}}{2 x^{2}}$

$\Rightarrow v+x \frac{d v}{d x}=\frac{2 v+v^{2}}{2}$

$\Rightarrow v+x \frac{d v}{d x}=v+\frac{v^{2}}{2}$

$\Rightarrow \frac{2}{v^{2}} d v=\frac{d x}{x}$

Integrating both sides, we get:

$$ \begin{align*} & 2 \cdot \frac{v^{-2+1}}{-2+1}=\log |x|+C \\ & \Rightarrow-\frac{2}{v}=\log |x|+C \\ & \Rightarrow-\frac{2}{\frac{y}{x}}=\log |x|+C \\ & \Rightarrow-\frac{2 x}{y}=\log |x|+C \\ & \text{ Now, } y=2 \text{ at } x=1 . \tag{2}\\ & \Rightarrow-1=\log (1)+C \\ & \Rightarrow C=-1 \end{align*} $$

Substituting $C=-1$ in equation (2), we get:

$ \begin{aligned} & -\frac{2 x}{y}=\log |x|-1 \\ & \Rightarrow \frac{2 x}{y}=1-\log |x| \\ & \Rightarrow y=\frac{2 x}{1-\log |x|},(x \neq 0, x \neq e) \end{aligned} $

This is the required solution of the given differential equation.

Solution

For solving the homogeneous equation of the form $\frac{d x}{d y}=h(\frac{x}{y})$, we need to make the substitution as $x=v y$.

Hence, the correct answer is $C$.

Solution

Function $F(x, y)$ is said to be the homogenous function of degree $n$, if $F(\lambda x, \lambda y)=\lambda^{n} F(x, y)$ for any non-zero constant $(\lambda)$.

Consider the equation given in alternativeD:

$y^{2} d x+(x^{2}-x y-y^{2}) d y=0$

$\Rightarrow \frac{d y}{d x}=\frac{-y^{2}}{x^{2}-x y-y^{2}}=\frac{y^{2}}{y^{2}+x y-x^{2}}$

Let $F(x, y)=\frac{y^{2}}{y^{2}+x y-x^{2}}$.

$\Rightarrow F(\lambda x, \lambda y)=\frac{(\lambda y)^{2}}{(\lambda y)^{2}+(\lambda x)(\lambda y)-(\lambda x)^{2}}$

$=\frac{\lambda^{2} y^{2}}{\lambda^{2}(y^{2}+x y-x^{2})}$

$=\lambda^{0}(\frac{y^{2}}{y^{2}+x y-x^{2}})$

$=\lambda^{0} \cdot F(x, y)$

Hence, the differential equation given in alternative $\mathbf{D}$ is a homogenous equation.