In each of the following differential equation show that it is homogeneous and solve it. $(x-y) \frac{d y}{d x}=x+3 y$

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In each of the following differential equation show that it is homogeneous and solve it. $(x-y) \frac{d y}{d x}=x+3 y$

Solution:

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}+3 \mathrm{y}}{\mathrm{x}-\mathrm{y}}$
$\begin{array}{l} \Rightarrow \frac{d y}{d x}=\frac{1+3 \frac{y}{x}}{1-\frac{y}{x}} \\ \Rightarrow \frac{d y}{d x}=f\left(\frac{y}{x}\right) \end{array}$
$\Rightarrow$ the given differential eq. is a homogenous eq.
Solution of the given differential eq. is:
Putting $y=v x$
$\begin{array}{l} \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\ v+x \frac{d v}{d x}=\frac{1+3 \frac{v x}{x}}{1-\frac{v x}{x}} \\ \Rightarrow x \frac{d v}{d x}=\frac{1+3 v}{1-v}-v=\frac{1+3 v-v+v^{2}}{1-v}=\frac{1+2 v+v^{2}}{1-v} \\ \Rightarrow \frac{1-v}{1+2 v+v^{2}} d v=\frac{d x}{x} \end{array}$
On integrating both sides we obtain:
$\begin{array}{l} \Rightarrow \int \frac{1-v}{1+2 v+v^{2}} d v=\int \frac{d x}{x}+c \\ \Rightarrow \int \frac{v-1}{1+2 v+v^{2}} d v=-\int \frac{d x}{x}+c \\ \Rightarrow \frac{\ln \left|1+2 v+v^{2}\right|}{2}=-\ln |x|+\ln c \end{array}$
By resubstituting the value of $y=v x$ we have
$\begin{array}{l} \Rightarrow \frac{\ln \left|1+2 \frac{y}{x}+\left(\frac{y}{x}\right)^{2}\right|}{2}=-\ln |x|+\ln c \\ \Rightarrow \log |x+y|+\frac{2 x}{(x+y)}=C \end{array}$