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Q.
The solution of differential equation $\frac{dy}{dx}=\frac{x-y}{x+y}$ is
Differential Equations
Solution:
$\frac{dy}{dx}=\frac{x-y}{x+y}=\frac{1-y/x}{1+y/x}\quad\ldots\left(i\right)$
Since, it is a homogeneous differential equation,
Put $y=vx$
$\Rightarrow \frac{dy}{dx}=v+x \frac{dv}{dx}$
Hence, eq. $(i)$ becomes
$v+x \frac{dv}{dx}=\frac{1-v}{1+v}$
$\Rightarrow x \frac{dv}{dx}=\frac{1-v}{1+v}-v$
$\Rightarrow x \frac{dv}{dx}=\frac{1-2v-v^{2}}{1+v}$
$\Rightarrow \frac{dx}{x}=\frac{-1}{2}\left(\frac{-2-2v}{1-2v-v^{2}}\right)dv$
On integration, we get
$\int \frac{1}{x}dx=\frac{-1}{2} \int \frac{-2-2v}{1-2v-v^{2}}dv$
$\Rightarrow logc_{1}-2log\,x=log \left(1-2v-v^{2}\right)$
$\Rightarrow log \frac{c_{1}}{x^{2}}=log\left(\frac{x^{2}-2yx-y^{2}}{x^{2}}\right)$
$\Rightarrow x^{2}-2yx-y^{2}=c_{1}$
$\Rightarrow x^{2}-2yx-y^{2}+c=0\,\left(c=-c_{1}\right)$