Question

The solution of differential equation $\frac{dy}{dx}=\frac{x-y}{x+y}$ is

• A. $x^2-y^2+2xy+c=0$
• B. $x^2-y^2-xy+c=0$
• C. $x^2-y^2+xy+c=0$
• D. $x^2-y^2-2xy+c=0$

Answer 1

Answer: (D)

$\frac{dy}{dx}=\frac{x-y}{x+y}=\frac{1-y/x}{1+y/x}\dots \left(i\right)$
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
$\int \frac{1}{x}dx=\frac{-1}{2}\int \frac{-2-2v}{1-2v-v^2}dv$
$\Rightarrow log{c}_1-2logx=log\left(1-2v-v^2\right)$
$\Rightarrow log\frac{c_1}{x^2}=log\left(\frac{x^2-2yx-y^2}{x^2}\right)$
$\Rightarrow x^2-2yx-y^2=c_1$
$\Rightarrow x^2-2yx-y^2+c=0\left(c=-c_1\right)$