Question

The general solution of $\frac{dy}{dx}=\frac{2x-y}{x+2y}$ is

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

Answer 1

Answer: (B)

Given equation is $\frac{dy}{dx}=\frac{2x-y}{x+2y}$ ..(i) Put $y=vx\Rightarrow \frac{dy}{dx}=v+\frac{xdv}{dx}$
$\therefore$ $v+x\frac{dv}{dx}=\frac{2-v}{1+2v}$
$\Rightarrow$ $\frac{xdv}{dx}=\frac{2-v-v(1+2v)}{1+2v}$
$\Rightarrow$ $\int \frac{1+2v}{2(1-v-v^2)}dv=\int \frac{1}{x}dx$
$\Rightarrow$ $\log k-\frac{1}{2}\log (1-v-v^2)=\log x$
$\Rightarrow$ $2\log k-\log (1-v-v^2)=2\log x$
$\Rightarrow$ $\log c=\log [x^2(1-v-v^2)]$
$\Rightarrow$ $c=x^2\left(1-\frac{y}{x}-\frac{y^2}{x^2}\right)$
$\Rightarrow$ $x^2-xy-y^2=c$