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

The correct option is B $x^2-xy-y^2=c$

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