Question

Solve the differential equation: $\frac{dy}{dx}+\frac{x-2y}{2x-y}=0$

• A. $(x+y{)}^3=c(x+y)$
• B. $(x-y{)}^3=c(x+y)$
• C. $(x+y{)}^3=c(x-y)$
• D. $(x-y{)}^3=c(x-y)$

Answer 1

Answer: (C)

$(x+y{)}^3=c(x-y)$

$\frac{dy}{dx}+\frac{x-2y}{2x-y}=0$
Substitute $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\therefore x\frac{dv}{dx}+v=\frac{x-2xv}{2x-xv}\Rightarrow \frac{dv}{dx}=\frac{-v^2+4v+1}{x(v-2)}\Rightarrow \frac{v-2}{-v^2+4v+1}\frac{dv}{dx}=\frac{1}{x}$
Integrating both sides w.r.t $x$ we get
$\int \frac{v-2}{-v^2+4v+1}\frac{dv}{dx}dx=\int \frac{1}{x}dx$
$\Rightarrow -\frac{1}{2}\log (v^2-4v+1)=\log x+c\Rightarrow {(x+y)}^3=c(x-y)$