Question

Solve the differential equation $y^2+x^2\frac{dy}{dx}=xy\frac{dy}{dx}$.

• A. $logxy=2-\frac{2y}{x}$
• B. $y=K{e}^{\frac{y}{x}}$
• C. $xy=K{e}^{2+\frac{2y}{x}}$
• D. $logKxy=2+\frac{2y}{x}$

Answer 1

Answer: (B)

$y=K{e}^{\frac{y}{x}}$

$y^2+x^2\frac{dy}{dx}=xy\frac{dy}{dx}$
Substitute $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\therefore x^2(x\frac{dv}{dx}+v)+x^2{v}^2=x^2(x\frac{dv}{dx}+v)v$
$\Rightarrow \frac{dv}{dx}=\frac{vx}{x(v-1)}\Rightarrow \frac{v-1}{v}\frac{dv}{dx}=\frac{1}{x}$
Integrating both sides w.r.t $x$ we get
$\int \frac{v-1}{v}\frac{dv}{dx}dx=\int \frac{1}{x}dx$
$\Rightarrow -\log v+v=\log x+logc\Rightarrow \log cy=\frac{y}{x}\Rightarrow cy=e^{\frac{y}{x}}\to y=K{e}^{\frac{y}{x}}$