Question

Solve: $\frac{dy}{dx}+\frac{y}{x}=\frac{y^2}{x^2}.$

• A. $\frac{1}{yx}=\frac{1}{{\left(2x\right)}^2}+c$
• B. $\frac{1}{y}=\frac{1}{{\left(4x\right)}^2}+c$
• C. $\frac{1}{yx}=\frac{1}{{\left(4x\right)}^2}+c$
• D. $\frac{1}{y}=\frac{1}{{\left(2x\right)}^2}+c$

Answer 1

The correct option is A $\frac{1}{yx}=\frac{1}{{\left(2x\right)}^2}+c$

$\frac{dy}{dx}+\frac{y}{x}=\frac{y^2}{x^2}$
$\Rightarrow \frac{dy}{dx}=\frac{y^2-xy}{x^2}$ ....(1)
which is a homogeneous differential eqn
Put $y=vx$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$
So, eqn (1) becomes
$v+x\frac{dv}{dx}=v^2-v$
$\Rightarrow x\frac{dv}{dx}=v^2-2v$
$\Rightarrow \left(\frac{1}{v^2-2v}\right)dv=\frac{1}{x}dx$
Integrating both sides,
$\int \left(\frac{1}{(v-1{)}^2-1}\right)dv=\log x+\log C$
$\Rightarrow \frac{1}{2}\log \left(\frac{v-2}{v}\right)=\log Cx$
$\Rightarrow \frac{v-2}{v}=C^2{x}^2$
$\Rightarrow 1-2\frac{x}{y}=c{x}^2$
$\Rightarrow \frac{1}{yx}=\frac{1}{{\left(2x\right)}^2}+c$