Question

The solution of differential equation $(x^2-xy)dy=(xy+y^2)dx,$ is
(where $c$ is integration constant)

• A. $|xy|=|c|e^{\frac{-x}{y}}$
• B. $|xy|=c{e}^{\frac{-y}{x}}$
• C. $y^2|x|=c{e}^{\frac{1}{|x|}}$
• D. $y|x|=c{e}^{\frac{1}{|x|}}$

Answer 1

The correct option is A $|xy|=|c|e^{\frac{-x}{y}}$

$(x^2-xy)dy=(xy+y^2)dx\Rightarrow \frac{dy}{dx}=\frac{xy+y^2}{x^2-xy}\cdots \left(i\right)$
This is a homogeneous D.E

We put $y=vx$ so that $\frac{dy}{dx}=v+x\frac{dv}{dx}.$

Then, equation $\left(i\right)$ becomes:
$v+x\frac{dv}{dx}=\frac{v{x}^2+v^2{x}^2}{x^2-v{x}^2}\Rightarrow x\frac{dv}{dx}=\frac{v+v^2}{1-v}-v$
$\Rightarrow \int \frac{1-v}{2v^2}dv=\int \frac{dx}{x}\Rightarrow \frac{1}{2}\int \frac{dv}{v^2}-\frac{1}{2}\int \frac{dv}{v}=\int \frac{dx}{x}$
$\Rightarrow -\frac{1}{2v}-\frac{1}{2}\ln |v|=\ln |x|+\ln |k|\Rightarrow -\frac{1}{v}=\ln |v|+2\ln |x|+2\ln |k|\Rightarrow \frac{-1}{v}=\ln \left|v{x}^2{k}^2\right|$
$\Rightarrow -\frac{x}{y}=\ln |\frac{y}{x}\cdot x^2\cdot k^2|\Rightarrow |xy|k^2=e^{\frac{-x}{y}}$
$\Rightarrow |xy|=|c|e^{\frac{-x}{y}}$
where $|c|=\frac{1}{k^2}=$constant.