Question

Solve:
$(x-y{)}^2\frac{dy}{dx}=a^2$

Answer 1

Given that:

${(x-y)}^2\frac{dy}{dx}=a^2$

$Letx-y=v$

$1-\frac{dy}{dx}=\frac{dv}{dx}$

$1-\frac{dv}{dx}=\frac{dy}{dx}$

$Then$

$v^2(1-\frac{dv}{dx})=a^2$

$(v^2-v^2\frac{dv}{dx})=a^2$

$\frac{dv}{dx}=\frac{v^2-a^2}{v^2}$

$\frac{dx}{dv}=\frac{v^2}{v^2-a^2}$

$dx=(\frac{v^2-a^2}{v^2-a^2}+\frac{a^2}{v^2-a^2})dv$

$dx=(1+\frac{a^2}{v^2-a^2})dv$

$On\mathrm{int}egratingbothside,weget$

$x=v+a^2\times \frac{1}{2a}\log \left(\frac{v-a}{v+a}\right)+c$

$x=x-y+a^2\times \frac{1}{2a}\log \left(\frac{x-y-a}{x-y+a}\right)+c$

$y=\frac{a}{2}\log \left(\frac{x-y-a}{x-y+a}\right)+c.$

This is the required solution.