Question

Solve that differential equation $(x-y)dy=(x+y+1)dx$.

Answer 1

$(x-y)dy=(x-y+1)dx$

$\frac{dy}{dx}=\frac{x-y+1}{x-y}$

Let

$x-y=v$

Differentiating above equation w.r.t. $x$, we have

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

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

Therefore,

$1-\frac{dv}{dx}=\frac{v+1}{v}$

$\Rightarrow \frac{dv}{dx}=\frac{v-v-1}{v}$

$\Rightarrow \frac{dv}{dx}=-\frac{1}{v}$

$\Rightarrow vdv=-dx$

Integrating both sides, we have

$\int vdv=-\int dx$

$\frac{v^2}{2}=-x$

$v^2=-2x$

Substituting the value of $v$ in above equation, we have

${(x-y)}^2+2x=0$

Hence the differential equation is ${(x-y)}^2+2x=0$.