Question

Solve the differential equation $(x-y)dy-(x+y)dx=0$

Answer 1

$(x-y)dy-(x+y)dx=0$

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

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

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

Let, $\frac{y}{x}=V,y=v-x$

$\frac{dy}{dx}=v+x.\frac{dv}{dx}$

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

$\therefore x.\frac{dv}{dx}=\frac{1+v}{1-v}-v=\frac{1+v-v+v^2}{1-v}=\frac{1+v^2}{1-v}$

$\therefore \frac{1-v}{1+v^2}dv=\frac{1}{x}.dx$

Integrating both side

$\therefore \int \frac{1}{1+v^2}dv-\frac{1}{2}\int \frac{2v}{1+v^2}dv=\int \frac{1}{x}dx$

$\therefore {\tan }^{-1}v-\frac{1}{2}\log |1+v^2|=\log x+\log c$

$\therefore {\tan }^{-1}\frac{y}{x}-\frac{1}{2}\log |1+\frac{y^2}{x^2}|=\log x+\log c$

$\therefore {\tan }^{-1}\frac{y}{x}=\log (x.c)+\frac{1}{2}\log |1+\frac{y^2}{x^2}|$

$\therefore {\tan }^{-1}\frac{y}{x}=\log (x.c.(1+\frac{y^2}{x^2}{)}^{\frac{1}{2}})$

$\therefore e^{\tan \frac{y}{x}}=\frac{x.c.(x^2+y^2{)}^{\frac{1}{2}}}{x}=(x62+y^2{)}^{\frac{1}{2}}.c$

$\therefore e^{\tan \frac{y}{x}}=(x^2+y^2{)}^{\frac{1}{2}}.c$