Question

Differential Equations ( x + y ) y' = x - y

Answer 1

Explanation:

`y'=(x-y)/(x+y) \\put ~y=vx\\diff.\\y'=v+xv'\\v+xv'=(x-vx)/(x+vx) \\v+xv'=(1-v)/(1+v) \\xv'=(1-v)/(1+v) -v\\xv'=(1-v-v-v^2)/(1+v) =(1-2v-v^2)/(1+v) \\separating ~the~variables~and~integrating\\\int (1+v)/(1-2v-v^2) dv=\int x~dx+c\\-(1)/(2) \int (-2-2v)/(1-2v-v^2) dv=(x^2)/(2) +c\\-(1)/(2) ln (1-2v-v^2)=(x^2)/(2) +c\\ln(1-2v-v^2)=-x^2-2c\\1-2v-v^2=e^(-x^2-2c) =e^(-2c)e^(x^2)=Ce^x^2\\put~v=(y)/(x) \\`

`1-2(y)/(x) -(y^2)/(x^2) =Ce^(-x^2)\\x^2-2xy-y^2=Cx^2e^(-x^2)`