Question

What is a solution to the differential equation `y'=xe^(x^2-lny^2)`?

Answer 1

`y =root3 (3/2e^(x^2) + C)`

Explanation:

`y'=xe^(x^2-lny^2)`

we can separate it out first

`y'=xe^(x^2)e^(-lny^2)`

`y'=xe^(x^2)1/e^(lny^2)`

`y'=xe^(x^2)1/y^2`

`y^2 y'=xe^(x^2)`

`int y^2 y' dx=int xe^(x^2) dx`

`int y^2 dy=int xe^(x^2) dx`

.....spotting the pattern on the RHS

`y^3/3 =int d/dx (1/2e^(x^2) ) dx`

`y^3/3 =1/2e^(x^2) + C`

`y^3 =3/2e^(x^2) + C`

`y =root3 (3/2e^(x^2) + C)`