Question

Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.

Answer 1

`\frac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-(2y)/((x+y^2)^2)`

`(\partial\left(4x^2y^3+(2y)/(x+y^2)\right))/(\partial x)=8xy^3-(2y)/((x+y^2)^2)`

so the ODE is indeed exact and there is a solution of the form
`F(x,y)=C`. We have

`(\partial F)/(\partial x)=2xy^4+\frac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)`

`(\partial F)/(\partial y)=4x^2y^3+(2y)/(x+y^2)=4x^2y^3+(2y)/(x+y^2)+f'(y)`

`f'(y)=0\implies f(y)=C`

`\implies F(x,y)=x^2y^3+\ln(x+y^2)=C`

With
`y(1)=2`, we have

`8+\ln9=C`

so

`\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}`