Question

Show that the equation is exact and find an implicit solution. y cos(xy) + 3x^2 + [x cos(xy) + 2y]y' = 0

Answer 1

We have

`(\partial(y\cos(xy)+3x^2))/(\partial y)=\cos(xy)-xy\sin(xy)`

`(\partial(x\cos(xy)+2y))/(\partial x)=\cos(xy)-xy\sin(xy)`

so the ODE is indeed exact. Then there's a solution of the form
`f(x,y)=C` such that

`(\partial f)/(\partial x)=y\cos(xy)+3x^2`

`\implies f(x,y)=\sin(xy)+x^3+g(y)`

Differentiating wrt
`y` gives

`(\partial f)/(\partial y)=x\cos(xy)+2y=x\cos(xy)+g'(y)`

`\implies g'(y)=2y\implies g(y)=y^2+C`

Then the solution to the ODE is

`f(x,y)=\boxed{\sin(xy)+x^3+y^2=C}`