Question

Show that (2xy3 + cos x)dx + (3x2 y2 − sin y)dy = 0 is exact, and find the solution. Find c if y(0) = π

Answer 1

We're looking for a solution of the form
`F(x,y)=C`. By the chain rule, this solution should have total differential

`\mathrm dF=(\partial F)/(\partial x)\,\mathrm dx+(\partial F)/(\partial y)\,\mathrm dy=0`

and the equation is exact if the mixed second-order partial derivatives of
`F`are equal, i.e.
`(\partial^2F)/(\partial x\partial y)=(\partial^2F)/(\partial y\partial x)`.

The given ODE is exact, since

`(\partial(2xy^3+\cos x))/(\partial y)=6xy^2`

`(\partial(3x^2y^2-\sin y))/(\partial x)=6xy^2`

Then

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

`(\partial F)/(\partial y)=3x^2y^2-\sin y=3x^2y^2+(\mathrm df)/(\mathrm dy)`

`(\mathrm df)/(\mathrm dy)=-\sin y\implies f(y)=\cos y+C`

`\implies x^2y^3+\sin x+\cos y=C`

With
`y(0)=\pi`, we get

`\cos\pi=C\implies C=-1`

`\implies\boxed{x^2y^3+\sin x+\cos y=-1}`