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

$\mathrm dF=\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0$

and the equation is exact if the mixed second-order partial derivatives of $F$are equal, i.e. $\frac{\partial^2F}{\partial x\partial y}=\dfrac{\partial^2F}{\partial y\partial x}$.

The given ODE is exact, since

$\dfrac{\partial(2xy^3+\cos x)}{\partial y}=6xy^2$

$\dfrac{\partial(3x^2y^2-\sin y)}{\partial x}=6xy^2$

Then

$\dfrac{\partial F}{\partial x}=2xy^3+\cos x\implies F(x,y)=x^2y^3+\sin x+f(y)$

$\dfrac{\partial F}{\partial y}=3x^2y^2-\sin y=3x^2y^2+\dfrac{\mathrm df}{\mathrm dy}$

$\dfrac{\mathrm df}{\mathrm dy}=-\sin y\implies f(y)=\cos y+C$

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

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

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

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