Question

Verify that the given differential equation is not exact. (−xy sin(x) + 2y cos(x)) dx + 2x cos(x) dy = 0 If the given DE is written in the form M(x, y) dx + N(x, y) dy = 0, one has My = Nx = . Since My and Nx equal, the equation is not exact. Multiply the given differential equation by the integrating factor μ(x, y) = xy and verify that the new equation is exact. If the new DE is written in the form M(x, y) dx + N(x, y) dy = 0, one has My = Nx = . Since My and Nx equal, the equation is exact. Solve.

Answer 1

The ODE

$M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0$

is exact if

$\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}$

We have

$M=-xy\sin x+2y\cos x\implies M_y=-x\sin x+2\cos x$

$N=2x\cos x\implies N_x=2\cos x-2x\sin x$

so the ODE is indeed not exact.

Multiplying both sides of the ODE by $\mu(x,y)=xy$ gives

$\mu M=-x^2y^2\sin x+2xy^2\cos x\implies(\mu M)_y=-2x^2y\sin x+4xy\cos x$

$\mu N=2x^2y\cos x\implies(\mu N)_x=4xy\cos x-2x^2y\sin x$

so that $(\mu M)_y=(\mu N)_x$, and the modified ODE is exact.

We're looking for a solution of the form

$\Psi(x,y)=C$

so that by differentiation, we should have

$\Psi_x\,\mathrm dx+\Psi_y\,\mathrm dy=0$

$\implies\begin{cases}\Psi_x=\mu M\\\Psi_y=\mu N\end{cases}$

Integrating both sides of the second equation with respect to $y$ gives

$\Psi_y=2x^2y\cos x\implies\Psi=x^2y^2\cos x+f(x)$

Differentiating both sides with respect to $x$ gives

$\Psi_x=-x^2y^2\sin x+2xy^2\cos x=2xy^2\cos x-x^2y^2\sin x+\dfrac{\mathrm df}{\mathrm dx}$

$\implies\dfrac{\mathrm df}{\mathrm dx}=0\implies f(x)=c$

for some constant $c$.

So the general solution to this ODE is

$x^2y^2\cos x+c=C$

or simply

$x^2y^2\cos x=C$