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.

The ODE[tex]M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0[/tex]is exact if[tex]\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}[/tex]We have[tex]M=-xy\sin x+2y\cos x\implies M_y=-x\sin x+2\cos x[/tex][tex]N=2x\cos x\implies N_x=2\cos x-2x\sin x[/tex]so the ODE is indeed not exact.Multiplying both sides of the ODE by [tex]\mu(x,y)=xy[/tex] gives[tex]\mu M=-x^2y^2\sin x+2xy^2\cos x\implies(\mu M)_y=-2x^2y\sin x+4xy\cos x[/tex][tex]\mu N=2x^2y\cos x\implies(\mu N)_x=4xy\cos x-2x^2y\sin x[/tex]so that [tex](\mu M)_y=(\mu N)_x[/tex], and the modified ODE is exact.We're looking for a solution of the form[tex]\Psi(x,y)=C[/tex]so that by differentiation, we should have[tex]\Psi_x\,\mathrm dx+\Psi_y\,\mathrm dy=0[/tex][tex]\implies\begin{cases}\Psi_x=\mu M\\\Psi_y=\mu N\end{cases}[/tex]Integrating both sides of the second equation with respect to [tex]y[/tex] gives[tex]\Psi_y=2x^2y\cos x\implies\Psi=x^2y^2\cos x+f(x)[/tex]Differentiating both sides with respect to [tex]x[/tex] gives[tex]\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}[/tex][tex]\implies\dfrac{\mathrm df}{\mathrm dx}=0\implies f(x)=c[/tex]for some constant [tex]c[/tex].So the general solution to this ODE is[tex]x^2y^2\cos x+c=C[/tex]or simply[tex]x^2y^2\cos x=C[/tex]