Question:
Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.
$$(\frac{sin y}{y} - 2e^{-x}sinx) + (\frac{cosy + 2e^{-x}cosx}{y})y' = 0, \ u = ye^x$$
Solution:
Check $$M_y = N_x$$
$$M_y = \frac{y cos y - sin y}{y^2} $$
$$N_x = \frac{2}{y}(-e^{-x}cos x -e^{-x}sinx)$$
$$M_y \neq N_x$$
So the equation is not exact.
Multiply both sides by u.
$$(e^x sin y - 2ysinx)+(cos ye^x +2 cosx)\frac{dy}{dx} = 0$$
$$M_y = e^xcosy - 2sinx$$
$$Nx = e^xcosy - 2sinx$$
$$M_y = N_x$$
So the equation is now exact.
So there exist a function $$\psi (x,y)$$ such that $$\psi_x = M, \psi_y = N.$$
$$\psi = \int_{}{}M dx = \int (e^x sin y - 2ysinx) dx = e^x siny +2 ycos x +h(y)$$
$$\psi_y = e^x cos y + 2 cos x +h'(y) = N = e^xcosy +2 cosx$$
$$So h'(y) = 0$$
$$H(y) = 0$$
$$\psi = e^x siny +2 ycos x$$
The solution is
$$ e^x siny +2 ycos x = c$$
