Let
\(\displaystyle M(x,y) = e^{2y} - y\cos xy\)
\(\displaystyle N(x,y) = 2xe^{2y} - x\cos xy + 2y\)
\(\displaystyle \frac{\partial M}{\partial y} = 2e^{2y} - \cos xy + xy \sin xy\)
\(\displaystyle \frac{\partial N}{\partial x} = 2e^{2y} - \cos xy + xy\sin xy\)
Since \(\displaystyle \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), there is a function \(\displaystyle f(x,y) = c\) such that:
\(\displaystyle \frac{\partial f}{\partial x} = e^{2y} - y\cos xy\)
and
\(\displaystyle \frac{\partial f}{\partial y} = 2xe^{2y} - x\cos xy + 2y\)
\(\displaystyle f(x,y) = \int (e^{2y} - y\cos xy) \ dx = xe^{2y} - \sin xy + g(y)\)
\(\displaystyle \frac{\partial f}{\partial y} = 2xe^{2y} - x\cos xy + g'(y)\)
\(\displaystyle 2xe^{2y} - x\cos xy + 2y = 2xe^{2y} - x\cos xy + g'(y)\)
This gives:
\(\displaystyle g'(y) = 2y\)
\(\displaystyle g(y) = y^2\)
\(\displaystyle f(x,y) = \int (e^{2y} - y\cos xy) \ dx = xe^{2y} - \sin xy + y^2\)
The \(\displaystyle \textcolor{indigo}{\bold{solution}}\) to the differential equation is:
\(\displaystyle \textcolor{blue}{xe^{2y(x)} - \sin xy(x) + y^2(x) = c}\)
