\[ y'(x)+y(x) \cos (x)-e^{2 x}=0 \] ✓ Mathematica : cpu = 0.401098 (sec), leaf count = 39
\[\left \{\left \{y(x)\to e^{-\sin (x)} \int _1^xe^{2 K[1]+\sin (K[1])}dK[1]+c_1 e^{-\sin (x)}\right \}\right \}\] ✓ Maple : cpu = 0.13 (sec), leaf count = 21
\[y \relax (x ) = \left (\int {\mathrm e}^{2 x +\sin \relax (x )}d x +c_{1}\right ) {\mathrm e}^{-\sin \relax (x )}\]
Hand solution
\begin {equation} \frac {dy}{dx}+y\relax (x) \cos \relax (x) =e^{2x}\tag {1} \end {equation}
Integrating factor \(\mu =e^{\int \cos \relax (x) dx}=e^{\sin \left ( x\right ) }\). Hence (1) becomes
\[ \frac {d}{dx}\left (e^{\sin \relax (x) }y\relax (x) \right ) =e^{\sin \relax (x) }e^{2x}\]
Integrating both sides
\begin {align*} e^{\sin \relax (x) }y\relax (x) & =\int e^{\sin \left ( x\right ) }e^{2x}+C\\ y\relax (x) & =e^{-\sin \relax (x) }\int e^{2x+\sin \left ( x\right ) }+Ce^{-\sin \relax (x) } \end {align*}