What is the solution for y^{''}+3y+2y=cos(e^x) ?

The solution for y^{''}+3y+2y=cos(e^x) is

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y^{''}+3y+2y=cos(e^x)

y^{^{''}} + 3 y + 2 y = cos (e^{x})

y = c_{1} cos (5 x) + c_{2} sin (5 x) + cos (5 x) \cdot \int - \frac{5 sin ( 5 x ) cos ( e ^{x} )}{5} d x + sin (5 x) \cdot \int \frac{5 cos ( 5 x ) cos ( e ^{x} )}{5} d x

Solution steps

y^{''} (x) + 3 y + 2 y = cos (e^{x})

Solve linear ODE:

y = c_{1} cos (5 x) + c_{2} sin (5 x) + cos (5 x) \cdot \int - \frac{5 sin ( 5 x ) cos ( e ^{x} )}{5} d x + sin (5 x) \cdot \int \frac{5 cos ( 5 x ) cos ( e ^{x} )}{5} d x

y = c_{1} cos (5 x) + c_{2} sin (5 x) + cos (5 x) \cdot \int - \frac{5 sin ( 5 x ) cos ( e ^{x} )}{5} d x + sin (5 x) \cdot \int \frac{5 cos ( 5 x ) cos ( e ^{x} )}{5} d x

y^{^{''}} + 15 y^{^{'}} + 7 y = sin (x) + cos (x)

(y^{^{''}}) - 2 y^{^{'}} + y = \frac{e ^{x}}{1 + x ^{2}}

y^{^{''}} + 2 y^{^{'}} + 2 y = t^{2}

y^{^{''}} - 2 y^{^{'}} + y = e - 2 x

y^{^{''}} - 3 y^{^{'}} + 5 y = 2 e^{t}

What is the solution for y^{''}+3y+2y=cos(e^x) ?
The solution for y^{''}+3y+2y=cos(e^x) is