What is the solution for y^{''}+4y^'=e^xcos(2x) ?

The solution for y^{''}+4y^'=e^xcos(2x) is y=c_{1}+c_{2}e^{-4x}+12/145 e^xsin(2x)+1/145 e^xcos(2x)

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y^{''}+4y^'=e^xcos(2x)

y^{^{''}} + 4 y^{^{'}} = e^{x} cos (2 x)

y = c_{1} + c_{2} e^{- 4 x} + \frac{12}{145} e^{x} sin (2 x) + \frac{1}{145} e^{x} cos (2 x)

Solution steps

y^{''} (x) + 4 y^{'} (x) = e^{x} cos (2 x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 x} + \frac{12}{145} e^{x} sin (2 x) + \frac{1}{145} e^{x} cos (2 x)

y = c_{1} + c_{2} e^{- 4 x} + \frac{12}{145} e^{x} sin (2 x) + \frac{1}{145} e^{x} cos (2 x)

What is the solution for y^{''}+4y^'=e^xcos(2x) ?
The solution for y^{''}+4y^'=e^xcos(2x) is y=c_{1}+c_{2}e^{-4x}+12/145 e^xsin(2x)+1/145 e^xcos(2x)