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

The solution for y^{''}-y^'-2y=e^xcos(x) is y=c_{1}e^{2x}+c_{2}e^{-x}+1/10 e^xsin(x)-3/10 e^xcos(x)

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

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

y = c_{1} e^{2 x} + c_{2} e^{- x} + \frac{1}{10} e^{x} sin (x) - \frac{3}{10} e^{x} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} + \frac{1}{10} e^{x} sin (x) - \frac{3}{10} e^{x} cos (x)

y = c_{1} e^{2 x} + c_{2} e^{- x} + \frac{1}{10} e^{x} sin (x) - \frac{3}{10} e^{x} cos (x)

What is the solution for y^{''}-y^'-2y=e^xcos(x) ?
The solution for y^{''}-y^'-2y=e^xcos(x) is y=c_{1}e^{2x}+c_{2}e^{-x}+1/10 e^xsin(x)-3/10 e^xcos(x)