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

The solution for y^{''}-4y=e^xcos(x) is y=c_{1}e^{2x}+c_{2}e^{-2x}+1/10 e^xsin(x)-1/5 e^xcos(x)

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

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

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{1}{10} e^{x} sin (x) - \frac{1}{5} e^{x} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{1}{10} e^{x} sin (x) - \frac{1}{5} e^{x} cos (x)

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{1}{10} e^{x} sin (x) - \frac{1}{5} e^{x} cos (x)

y^{^{''}} + 4 y^{^{'}} - 2 y = 8 sin (2 x)

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 t + 12 e^{- t}

y^{^{''}} - 64 y = cosh (8 x)

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + 6 y = x e^{2 x}

y^{^{''}} - 2 y^{^{'}} + y = \frac{5 e ^{t}}{1 + t ^{2}}

What is the solution for y^{''}-4y=e^xcos(x) ?
The solution for y^{''}-4y=e^xcos(x) is y=c_{1}e^{2x}+c_{2}e^{-2x}+1/10 e^xsin(x)-1/5 e^xcos(x)