What is the solution for y^{''}+4y=20(e^x-cos(2x)) ?

The solution for y^{''}+4y=20(e^x-cos(2x)) is y=c_{1}cos(2x)+c_{2}sin(2x)+4e^x-5xsin(2x)

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y^{''}+4y=20(e^x-cos(2x))

y^{^{''}} + 4 y = 20 (e^{x} - cos (2 x))

y = c_{1} cos (2 x) + c_{2} sin (2 x) + 4 e^{x} - 5 x sin (2 x)

Solution steps

y^{''} (x) + 4 y = 20 (e^{x} - cos (2 x))

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + 4 e^{x} - 5 x sin (2 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + 4 e^{x} - 5 x sin (2 x)

y^{^{''}} - 10 y^{^{'}} + 25 y = (x + 1) e^{x}

y^{^{''}} - 6 y^{^{'}} + 8 y = x

y^{^{'''}} - 5 y^{^{''}} = 4 x^{2} - 3 x + 6 e^{5 x}

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

4 y^{^{''}} + 4 y^{^{'}} + y = e^{- \frac{x}{2}}

What is the solution for y^{''}+4y=20(e^x-cos(2x)) ?
The solution for y^{''}+4y=20(e^x-cos(2x)) is y=c_{1}cos(2x)+c_{2}sin(2x)+4e^x-5xsin(2x)