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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 9 y = (x^{2} - 9) sin (3 x)

y^{^{''}} - 36 y = 8, y (0) = 1, y^{^{'}} (0) = 0

y^{^{'''}} - 36 y^{^{'}} = t e^{- 6 t} + 7 cos (6 t)

2 y^{^{''}} + 3 y^{^{'}} + y = g (t)

y^{^{''}} + y^{^{'}} = - 4 x + 2

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