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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} - 2 y = 2 + 5 \cdot cos (x)

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

y^{^{''}} - 3 y^{^{'}} + 2 y = 2 x

x^{^{''}} - 4 x^{^{'}} + 5 x = cos (t)

y^{^{''}} + 3 y^{^{'}} + 2 y = 4 e^{2 x}, y (0) = - 3, y^{^{'}} (0) = 5

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