What is the solution for y^{''}+5y^'+4y=2x+e^{-2x} ?

The solution for y^{''}+5y^'+4y=2x+e^{-2x} is y=c_{1}e^{-x}+c_{2}e^{-4x}-5/8+(x-e^{-2x})/2

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

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

y = c_{1} e^{- x} + c_{2} e^{- 4 x} - \frac{5}{8} + \frac{x - e ^{- 2 x}}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 4 x} - \frac{5}{8} + \frac{x - e ^{- 2 x}}{2}

y = c_{1} e^{- x} + c_{2} e^{- 4 x} - \frac{5}{8} + \frac{x - e ^{- 2 x}}{2}

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 e^{5 x}, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} - 2 y^{^{'}} + 5 y = 20 sin (t)

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

y^{^{''}} + 4 y = 6 sin (2 t), y (0) = 2, y^{^{'}} (0) = 1

10 y^{^{''}} + 1000 y = 160 cos (8 t)

What is the solution for y^{''}+5y^'+4y=2x+e^{-2x} ?
The solution for y^{''}+5y^'+4y=2x+e^{-2x} is y=c_{1}e^{-x}+c_{2}e^{-4x}-5/8+(x-e^{-2x})/2