What is the solution for y^{''}+4y^'+4y=3e^{-2x} ?

The solution for y^{''}+4y^'+4y=3e^{-2x} is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(3e^{-2x}x^2)/2

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

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

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{3 e ^{- 2 x} x ^{2}}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{3 e ^{- 2 x} x ^{2}}{2}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{3 e ^{- 2 x} x ^{2}}{2}

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\frac{d}{d x} (e^{s e c (x)})

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x + 7, 0.5 (x + 7)

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What is the solution for y^{''}+4y^'+4y=3e^{-2x} ?
The solution for y^{''}+4y^'+4y=3e^{-2x} is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(3e^{-2x}x^2)/2