What is the solution for y^{''}+4y^'+4y=8e^{-2x} ?

The solution for y^{''}+4y^'+4y=8e^{-2x} is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+4x^2e^{-2x}

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y^{''}+4y^'+4y=8e^{-2x}

y^{^{''}} + 4 y^{^{'}} + 4 y = 8 e^{- 2 x}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + 4 x^{2} e^{- 2 x}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = 8 e^{- 2 x}

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + 4 x^{2} e^{- 2 x}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + 4 x^{2} e^{- 2 x}

y^{^{''}} - 8 y^{^{'}} + 16 y = e^{4 x} ln (9 + x^{2})

\frac{d ^{2} y}{d x ^{2}} - 8 \frac{d y}{d x} + 16 y = e^{4 x}

y^{^{''}} + y = 5 x sin (x)

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

y^{^{''}} - y^{^{'}} - 2 y = cosh (2 t)

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