What is the solution for y^{''}+4y^'+4y=x^2*e^{-2x}+1 ?

The solution for y^{''}+4y^'+4y=x^2*e^{-2x}+1 is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(e^{-2x}x^4)/(12)+1/4

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

y^{^{''}} + 4 y^{^{'}} + 4 y = x^{2} \cdot e^{- 2 x} + 1

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{- 2 x} x ^{4}}{12} + \frac{1}{4}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = x^{2} \cdot e^{- 2 x} + 1

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{- 2 x} x ^{4}}{12} + \frac{1}{4}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{- 2 x} x ^{4}}{12} + \frac{1}{4}

y^{^{''}} + 7 y = 5

x^{^{''}} + x^{^{'}} - 2 x = 8 sin (2 t)

y^{^{''}} - 5 y^{^{'}} + 6 y = (12 x - 7) e^{- x}

y^{^{''}} + 25 y = 20 sin (5 x) + x cos (x)

2 y^{^{''}} - 13 y^{^{'}} + 20 y = 40 x^{2} + 8 x + 10

What is the solution for y^{''}+4y^'+4y=x^2*e^{-2x}+1 ?
The solution for y^{''}+4y^'+4y=x^2*e^{-2x}+1 is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(e^{-2x}x^4)/(12)+1/4