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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(d^{2} + 3 d + 2) y = e^{e^{x}}

(d^{2} - 6 d + 9) y = 2 e^{3} x

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

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 e^{x}

(d^{2} - 1) \cdot y = x \cdot sin (x) + x^{2} \cdot e^{x}

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