What is the solution for (D^2-4D+4)y=e^x ?

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

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(D^2-4D+4)y=e^x

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} + 2 y = 12 e^{- 2 t}

3 y^{^{''}} + 12 y^{^{'}} + 39 y = 90 sin (3 t)

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

y^{^{''}} - 2 y^{^{'}} + 6 = 0

y^{^{''}} + 5 y^{^{'}} + 4 y = 20

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