What is the solution for (D^2-4D+4)y=x^3 ?

The solution for (D^2-4D+4)y=x^3 is y=c_{1}e^{2x}+c_{2}xe^{2x}+(x^3)/4+(3x^2)/4+(9x)/8+3/4

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(D^2-4D+4)y=x^3

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

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{x ^{3}}{4} + \frac{3 x ^{2}}{4} + \frac{9 x}{8} + \frac{3}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{x ^{3}}{4} + \frac{3 x ^{2}}{4} + \frac{9 x}{8} + \frac{3}{4}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{x ^{3}}{4} + \frac{3 x ^{2}}{4} + \frac{9 x}{8} + \frac{3}{4}

y^{^{''}} - y^{^{'}} - 2 y = sin (2 x), y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} + y = cos (4 x)

y^{^{''}} - 8 y^{^{'}} + 16 y = \frac{e ^{- 4 x}}{1 + x ^{2}}

y^{^{''''}} - 4 y^{^{''}} = t^{2} - e^{t}

y^{^{''}} - 2 y^{^{'}} + y = \frac{9 e ^{t}}{1 + t ^{2}}

What is the solution for (D^2-4D+4)y=x^3 ?
The solution for (D^2-4D+4)y=x^3 is y=c_{1}e^{2x}+c_{2}xe^{2x}+(x^3)/4+(3x^2)/4+(9x)/8+3/4