What is the solution for y^{''}-4y^'+4y=x^3e^{2x} ?

The solution for y^{''}-4y^'+4y=x^3e^{2x} is y=c_{1}e^{2x}+c_{2}xe^{2x}+(e^{2x}x^5)/(20)

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y^{''}-4y^'+4y=x^3e^{2x}

y^{^{''}} - 4 y^{^{'}} + 4 y = x^{3} e^{2 x}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{e ^{2 x} x ^{5}}{20}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{e ^{2 x} x ^{5}}{20}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{e ^{2 x} x ^{5}}{20}

y^{^{''}} + 4 y = - 12 sin (2 x), y (0) = 1.8, y^{^{'}} (0) = 5

y^{^{''}} - 11 y^{^{'}} + 24 y = 13 + 24 x

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + y^{^{'}} - 2 y = - 260 sin (3 t)

y^{^{''}} - 9 y = 81 x^{3}

What is the solution for y^{''}-4y^'+4y=x^3e^{2x} ?
The solution for y^{''}-4y^'+4y=x^3e^{2x} is y=c_{1}e^{2x}+c_{2}xe^{2x}+(e^{2x}x^5)/(20)