What is the solution for y^{''}-4y^'+4y=t^3*e^{2t} ?

The solution for y^{''}-4y^'+4y=t^3*e^{2t} is y=c_{1}e^{2t}+c_{2}te^{2t}+(e^{2t}t^5)/(20)

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y^{''}-4y^'+4y=t^3*e^{2t}

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{3} \cdot e^{2 t}

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

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = t^{3} \cdot e^{2 t}

Solve linear ODE:

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

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

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

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} + 6 x = 0

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

2 y^{^{''}} + y^{^{'}} - y = 1, y^{^{'}} (0) = 0

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

What is the solution for y^{''}-4y^'+4y=t^3*e^{2t} ?
The solution for y^{''}-4y^'+4y=t^3*e^{2t} is y=c_{1}e^{2t}+c_{2}te^{2t}+(e^{2t}t^5)/(20)