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

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

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

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

y^{^{''}} - 3 y^{^{'}} + 2 y = cos (x), y (0) = 0, y^{^{'}} (0) = 2

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

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

y^{^{''}} - 5 y^{^{'}} = 2 x^{3} - 4 x^{2} - x + 6

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