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

The solution for y^{''}-4y^'+4y=t^2e^{2t} is y=c_{1}e^{2t}+c_{2}te^{2t}+(e^{2t}t^4)/(12)

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

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

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t} t ^{4}}{12}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t} t ^{4}}{12}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t} t ^{4}}{12}

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

(d^{3} - 3 d) y = 9 x^{2}

y^{^{''}} + y^{^{'}} - 12 y = 4 e^{x} sin (2 x) + 4 e^{3 x} + 2

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + 2 y = 5 e^{3 x}

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - 15 y = (x + 1) e^{x}

What is the solution for y^{''}-4y^'+4y=t^2e^{2t} ?
The solution for y^{''}-4y^'+4y=t^2e^{2t} is y=c_{1}e^{2t}+c_{2}te^{2t}+(e^{2t}t^4)/(12)