What is the solution for y^{''}-4y^'+4y=2t^6e^{2t} ?

The solution for y^{''}-4y^'+4y=2t^6e^{2t} is y=c_{1}e^{2t}+c_{2}te^{2t}+(e^{2t}t^8)/(28)

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y^{''}-4y^'+4y=2t^6e^{2t}

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 t^{6} e^{2 t}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t} t ^{8}}{28}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = 2 t^{6} e^{2 t}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t} t ^{8}}{28}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t} t ^{8}}{28}

y^{^{''}} - 2 y^{^{'}} + y = t^{- 7} e^{t}

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

y^{^{''}} - 6 y^{^{'}} + 8 y = 2

y^{^{''}} + 3 y^{^{'}} - 4 = 0

y^{^{''}} + 4 y + 4 = x^{2} + 2 sin (x) cos (x)

What is the solution for y^{''}-4y^'+4y=2t^6e^{2t} ?
The solution for y^{''}-4y^'+4y=2t^6e^{2t} is y=c_{1}e^{2t}+c_{2}te^{2t}+(e^{2t}t^8)/(28)