What is the solution for y^{''}-4y^'+4y=t^{-8}e^{2t} ?

The solution for y^{''}-4y^'+4y=t^{-8}e^{2t} is y=c_{1}e^{2t}+c_{2}te^{2t}+(e^{2t})/(42t^6)

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y^{''}-4y^'+4y=t^{-8}e^{2t}

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{- 8} e^{2 t}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t}}{42 t ^{6}}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = t^{- 8} e^{2 t}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t}}{42 t ^{6}}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t}}{42 t ^{6}}

\frac{d ^{2} x}{d t ^{2}} + k \frac{d x}{d t} + a = 0

3 \frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 30 y = 15 sin (x)

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

y^{^{''}} - y^{^{'}} + 9 y = 3 sin (3 x)

y^{^{''}} + y^{^{'}} = 4

What is the solution for y^{''}-4y^'+4y=t^{-8}e^{2t} ?
The solution for y^{''}-4y^'+4y=t^{-8}e^{2t} is y=c_{1}e^{2t}+c_{2}te^{2t}+(e^{2t})/(42t^6)