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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

t^{2} \cdot e^{- 3 t}

\frac{d}{d x} (\frac{x}{10} - \frac{10}{x})

\int 20 θ sec^{2} (θ) d θ

\int \frac{\frac{d}{1} ( x )}{x} d x

\frac{d}{d x} (\frac{x}{a ^{2} + x ^{2}})

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