What is the solution for y^{''}-4y^'+4y=t^{-7}e^{2t} ?

The solution for y^{''}-4y^'+4y=t^{-7}e^{2t} is y=c_{1}e^{2t}+c_{2}te^{2t}+(e^{2t})/(30t^5)

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y^{''}-4y^'+4y=t^{-7}e^{2t}

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{- 7} e^{2 t}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t}}{30 t ^{5}}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = t^{- 7} e^{2 t}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t}}{30 t ^{5}}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{e ^{2 t}}{30 t ^{5}}

\int \frac{1}{( 1 - y ) ( 2 - y )} d y

f (s) = 2 t e^{- 5 t}

f (x) = x^{2} + 2 x, [1, 2]

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

x \to 1 - lim (1 - e^{\frac{1}{x - 1}})

What is the solution for y^{''}-4y^'+4y=t^{-7}e^{2t} ?
The solution for y^{''}-4y^'+4y=t^{-7}e^{2t} is y=c_{1}e^{2t}+c_{2}te^{2t}+(e^{2t})/(30t^5)