What is the solution for y^{''}-10y^'+25y=t^{-8}e^{5t} ?

The solution for y^{''}-10y^'+25y=t^{-8}e^{5t} is y=c_{1}e^{5t}+c_{2}te^{5t}+(e^{5t})/(42t^6)

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y^{''}-10y^'+25y=t^{-8}e^{5t}

y^{^{''}} - 10 y^{^{'}} + 25 y = t^{- 8} e^{5 t}

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{e ^{5 t}}{42 t ^{6}}

Solution steps

y^{''} (t) - 10 y^{'} (t) + 25 y = t^{- 8} e^{5 t}

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{e ^{5 t}}{42 t ^{6}}

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{e ^{5 t}}{42 t ^{6}}

\int 7 x e^{5 x} d x

\int (2 - 3 x) cos (4 x) d x

\frac{\partial}{\partial x} (x^{2} + y^{2})

\int sech (x) d x

What is the solution for y^{''}-10y^'+25y=t^{-8}e^{5t} ?
The solution for y^{''}-10y^'+25y=t^{-8}e^{5t} is y=c_{1}e^{5t}+c_{2}te^{5t}+(e^{5t})/(42t^6)