What is the solution for y^{''}-10y^'+25y=t^{-7}e^{5t} ?

The solution for y^{''}-10y^'+25y=t^{-7}e^{5t} is y=c_{1}e^{5t}+c_{2}te^{5t}+(e^{5t})/(30t^5)

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y^{''}-10y^'+25y=t^{-7}e^{5t}

y^{^{''}} - 10 y^{^{'}} + 25 y = t^{- 7} e^{5 t}

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

Solution steps

y^{''} (t) - 10 y^{'} (t) + 25 y = t^{- 7} e^{5 t}

Solve linear ODE:

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

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

\int \frac{6 x ^{4} + 23 x ^{2} + 4 x + 15}{x ^{2} + 3} d x

\frac{d}{d x} (x^{2} sin^{7} (x) + x cos^{- 2} (x))

4 y^{^{''}} + 2 y^{^{'}} + 2 y = 0

y = cos (x), at x = \frac{π}{2}

\frac{d}{d x} (x + x x)

What is the solution for y^{''}-10y^'+25y=t^{-7}e^{5t} ?
The solution for y^{''}-10y^'+25y=t^{-7}e^{5t} is y=c_{1}e^{5t}+c_{2}te^{5t}+(e^{5t})/(30t^5)