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

The solution for y^{''}-10y^'+25y=t^{-5}e^{5t} is y=c_{1}e^{5t}+c_{2}te^{5t}+(e^{5t})/(12t^3)

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

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

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{e ^{5 t}}{12 t ^{3}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{e ^{5 t}}{12 t ^{3}}

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{e ^{5 t}}{12 t ^{3}}

\frac{x}{x ^{2} - 4}

\frac{d}{d θ} (\frac{1}{2 cos ( θ ) + 3 sin ( θ )})

\frac{d}{d x} (sin (sin (2 x)))

\int - 2 d x

\int (3 x^{5} - 5 x + 6) d x

What is the solution for y^{''}-10y^'+25y=t^{-5}e^{5t} ?
The solution for y^{''}-10y^'+25y=t^{-5}e^{5t} is y=c_{1}e^{5t}+c_{2}te^{5t}+(e^{5t})/(12t^3)