What is the solution for y^{''}-10y^'+25y=t^{-4}e^{5t} ?

The solution for y^{''}-10y^'+25y=t^{-4}e^{5t} is y=c_{1}e^{5t}+c_{2}te^{5t}+(e^{5t})/(6t^2)

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y^{''}-10y^'+25y=t^{-4}e^{5t}

y^{^{''}} - 10 y^{^{'}} + 25 y = t^{- 4} e^{5 t}

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

Solution steps

y^{''} (t) - 10 y^{'} (t) + 25 y = t^{- 4} e^{5 t}

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + 17 y = 289 x^{2} + 9

y^{^{'''}} + 3 y^{^{''}} - 2 y = x^{2} - 3 x + 2

- x^{^{''}} - 4 x = t^{2}

y^{^{''}} + 3 y^{^{'}} + 2 y = cos (x)

x^{^{''}} + 3 x^{^{'}} + 33 x = 9 cos (5 t)

What is the solution for y^{''}-10y^'+25y=t^{-4}e^{5t} ?
The solution for y^{''}-10y^'+25y=t^{-4}e^{5t} is y=c_{1}e^{5t}+c_{2}te^{5t}+(e^{5t})/(6t^2)