What is the solution for y^{''}-10y^'+25y=t^{-6}e^{5t} ?

The solution for y^{''}-10y^'+25y=t^{-6}e^{5t} is y=c_{1}e^{5t}+c_{2}te^{5t}+(e^{5t})/(20t^4)

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y^{''}-10y^'+25y=t^{-6}e^{5t}

y^{^{''}} - 10 y^{^{'}} + 25 y = t^{- 6} e^{5 t}

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{e ^{5 t}}{20 t ^{4}}

Solution steps

y^{''} (t) - 10 y^{'} (t) + 25 y = t^{- 6} e^{5 t}

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{e ^{5 t}}{20 t ^{4}}

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{e ^{5 t}}{20 t ^{4}}

y^{^{''}} - y^{^{'}} - 2 y = x e^{4 x}

q^{^{''}} + 100 q^{^{'}} + 2500 q = 40, q (0) = 0

(D^{2} + 5) y = cos (5) x

(D^{2} + 4 D + 4) y = 5 e^{3 x}

3 y^{^{'}} + 2 y^{^{''}} = - 1000 cos (3.14 \cdot \frac{t}{0.2})

What is the solution for y^{''}-10y^'+25y=t^{-6}e^{5t} ?
The solution for y^{''}-10y^'+25y=t^{-6}e^{5t} is y=c_{1}e^{5t}+c_{2}te^{5t}+(e^{5t})/(20t^4)