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

The solution for y^{'''}+10y^{''}+25y^'=e^t is y=c_{1}+c_{2}e^{-5t}+c_{3}te^{-5t}+1/36 e^t

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

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

y = c_{1} + c_{2} e^{- 5 t} + c_{3} t e^{- 5 t} + \frac{1}{36} e^{t}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 5 t} + c_{3} t e^{- 5 t} + \frac{1}{36} e^{t}

y = c_{1} + c_{2} e^{- 5 t} + c_{3} t e^{- 5 t} + \frac{1}{36} e^{t}

y^{^{''}} - 2 y^{^{'}} + 5 y = 1 + x, y (0) = 0, y^{^{'}} (0) = 4

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

y^{^{''}} + 2 y - 1 = 0

y^{^{''}} + 5 y^{^{'}} + 6 y = 6 t + 17, y (0) = 3, y^{^{'}} (0) = - 4

y^{^{''}} - 3 y = e^{x} sin (5 x)

What is the solution for y^{'''}+10y^{''}+25y^'=e^t ?
The solution for y^{'''}+10y^{''}+25y^'=e^t is y=c_{1}+c_{2}e^{-5t}+c_{3}te^{-5t}+1/36 e^t