What is the solution for y^{'''}+10y^{''}+25y^'=e^x ?

The solution for y^{'''}+10y^{''}+25y^'=e^x is y=c_{1}+c_{2}e^{-5x}+c_{3}xe^{-5x}+1/36 e^x

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y^{'''}+10y^{''}+25y^'=e^x

y^{^{'''}} + 10 y^{^{''}} + 25 y^{^{'}} = e^{x}

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

Solution steps

y^{'''} (x) + 10 y^{''} (x) + 25 y^{'} (x) = e^{x}

Solve linear ODE:

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

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

y^{^{''}} + 4 y = (x^{2} - 3) sin (2 x)

y^{^{''}} - 5 y^{^{'}} - 6 y = 2 cos (3 x) + 4 e^{2 x}

(D^{2} - D - 2) y = e^{3 x}

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

y^{^{''''}} + 4 y = cos (2 x)

What is the solution for y^{'''}+10y^{''}+25y^'=e^x ?
The solution for y^{'''}+10y^{''}+25y^'=e^x is y=c_{1}+c_{2}e^{-5x}+c_{3}xe^{-5x}+1/36 e^x