What is the solution for y^{''}+10y^'+25y=e^{-5}x ?

The solution for y^{''}+10y^'+25y=e^{-5}x is y=c_{1}e^{-5x}+c_{2}xe^{-5x}+x/(25e^5)-2/(125e^5)

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y^{''}+10y^'+25y=e^{-5}x

y^{^{''}} + 10 y^{^{'}} + 25 y = e^{- 5} x

y = c_{1} e^{- 5 x} + c_{2} x e^{- 5 x} + \frac{x}{25 e ^{5}} - \frac{2}{125 e ^{5}}

Solution steps

y^{''} (x) + 10 y^{'} (x) + 25 y = e^{- 5} x

Solve linear ODE:

y = c_{1} e^{- 5 x} + c_{2} x e^{- 5 x} + \frac{x}{25 e ^{5}} - \frac{2}{125 e ^{5}}

y = c_{1} e^{- 5 x} + c_{2} x e^{- 5 x} + \frac{x}{25 e ^{5}} - \frac{2}{125 e ^{5}}

y^{^{''}} + 3 y^{^{'}} + 2 y = 12 e^{x}

y, (d^{3} - d) y = 2 cos (t)

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

(d^{2} + 3 d + 2) y = e^{e^{x}}

(d^{2} - 6 d + 9) y = 2 e^{3} x

What is the solution for y^{''}+10y^'+25y=e^{-5}x ?
The solution for y^{''}+10y^'+25y=e^{-5}x is y=c_{1}e^{-5x}+c_{2}xe^{-5x}+x/(25e^5)-2/(125e^5)