What is the solution for y^{''}+10y^'+25y=-50e^{-5x} ?

The solution for y^{''}+10y^'+25y=-50e^{-5x} is y=c_{1}e^{-5x}+c_{2}xe^{-5x}-25x^2e^{-5x}

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y^{''}+10y^'+25y=-50e^{-5x}

y^{^{''}} + 10 y^{^{'}} + 25 y = - 50 e^{- 5 x}

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

Solution steps

y^{''} (x) + 10 y^{'} (x) + 25 y = - 50 e^{- 5 x}

Solve linear ODE:

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

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

\int \frac{e ^{x}}{( e ^{x} - 5 ) ( e ^{x} - 11 )} d x

\int_{2}^{6} (x^{2} - 7 x + 6) d x

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What is the solution for y^{''}+10y^'+25y=-50e^{-5x} ?
The solution for y^{''}+10y^'+25y=-50e^{-5x} is y=c_{1}e^{-5x}+c_{2}xe^{-5x}-25x^2e^{-5x}