What is the solution for y^{''}-3y^'-4y=10e^{-x}+4x+3 ?

The solution for y^{''}-3y^'-4y=10e^{-x}+4x+3 is y=c_{1}e^{4x}+c_{2}e^{-x}-2xe^{-x}-x

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y^{''}-3y^'-4y=10e^{-x}+4x+3

y^{^{''}} - 3 y^{^{'}} - 4 y = 10 e^{- x} + 4 x + 3

y = c_{1} e^{4 x} + c_{2} e^{- x} - 2 x e^{- x} - x

Solution steps

y^{''} (x) - 3 y^{'} (x) - 4 y = 10 e^{- x} + 4 x + 3

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- x} - 2 x e^{- x} - x

y = c_{1} e^{4 x} + c_{2} e^{- x} - 2 x e^{- x} - x

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y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 1, y (1) = 0

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What is the solution for y^{''}-3y^'-4y=10e^{-x}+4x+3 ?
The solution for y^{''}-3y^'-4y=10e^{-x}+4x+3 is y=c_{1}e^{4x}+c_{2}e^{-x}-2xe^{-x}-x