What is the solution for y^{''}+4y^'-5y=24e^x ?

The solution for y^{''}+4y^'-5y=24e^x is y=c_{1}e^x+c_{2}e^{-5x}+4xe^x

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y^{''}+4y^'-5y=24e^x

y^{^{''}} + 4 y^{^{'}} - 5 y = 24 e^{x}

y = c_{1} e^{x} + c_{2} e^{- 5 x} + 4 x e^{x}

Solution steps

y^{''} (x) + 4 y^{'} (x) - 5 y = 24 e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 5 x} + 4 x e^{x}

y = c_{1} e^{x} + c_{2} e^{- 5 x} + 4 x e^{x}

What is the solution for y^{''}+4y^'-5y=24e^x ?
The solution for y^{''}+4y^'-5y=24e^x is y=c_{1}e^x+c_{2}e^{-5x}+4xe^x