What is the solution for y^{''}+5y^'-14y=e^{(4x)} ?

The solution for y^{''}+5y^'-14y=e^{(4x)} is y=c_{1}e^{2x}+c_{2}e^{-7x}+1/22 e^{4x}

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y^{''}+5y^'-14y=e^{(4x)}

y^{^{''}} + 5 y^{^{'}} - 14 y = e^{(4 x)}

y = c_{1} e^{2 x} + c_{2} e^{- 7 x} + \frac{1}{22} e^{4 x}

Solution steps

y^{''} (x) + 5 y^{'} (x) - 14 y = e^{(4 x)}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 7 x} + \frac{1}{22} e^{4 x}

y = c_{1} e^{2 x} + c_{2} e^{- 7 x} + \frac{1}{22} e^{4 x}

(- 1.2) (3.3)

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What is the solution for y^{''}+5y^'-14y=e^{(4x)} ?
The solution for y^{''}+5y^'-14y=e^{(4x)} is y=c_{1}e^{2x}+c_{2}e^{-7x}+1/22 e^{4x}