What is the solution for y^{''}-y^'-2y=3e^{(2x)}-e^{(-x)} ?

The solution for y^{''}-y^'-2y=3e^{(2x)}-e^{(-x)} is y=c_{1}e^{2x}+c_{2}e^{-x}+e^{2x}x+(e^{-x}x)/3

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y^{''}-y^'-2y=3e^{(2x)}-e^{(-x)}

y^{^{''}} - y^{^{'}} - 2 y = 3 e^{(2 x)} - e^{(- x)}

y = c_{1} e^{2 x} + c_{2} e^{- x} + e^{2 x} x + \frac{e ^{- x} x}{3}

Solution steps

y^{''} (x) - y^{'} (x) - 2 y = 3 e^{(2 x)} - e^{(- x)}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} + e^{2 x} x + \frac{e ^{- x} x}{3}

y = c_{1} e^{2 x} + c_{2} e^{- x} + e^{2 x} x + \frac{e ^{- x} x}{3}

What is the solution for y^{''}-y^'-2y=3e^{(2x)}-e^{(-x)} ?
The solution for y^{''}-y^'-2y=3e^{(2x)}-e^{(-x)} is y=c_{1}e^{2x}+c_{2}e^{-x}+e^{2x}x+(e^{-x}x)/3