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}

D (r) = \frac{27 ( r ) + 13}{26}

y^{^{''}} + 16 y = 4 (u \frac{7}{4} π)

y^{^{'''}} - y^{^{''}} - 4 y^{^{'}} + 4 y = - 20 t^{2} + 52 t - 6

y^{^{'''}} - 2 y^{^{''}} + 5 y^{^{'}} = - 24 e^{3 x}

y^{^{''}} + 4 y = 4 sec^{3} (2 t)

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