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

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

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y^{''}-y^'-2y=e^{2x}

y^{^{''}} - y^{^{'}} - 2 y = e^{2 x}

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

Solution steps

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

Solve linear ODE:

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

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

\int x^{3} x - 2 d x

\frac{d}{d x} (x^{3} - 5 x^{2} + 4 x - 20 - 5 e^{3 x})

\frac{d}{d t} (tan (cos (2 t)))

\frac{d}{d x} (x^{5} + 2 x^{3} - x^{2})

\frac{d}{d x} (2 + cot^{2} (x))

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