What is the solution for 2y^{''}-2y^'-4y=2e^{2x} ?

The solution for 2y^{''}-2y^'-4y=2e^{2x} is y=c_{1}e^{2x}+c_{2}e^{-x}+(e^{2x}x)/3

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2y^{''}-2y^'-4y=2e^{2x}

2 y^{^{''}} - 2 y^{^{'}} - 4 y = 2 e^{2 x}

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

Solution steps

2 y^{''} (x) - 2 y^{'} (x) - 4 y = 2 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}

y^{^{''}} + y = 2 sin (2 t)

y^{^{''}} - y^{^{'}} - 2 y = 6 x + 6 e^{- x}

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{t}}{( 1 + t ^{2} )}

D (x) = - 6 x + 4200

y^{^{''}} + 4 y = - 2, y (\frac{π}{8}) = \frac{1}{2}, y^{^{'}} (\frac{π}{8}) = 2

What is the solution for 2y^{''}-2y^'-4y=2e^{2x} ?
The solution for 2y^{''}-2y^'-4y=2e^{2x} is y=c_{1}e^{2x}+c_{2}e^{-x}+(e^{2x}x)/3