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

The solution for y^{''}+y^'-2y=2xe^{2x} is y=c_{1}e^x+c_{2}e^{-2x}+(e^{2x}x)/2-5/8 e^{2x}

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y^{''}+y^'-2y=2xe^{2x}

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

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{e ^{2 x} x}{2} - \frac{5}{8} e^{2 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{e ^{2 x} x}{2} - \frac{5}{8} e^{2 x}

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{e ^{2 x} x}{2} - \frac{5}{8} e^{2 x}

y^{^{''}} - 5 y^{^{'}} + 6 y = g (t)

2 y^{^{''}} - 4 y^{^{'}} + 4 y = e^{x} \cdot sec (x)

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 6 y = 1 + e^{- x}

y^{^{''}} - 7 y^{^{'}} + 12 = 0

y^{^{''}} - 2 y^{^{'}} - 3 = 0

What is the solution for y^{''}+y^'-2y=2xe^{2x} ?
The solution for y^{''}+y^'-2y=2xe^{2x} is y=c_{1}e^x+c_{2}e^{-2x}+(e^{2x}x)/2-5/8 e^{2x}