What is the solution for y^{''}+2y^'=2x+5-e^{2x} ?

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

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

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

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

Solution steps

y^{''} (x) + 2 y^{'} (x) = 2 x + 5 - e^{2 x}

Solve linear ODE:

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

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

y^{^{''''}} + y^{^{'''}} = 1 - x^{2} e^{- x}

y^{^{''}} - 5 y^{^{'}} + 6 y = 8 x e^{x}

y^{^{''}} + 2 y^{^{'}} + y = 4 cos (2 x)

y^{^{''}} + 3 y^{^{'}} = 1 - e^{- 3 x}

y^{^{''}} + 16 y = 6 x, y (0) = 1, y^{^{'}} (0) = 4

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