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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

9 y^{^{''}} + 6 y^{^{'}} + y = x^{2}

y^{^{''}} + 10 y^{^{'}} + 26 y = 2 e^{- 5 x} cos (x)

(D^{2} + 3 D + 2) y = 40 e^{3 x}

y^{^{''}} - y^{^{'}} - 6 y = sin (3 x)

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

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