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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} = cos (t), y (0) = 0

y^{^{''}} + 7 y^{^{'}} = cos (2 x) + 8 e^{- 5 x}

y^{^{''}} + 3 y^{^{'}} - 4 y = 2, y^{^{'}} (0) = 0, y (0) = 0

y^{^{''}} - 3 y^{^{'}} - 4 y = 3 e^{2} t + 2 sin (t)

y^{^{''}} + 3 y^{^{'}} - 4 y = x^{2} e^{x}

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