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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int (x^{2} - \frac{3}{x} + cos (x) - \frac{2}{x}) d x

y = x^{2} - 7

\frac{\partial}{\partial y} (x cos (y) - y cos (x))

sin (t + c)

\int cos^{4} (\frac{x}{5}) d x

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