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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

f (x) = x^{- \frac{2}{5}} + x^{\frac{2}{5}}

\frac{\partial}{\partial y} (\frac{y ^{2}}{1 + x ^{2}})

\int \frac{x - 2}{x ^{2} - 7 x + 12} d x

\frac{d y}{d x} = \frac{x ( 1 + y ^{2} )}{4 y}

\frac{d}{d x} (\frac{x}{2} + \frac{3}{2} + \frac{4}{x + 1})

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