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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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