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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

y = c_{1} e^{2 x} + c_{2} - \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}e^{2x}+c_{2}-(x^2)/4-x/4