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}

\int \frac{1}{9 + 4 x ^{2}} d x

\frac{d}{d x} (\frac{9}{2 x + 1})

\int_{1 - 3 x}^{1} \frac{u ^{3}}{1 + u ^{2}} d u

\int 16 + 6 x - x^{2} d x

n \to \infty lim (5)

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