What is the solution for y^{''}-3y^'=x ?

The solution for y^{''}-3y^'=x is y=c_{1}e^{3x}+c_{2}-(x^2)/6-x/9

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y^{''}-3y^'=x

y^{^{''}} - 3 y^{^{'}} = x

y = c_{1} e^{3 x} + c_{2} - \frac{x ^{2}}{6} - \frac{x}{9}

Solution steps

y^{''} (x) - 3 y^{'} (x) = x

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} - \frac{x ^{2}}{6} - \frac{x}{9}

y = c_{1} e^{3 x} + c_{2} - \frac{x ^{2}}{6} - \frac{x}{9}

What is the solution for y^{''}-3y^'=x ?
The solution for y^{''}-3y^'=x is y=c_{1}e^{3x}+c_{2}-(x^2)/6-x/9