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}

y^{^{''}} + 12 y^{^{'}} + 100 y = 48 sin (10 t)

x^{^{''}} + 25 x = 102 cos (t)

y^{^{''}} - y^{^{'}} = 3 - 4 x

y^{^{''}} + 3 y + 2 y = cos (e^{x})

y^{^{''}} + 15 y^{^{'}} + 7 y = sin (x) + cos (x)

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