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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

6 \frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} - 2 y = 3 x - 2

y^{^{''}} + 4 y^{^{'}} + 5 y = t

y^{^{''}} - y^{^{'}} - 2 y = 4 x e^{3 x}

y^{^{''}} + 9 y = cos (t) \cdot H (t - π)

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 e^{3 x}

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