What is the solution for y^{''}+3y^'+6*x=0 ?

The solution for y^{''}+3y^'+6*x=0 is y=c_{1}+c_{2}e^{-3x}-x^2+(2x)/3

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y^{''}+3y^'+6*x=0

y^{^{''}} + 3 y^{^{'}} + 6 \cdot x = 0

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

Solution steps

y^{''} (x) + 3 y^{'} (x) + 6 \cdot x = 0

Solve linear ODE:

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

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

y^{^{''}} - 8 y^{^{'}} + 12 y = 2 x (e^{2 x} - e^{- 2 x})

4 q^{^{''}} + 8 q^{^{'}} + 80 q = 88 t

3 y^{^{''}} - 8 y^{^{'}} - 3 y = 5

y^{^{''}} + 4 y^{^{'}} - 5 = 2 t^{2} - 3 t + 6

y^{^{''}} + 25 y = 6 x, y (0) = 1, y^{^{'}} (0) = 5

What is the solution for y^{''}+3y^'+6*x=0 ?
The solution for y^{''}+3y^'+6*x=0 is y=c_{1}+c_{2}e^{-3x}-x^2+(2x)/3