What is the solution for 3y^{''}-2y^'-y=x^2 ?

The solution for 3y^{''}-2y^'-y=x^2 is y=c_{1}e^x+c_{2}e^{-x/3}-x^2+4x-14

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

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

y = c_{1} e^{x} + c_{2} e^{- \frac{x}{3}} - x^{2} + 4 x - 14

Solution steps

3 y^{''} (x) - 2 y^{'} (x) - y = x^{2}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- \frac{x}{3}} - x^{2} + 4 x - 14

y = c_{1} e^{x} + c_{2} e^{- \frac{x}{3}} - x^{2} + 4 x - 14

D (t) = t + 5

y^{^{''}} + 2 y^{^{'}} - y = t^{- 1} e^{t}

y^{^{''}} - 4 y^{^{'}} - 12 y = 4

y^{^{''}} - 2 y^{^{'}} + 2 y = sin (t), y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 5 y^{^{'}} + 6 y = 3 x^{2} + 5 x

What is the solution for 3y^{''}-2y^'-y=x^2 ?
The solution for 3y^{''}-2y^'-y=x^2 is y=c_{1}e^x+c_{2}e^{-x/3}-x^2+4x-14