What is the solution for y^{''}+3y^'+2y=x^2-1 ?

The solution for y^{''}+3y^'+2y=x^2-1 is y=c_{1}e^{-x}+c_{2}e^{-2x}+(x^2)/2-(3x)/2+5/4

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y^{''}+3y^'+2y=x^2-1

y^{^{''}} + 3 y^{^{'}} + 2 y = x^{2} - 1

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{x ^{2}}{2} - \frac{3 x}{2} + \frac{5}{4}

Solution steps

y^{''} (x) + 3 y^{'} (x) + 2 y = x^{2} - 1

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{x ^{2}}{2} - \frac{3 x}{2} + \frac{5}{4}

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{x ^{2}}{2} - \frac{3 x}{2} + \frac{5}{4}

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

y^{^{''}} - 4 y^{^{'}} + 5 y = \frac{2 e ^{2 x}}{sin ( x )}

3 y^{^{''}} - 6 y^{^{'}} + 6 y = e^{x} \cdot sec (x)

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

y^{^{''}} + 9 y = x e^{3 x}

What is the solution for y^{''}+3y^'+2y=x^2-1 ?
The solution for y^{''}+3y^'+2y=x^2-1 is y=c_{1}e^{-x}+c_{2}e^{-2x}+(x^2)/2-(3x)/2+5/4