What is the solution for y^{''}+4y^'+3y=x^2-2 ?

The solution for y^{''}+4y^'+3y=x^2-2 is y=c_{1}e^{-x}+c_{2}e^{-3x}+(x^2)/3-(8x)/9+8/27

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

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

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{x ^{2}}{3} - \frac{8 x}{9} + \frac{8}{27}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{x ^{2}}{3} - \frac{8 x}{9} + \frac{8}{27}

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{x ^{2}}{3} - \frac{8 x}{9} + \frac{8}{27}

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

y^{^{''}} + 2 y^{^{'}} + y = sin (x) + 7 cos (2 x)

y^{^{''}} + 4 y^{^{'}} + 4 y = (7 + x) e - 2 x

y^{^{''}} - y^{^{'}} - 6 y = e^{t}

\frac{d ^{2} y}{d x ^{2}} + 8 \frac{d y}{d x} + 25 y = 16 sin (x)

What is the solution for y^{''}+4y^'+3y=x^2-2 ?
The solution for y^{''}+4y^'+3y=x^2-2 is y=c_{1}e^{-x}+c_{2}e^{-3x}+(x^2)/3-(8x)/9+8/27