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

The solution for y^{''}+4y^'=3x^2-4 is y=c_{1}+c_{2}e^{-4x}+(x^3)/4-(3x^2)/(16)-(29x)/(32)

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

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

y = c_{1} + c_{2} e^{- 4 x} + \frac{x ^{3}}{4} - \frac{3 x ^{2}}{16} - \frac{29 x}{32}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 x} + \frac{x ^{3}}{4} - \frac{3 x ^{2}}{16} - \frac{29 x}{32}

y = c_{1} + c_{2} e^{- 4 x} + \frac{x ^{3}}{4} - \frac{3 x ^{2}}{16} - \frac{29 x}{32}

y^{^{''}} + 4 y^{^{'}} + 29 = 0

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

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} - 3 y = 3 x e^{2 x}

y^{^{''}} - 2 y^{^{'}} + y = 3 x^{2} + 6 + 4 e x

x^{^{''}} + x = 1, x (0) = 2, x^{^{'}} (0) = 1

What is the solution for y^{''}+4y^'=3x^2-4 ?
The solution for y^{''}+4y^'=3x^2-4 is y=c_{1}+c_{2}e^{-4x}+(x^3)/4-(3x^2)/(16)-(29x)/(32)