What is the solution for y^{''}+4y^'=8x^2 ?

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

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y^{''}+4y^'=8x^2

y^{^{''}} + 4 y^{^{'}} = 8 x^{2}

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

Solution steps

y^{''} (x) + 4 y^{'} (x) = 8 x^{2}

Solve linear ODE:

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

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

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

(d^{2} - 2 d + 1) \cdot y = e^{x} \cdot x^{2}

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = e^{x} - x + 16

(d^{2} + 4 d + 5) \cdot y = - 2 cos (x)

(d^{3} - 3 d + 2) \cdot y = 540 x^{3} \cdot e^{- x}

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