What is the solution for y^{''}+2y^'-8y=8x^2-3,y(0)=-1,y^'(0)=-3 ?

The solution for y^{''}+2y^'-8y=8x^2-3,y(0)=-1,y^'(0)=-3 is y=(-1-1/12)e^{2x}+1/12 e^{-4x}-x^2-x/2

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y^{''}+2y^'-8y=8x^2-3,y(0)=-1,y^'(0)=-3

y^{^{''}} + 2 y^{^{'}} - 8 y = 8 x^{2} - 3, y (0) = - 1, y^{^{'}} (0) = - 3

y = (- 1 - \frac{1}{12}) e^{2 x} + \frac{1}{12} e^{- 4 x} - x^{2} - \frac{x}{2}

Solution steps

y^{''} (x) + 2 y^{'} (x) - 8 y = 8 x^{2} - 3

Solve linear ODE:

y = (- 1 - \frac{1}{12}) e^{2 x} + \frac{1}{12} e^{- 4 x} - x^{2} - \frac{x}{2}

y = (- 1 - \frac{1}{12}) e^{2 x} + \frac{1}{12} e^{- 4 x} - x^{2} - \frac{x}{2}

y^{^{''}} + y = sec^{3} (x), y (0) = 2, y^{^{'}} (0) = \frac{5}{2}

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + 3 y = 27 x

9 y^{^{''}} + 6 y^{^{'}} + 8 y = 37 sin (x)

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{t}}{t}

a y^{^{''}} + b y^{^{'}} + c = 0

What is the solution for y^{''}+2y^'-8y=8x^2-3,y(0)=-1,y^'(0)=-3 ?
The solution for y^{''}+2y^'-8y=8x^2-3,y(0)=-1,y^'(0)=-3 is y=(-1-1/12)e^{2x}+1/12 e^{-4x}-x^2-x/2