What is the solution for y^{''}+2y^'-8y=1-2x^2 ?

The solution for y^{''}+2y^'-8y=1-2x^2 is y=c_{1}e^{2x}+c_{2}e^{-4x}+(x^2)/4+x/8-1/32

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y^{''}+2y^'-8y=1-2x^2

y^{^{''}} + 2 y^{^{'}} - 8 y = 1 - 2 x^{2}

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} + \frac{x ^{2}}{4} + \frac{x}{8} - \frac{1}{32}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} + \frac{x ^{2}}{4} + \frac{x}{8} - \frac{1}{32}

y = c_{1} e^{2 x} + c_{2} e^{- 4 x} + \frac{x ^{2}}{4} + \frac{x}{8} - \frac{1}{32}

y^{^{''}} + 64 y = csc (8 x)

y^{^{''}} + y = t, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 9 y = 8 sec (3 x)

y^{^{''}} + 16 y = 3 e^{8 x}

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

What is the solution for y^{''}+2y^'-8y=1-2x^2 ?
The solution for y^{''}+2y^'-8y=1-2x^2 is y=c_{1}e^{2x}+c_{2}e^{-4x}+(x^2)/4+x/8-1/32