What is the solution for y^{''}-4y^'-8y=8x^2+8x+1 ?

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

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y^{''}-4y^'-8y=8x^2+8x+1

y^{^{''}} - 4 y^{^{'}} - 8 y = 8 x^{2} + 8 x + 1

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

Solution steps

y^{''} (x) - 4 y^{'} (x) - 8 y = 8 x^{2} + 8 x + 1

Solve linear ODE:

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

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

y^{^{''}} - 9 y = 63

D, D (x) = e^{2 x^{3} + 4}

y^{^{''}} + 3 y^{^{'}} - 4 y = 12

y^{^{''}} + 2 y^{^{'}} + y = x^{2} e^{x}

y^{^{''}} - 6 y^{^{'}} + 5 y = - 1 + 5 x

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