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

The solution for y^{''}+2y^'-4y=8x^2 is y=c_{1}e^{(-1+sqrt(5))x}+c_{2}e^{(-1-sqrt(5))x}-2x^2-2x-2

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

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

y = c_{1} e^{(- 1 + 5) x} + c_{2} e^{(- 1 - 5) x} - 2 x^{2} - 2 x - 2

Solution steps

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

Solve linear ODE:

y = c_{1} e^{(- 1 + 5) x} + c_{2} e^{(- 1 - 5) x} - 2 x^{2} - 2 x - 2

y = c_{1} e^{(- 1 + 5) x} + c_{2} e^{(- 1 - 5) x} - 2 x^{2} - 2 x - 2

8 y^{^{''}} - 32 y = 16 cos (x)

y^{^{''}} - 6 y^{^{'}} + 5 y = x, y (0) = e^{x}

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

\frac{d ^{2} y}{d x ^{2}} + y = sin (2 x)

y^{^{''}} + 9 y = cos (3 t), y (0) = 2, y^{^{'}} (0) = 5

What is the solution for y^{''}+2y^'-4y=8x^2 ?
The solution for y^{''}+2y^'-4y=8x^2 is y=c_{1}e^{(-1+sqrt(5))x}+c_{2}e^{(-1-sqrt(5))x}-2x^2-2x-2