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

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

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

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

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + x^{2} + x

Solution steps

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

Solve linear ODE:

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + x^{2} + x

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + x^{2} + x

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

(D^{2} + 1) y = sec (x) tan (x)

0.5 q^{^{''}} + 6 q^{^{'}} + \frac{1}{0.02} q = 24 sin (10 t)

\frac{d ^{2} y}{d x ^{2}} - y = x \cdot e^{x}

y^{^{''}} - 3 y^{^{'}} + 2 y = 13 e^{3 x}

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