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

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

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

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

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + x^{2} + x + 2

Solution steps

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

Solve linear ODE:

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + x^{2} + x + 2

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + x^{2} + x + 2

y^{^{''}} + 9 y = cos (3 t), y (0) = 5, y^{^{'}} (0) = 4

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 t^{2} + 4 + e^{2 t} + t sin (2 t)

y^{^{''}} + 9 y = 4 (\frac{\frac{u ^{5}}{3} π - u ^{13}}{3 π})

y^{^{''}} + y^{^{'}} = sinh (t)

y^{^{''}} - 2 \cdot y^{^{'}} + 2 \cdot y = cos (t), y (0) = 1, y^{^{'}} (0) = 0

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