What is the solution for y^{''}-2y^'+2y=x^3-5 ?

The solution for y^{''}-2y^'+2y=x^3-5 is y=e^x(c_{1}cos(x)+c_{2}sin(x))+(x^3)/2+(3x^2)/2+(3x)/2-5/2

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y^{''}-2y^'+2y=x^3-5

y^{^{''}} - 2 y^{^{'}} + 2 y = x^{3} - 5

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{x ^{3}}{2} + \frac{3 x ^{2}}{2} + \frac{3 x}{2} - \frac{5}{2}

Solution steps

y^{''} (x) - 2 y^{'} (x) + 2 y = x^{3} - 5

Solve linear ODE:

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{x ^{3}}{2} + \frac{3 x ^{2}}{2} + \frac{3 x}{2} - \frac{5}{2}

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{x ^{3}}{2} + \frac{3 x ^{2}}{2} + \frac{3 x}{2} - \frac{5}{2}

y^{^{''}} - 2 y + y = 3 sin (2 x) + 2 e^{x}

y^{^{'''}} - 7 y^{^{'}} - 6 y = sin (2 x) - 5

2 y^{^{''}} - y^{^{'}} - 6 y = - 7 e^{2 x}, y (0) = 5, y^{^{'}} (0) = 2

y^{^{''}} + y^{^{'}} + y = 1, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 3 y^{^{'}} - 10 = 0, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}-2y^'+2y=x^3-5 ?
The solution for y^{''}-2y^'+2y=x^3-5 is y=e^x(c_{1}cos(x)+c_{2}sin(x))+(x^3)/2+(3x^2)/2+(3x)/2-5/2