What is the solution for y^{''}-2y^'+5y=10x^2-3x-3 ?

The solution for y^{''}-2y^'+5y=10x^2-3x-3 is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))+2x^2+x-1

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y^{''}-2y^'+5y=10x^2-3x-3

y^{^{''}} - 2 y^{^{'}} + 5 y = 10 x^{2} - 3 x - 3

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

Solution steps

y^{''} (x) - 2 y^{'} (x) + 5 y = 10 x^{2} - 3 x - 3

Solve linear ODE:

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

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

y^{^{''}} + 4 y = 3 cos (x)

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

y^{^{''}} + 4 y = 8 tan (t)

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

y^{^{''}} + y = t, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}-2y^'+5y=10x^2-3x-3 ?
The solution for y^{''}-2y^'+5y=10x^2-3x-3 is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))+2x^2+x-1