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

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

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y^{''}+2y^'+2y=x^2

y^{^{''}} + 2 y^{^{'}} + 2 y = x^{2}

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

Solution steps

y^{''} (x) + 2 y^{'} (x) + 2 y = x^{2}

Solve linear ODE:

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

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

(d^{2} - 4 d + 1) y = e^{2} x sin (x)

(d^{2} + 3 d + 2) y = e^{- x} + cos (x)

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

(d^{2} - 2 d + 1) \cdot y = a^{x} \cdot x^{2}

(d^{3} - 5 d^{2} + 8 d - 4) \cdot y = e^{2 x} + 2 e^{x} + 3 e^{- x} + 2

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