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}

y^{^{''}} - 5 y^{^{'}} + 7 y = 2

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + 2 y = 3 e^{x} sec (x)

y^{^{''}} - 25 y = 6 sin (x)

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

y^{^{''}} + 2 y^{^{'}} + 6 y = e^{- 2 t}, y (0) = 2, y^{^{'}} (0) = 1

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