What is the solution for y^{''}-2y^'+2y=x^2+1 ?

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

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y^{''}-2y^'+2y=x^2+1

y^{^{''}} - 2 y^{^{'}} + 2 y = x^{2} + 1

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

Solution steps

y^{''} (x) - 2 y^{'} (x) + 2 y = x^{2} + 1

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} = \frac{1}{e ^{x} + 1}

y^{^{''}} - 10 y^{^{'}} + 25 y = 3 e^{5 t}

y^{^{''}} + y = H (t - 1)

y^{^{''}} + y^{^{'}} = e^{2 t} + 3 + e^{t}

y^{^{''}} - y^{^{'}} - 6 y = 51 cos (3 x)

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