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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} - 2 y = - 1 + 5 x^{2} + x^{3}

y^{^{''}} - 8 y^{^{'}} + 15 y = t e^{3 t}

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

y^{^{''}} + k^{2} y = a cos (b x), y (0) = 0, y^{^{'}} (0) = 0

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

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