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

The solution for 2y^{''}-2y^'+y=1+x+x^2 is y=e^{x/2}(c_{1}cos(x/2)+c_{2}sin(x/2))+x^2+5x+7

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

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

y = e^{\frac{x}{2}} (c_{1} cos (\frac{x}{2}) + c_{2} sin (\frac{x}{2})) + x^{2} + 5 x + 7

Solution steps

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

Solve linear ODE:

y = e^{\frac{x}{2}} (c_{1} cos (\frac{x}{2}) + c_{2} sin (\frac{x}{2})) + x^{2} + 5 x + 7

y = e^{\frac{x}{2}} (c_{1} cos (\frac{x}{2}) + c_{2} sin (\frac{x}{2})) + x^{2} + 5 x + 7

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

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

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

y^{^{''}} + y = 5 x \cdot sin (x)

3 x^{^{''}} + 48 x = 60 e^{- 4 t} cos (4 t)

What is the solution for 2y^{''}-2y^'+y=1+x+x^2 ?
The solution for 2y^{''}-2y^'+y=1+x+x^2 is y=e^{x/2}(c_{1}cos(x/2)+c_{2}sin(x/2))+x^2+5x+7