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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} - 5 y = 2 e^{2 t}

y^{^{''}} + y = csc (x) cot (x)

y^{^{''}} + 4 y = 16 t sin (2 t)

y^{^{''}} + 2 y^{^{'}} - 8 y = 4 e^{- 3 x} - e^{- x}

4 \frac{d ^{2} y}{d x ^{2}} - y = x \cdot e^{\frac{x}{2}}

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