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

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

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

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

y = c_{1} e^{2 x} + c_{2} - \frac{x ^{2}}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} - \frac{x ^{2}}{2}

y = c_{1} e^{2 x} + c_{2} - \frac{x ^{2}}{2}

y^{^{''}} + 2 y^{^{'}} + y = 3 t + 1 e^{- t}

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

y^{^{''}} + 4 y = 5 x e^{2 x}

y^{^{''}} + 5 y^{^{'}} + 6 y = 2 e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

2 y^{^{''}} - 13 y^{^{'}} + 20 y = 4 x^{2} + 8 x + 10

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