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

The solution for y^{''}-2y^'+1=x^2 is y=c_{1}e^{2x}+c_{2}-(x^3)/6-(x^2)/4+x/4

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

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

y = c_{1} e^{2 x} + c_{2} - \frac{x ^{3}}{6} - \frac{x ^{2}}{4} + \frac{x}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} - \frac{x ^{3}}{6} - \frac{x ^{2}}{4} + \frac{x}{4}

y = c_{1} e^{2 x} + c_{2} - \frac{x ^{3}}{6} - \frac{x ^{2}}{4} + \frac{x}{4}

y^{^{''}} - 4 y^{^{'}} + 3 y = 3 t e^{t}, y (0) = 1, y^{^{'}} (0) = - 5

y^{^{''}} - 3 y^{^{'}} - 4 y = 8 e^{x} cos (2 x)

y^{^{''}} + 6 y^{^{'}} + 5 y = - 1020 sin (3 t)

y^{^{''}} + 2 y^{^{'}} - 8 y = 4 a

(D^{2} + 4 D + 8) y = 4 sin (2 x)

What is the solution for y^{''}-2y^'+1=x^2 ?
The solution for y^{''}-2y^'+1=x^2 is y=c_{1}e^{2x}+c_{2}-(x^3)/6-(x^2)/4+x/4