What is the solution for y^{''}-4y^'+4y=x^2 ?

The solution for y^{''}-4y^'+4y=x^2 is y=c_{1}e^{2x}+c_{2}xe^{2x}+(x^2)/4+x/2+3/8

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y^{''}-4y^'+4y=x^2

y^{^{''}} - 4 y^{^{'}} + 4 y = x^{2}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{x ^{2}}{4} + \frac{x}{2} + \frac{3}{8}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = x^{2}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{x ^{2}}{4} + \frac{x}{2} + \frac{3}{8}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{x ^{2}}{4} + \frac{x}{2} + \frac{3}{8}

(x + 1)^{\frac{1}{2}}

y^{^{'}} = 6 x (y - 1)^{\frac{2}{3}}

k = 0 \sum \infty \frac{k}{2 ^{k}} x^{k}

y^{^{''}} + 214 y^{^{'}} + 14 y = 0

y^{^{'}} + \frac{3}{x} y = x^{2} y^{2}

What is the solution for y^{''}-4y^'+4y=x^2 ?
The solution for y^{''}-4y^'+4y=x^2 is y=c_{1}e^{2x}+c_{2}xe^{2x}+(x^2)/4+x/2+3/8