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

The solution for y^{''}-4y^'+4y=x^2+2 is y=c_{1}e^{2x}+c_{2}xe^{2x}+(x^2)/4+x/2+7/8

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

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

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{x ^{2}}{4} + \frac{x}{2} + \frac{7}{8}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{x ^{2}}{4} + \frac{x}{2} + \frac{7}{8}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{x ^{2}}{4} + \frac{x}{2} + \frac{7}{8}

What is the solution for y^{''}-4y^'+4y=x^2+2 ?
The solution for y^{''}-4y^'+4y=x^2+2 is y=c_{1}e^{2x}+c_{2}xe^{2x}+(x^2)/4+x/2+7/8