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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} + 10 y^{^{''}} + 25 y^{^{'}} = e x

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

2 y^{^{''}} + 2 y^{^{'}} + 2 y = sin (t) + 1

u^{^{''}} + a u^{^{'}} (t) = b

6 y^{^{''}} + 3 y^{^{'}} - 30 y = 16 cos (x)

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