What is the solution for y^{''}-4y^'+4y=xe^{2x}+1 ?

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

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y^{''}-4y^'+4y=xe^{2x}+1

y^{^{''}} - 4 y^{^{'}} + 4 y = x e^{2 x} + 1

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

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = x e^{2 x} + 1

Solve linear ODE:

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

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

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

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

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

y^{^{'''}} + 6 y^{^{''}} + 12 y^{^{'}} + 8 y = 9 e^{- 2 x}

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

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