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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} + 5 x^{^{'}} + 6 x = 12 e^{- 5 t} + 12 e^{- t} + 24

y^{^{''}} + 6 y^{^{'}} + 5 y = 20

y^{^{''}} + y = e^{3 t} - 4 t cos (5 t)

y^{^{''''}} + 8 y^{^{''}} + 16 y = cos (x)

y^{^{''}} - 2 y^{^{'}} + y = 4 x^{2} - 6 + x^{- 1} e^{x}

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