What is the solution for y^{''}-4y^'+4y=3xe^{2x} ?

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

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y^{''}-4y^'+4y=3xe^{2x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 3 x e^{2 x}

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

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = 3 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}}{2}

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

y^{^{''}} - 3 y^{^{'}} - 4 y = 4 t^{2} - 1

y^{^{''}} + 11 y^{^{'}} + 24 y = 4 sin (4 t) + 2 e^{- 8 t}

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

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

y^{^{''}} + y = 2 sin (2 t), y (0) = 12, y^{^{'}} (0) = 0

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