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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - a \cdot y + b = 0

y^{^{''}} + 16 y^{^{'}} + 100 y = 10, y (0) = 0, y^{^{'}} (0) = 0

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

y^{^{''}} + π^{2} y = \frac{5}{ln ( 10 )} sin (x ln (2))

\frac{d ^{2} y}{d t ^{2}} + 6 \frac{d y}{d t} + 9 y = e^{- 3 t}

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