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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

f (x) = cot (x^{2})

\int_{- 1}^{0} x^{3} - x^{2} - 2 x d x

\int x^{4} cos (2 x) d x

9 - x^{2}, y = 0

\frac{d}{d x} (x sin (6 x))

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