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

The solution for y^{''}+4y^'+4y=e^{2x} is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+1/16 e^{2x}

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

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

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{1}{16} e^{2 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{1}{16} e^{2 x}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{1}{16} e^{2 x}

\int \frac{1}{x + 1} d x

x e^{x^{2}}

\int_{1}^{x} sec (t) d t

\frac{d}{d x} (\frac{10}{27 x ^{\frac{8}{3}}})

f (x) = (1 + 4 x)^{9}, (0, 1)

What is the solution for y^{''}+4y^'+4y=e^{2x} ?
The solution for y^{''}+4y^'+4y=e^{2x} is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+1/16 e^{2x}