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}+(e^{-2x}x^2)/2

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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{e ^{- 2 x} x ^{2}}{2}

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{e ^{- 2 x} x ^{2}}{2}

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

x = 6 y^{2}, x = 4 + 2 y^{2}

f (x) = x (x^{6} - 1), at x = 1

\int e^{x} \cdot cos (3 x) d x

\int \frac{6 x ^{2} - 15 x + 22}{( x + 3 ) ( x ^{2} + 2 ) ^{2}} d x

y = \frac{1}{x}, (4, \frac{1}{2})

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}+(e^{-2x}x^2)/2