What is the solution for y^{''}+4y^'+4y=(5+x)e-2x ?

The solution for y^{''}+4y^'+4y=(5+x)e-2x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+((e-2)x)/4+(4e+2)/4

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y^{''}+4y^'+4y=(5+x)e-2x

y^{^{''}} + 4 y^{^{'}} + 4 y = (5 + x) e - 2 x

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{( e - 2 ) x}{4} + \frac{4 e + 2}{4}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = (5 + x) e - 2 x

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{( e - 2 ) x}{4} + \frac{4 e + 2}{4}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{( e - 2 ) x}{4} + \frac{4 e + 2}{4}

y^{^{'''}} - 3 y^{^{'}} + 2 y = 8 e^{2 x}

y^{^{''}} - 4 y^{^{'}} = 8 x e^{3 x}

y^{^{''}} - 6 y^{^{'}} + 9 y = 4 e^{x}

y^{^{''}} - 4 y^{^{'}} + 7 y = t e^{t}

y^{^{''}} + 9 y = sin (2 x), y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}+4y^'+4y=(5+x)e-2x ?
The solution for y^{''}+4y^'+4y=(5+x)e-2x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+((e-2)x)/4+(4e+2)/4