What is the solution for y^{''}+4y^'+4y=(7+x)e-2x ?

The solution for y^{''}+4y^'+4y=(7+x)e-2x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+((e-2)x)/4+(6e+2)/4

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y^{''}+4y^'+4y=(7+x)e-2x

y^{^{''}} + 4 y^{^{'}} + 4 y = (7 + x) e - 2 x

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{( e - 2 ) x}{4} + \frac{6 e + 2}{4}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = (7 + 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{6 e + 2}{4}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{( e - 2 ) x}{4} + \frac{6 e + 2}{4}

y^{^{''}} - y^{^{'}} - 6 y = e^{t}

\frac{d ^{2} y}{d x ^{2}} + 8 \frac{d y}{d x} + 25 y = 16 sin (x)

y^{^{''''}} - y = cos (x)

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

y^{^{''}} + 10 y^{^{'}} + 9 y = 6 cos (3 t + 1.5708) H (t)

What is the solution for y^{''}+4y^'+4y=(7+x)e-2x ?
The solution for y^{''}+4y^'+4y=(7+x)e-2x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+((e-2)x)/4+(6e+2)/4