What is the solution for y^{''}+3y^'+2y= 1/(1+e) ?

The solution for y^{''}+3y^'+2y= 1/(1+e) is y=c_{1}e^{-t}+c_{2}e^{-2t}+1/(2(1+e))

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y^{''}+3y^'+2y= 1/(1+e)

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{1}{1 + e}

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

Solution steps

y^{''} (t) + 3 y^{'} (t) + 2 y = \frac{1}{1 + e}

Solve linear ODE:

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

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

\int 1 + x d x

\int x^{5} (1 + x^{3})^{\frac{3}{2}} d x

y^{^{'}} = 8 y^{2} + x y^{2}, y (0) = 1

4 (sin (t) y^{^{'}} + cos (t) y) = cos (t) sin^{5} (t)

\frac{2 x - 1}{x + 3}, at x = 4

What is the solution for y^{''}+3y^'+2y= 1/(1+e) ?
The solution for y^{''}+3y^'+2y= 1/(1+e) is y=c_{1}e^{-t}+c_{2}e^{-2t}+1/(2(1+e))