Answer
y = $C_1e^{-\frac{1}{2}x}$ + $C_2xe^{-\frac{1}{2}x}$
Work Step by Step
Question: 4y"+4y'+y=0
We know:
y = $e^{rx}$
y' = $re^{rx}$
y" = $r^{2}e^{rx}$
This results in $(4r^{2}+4r+1)e^{rx}$=0
The corresponding characteristic equation is:
$4r^{2}+4r+1=0$
Solving the characteristic equation using the quadratic formula:
$\frac{-b\frac{+}{-}\sqrt {b^{2}-4*a*c}}{2*a}$ $\frac{-4\frac{+}{-}\sqrt {4^{2}-4*4*1}}{2*4}$
$r=-\frac{1}{2}$ (twice)
Because we have to same solution twice we need to add an extra x to one of the solutions
Adding the constants gives the following general solution:
y = $C_1e^{-\frac{1}{2}x}$ + $C_2xe^{-\frac{1}{2}x}$
