## Calculus: Early Transcendentals 8th Edition

y = $C_1e^{-\frac{1}{2}x}$ + $C_2xe^{-\frac{1}{2}x}$
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 = $C_1e^{-\frac{1}{2}x}$ + $C_2xe^{-\frac{1}{2}x}$