# Auxillary Equation

• Jun 9th 2009, 05:56 AM
Redeemer_Pie
Auxillary Equation
Hi all i need some help with understanding auxillary equation based on boundary values

An example was:

y" + 2y' + y

with y(0) = 1, y(1) = 3

thus putting it into the auxillary equation and finding the roots:

r1 = r2 = -1

putting it into the formula gave

y = C1e^(-x) + C2xe^(-x) note that C1 is C with subscript one and so on :3

Heres the part i dont get. In my book it says

since y(0) = 1 we have 1 = y(0) = C1

Further, as y(1) = 3 we have

3 = y(1) = e^(-1) + C2e^(-1)

So C2 = 3e - 1 and therefore:

y = e^(-x) + (3e-1)xe

Can someone help me with this (could you also explain why C1 disappeared)? :(

thanks
• Jun 9th 2009, 06:19 AM
Ruun
It didn't dissappear. If \$\displaystyle C_1=1\$, then \$\displaystyle C_1e^{-x}=e^{-x}\$

(Hi)

-edit- forget the explanations:

\$\displaystyle y(0)\$ and \$\displaystyle y(1)\$ are given to know the values of \$\displaystyle C_1\$ and \$\displaystyle C_2\$. The solution is fome function \$\displaystyle y(x)\$ so plugging \$\displaystyle x=0\$ to obtain \$\displaystyle y(0)\$ and the same to \$\displaystyle x=1\$ for \$\displaystyle y(1)\$, you're able, as you've done, to compute the values of \$\displaystyle C_1\$ and \$\displaystyle C_2\$-
• Jun 9th 2009, 06:47 AM
Redeemer_Pie
thanks again mate :)