User:Egm4313.s12.team13.fsp.r2.9

With the given initial conditions, find and plot the solution to

${\displaystyle \lambda ^{2}+4\lambda +13=0\!}$

The initial conditions are:

${\displaystyle y(0)=1\qquad y'(0)=0\!}$

It important to remember, there is no excitation in this problem,

${\displaystyle i.e.\qquad r(x)=0\ }$

According to Kreyszig, this solution can be found by using the quadratic to find the roots. These roots come out to be

${\displaystyle \lambda _{1}=-2+3i\qquad \lambda _{2}=-2-3i}$

We can now write the general solution to the problem as

${\displaystyle y_{h}=e^{-2x}[Acos(3x)+Bsin(3x)]\!}$

it is now possible to apply the initial conditions to this solution, since we have two, it is necessary to take the derivative of the general solution.

${\displaystyle y(0)=e^{0}[A\cos(0)+B\sin(0)]=1\color {Red}\rightarrow A=1}$

Using the second pair of initial conditions yields

${\displaystyle y'(0)=0=-2\color {Red}(1)\color {Black}e^{0}\cos(0)+3\color {Red}C_{2}\color {Black}e^{0}\ cos(0)\color {Red}\rightarrow C_{2}=2/3}$

${\displaystyle y_{h}=e^{-2x}[cos(3x)+{\frac {2}{3}}sin(3x)]\!}$

File:Exponentialgraph.jpg