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

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

$$ \lambda^2 + 4\lambda + 13 = 0 \!$$

The initial conditions are:

$$ y(0) = 1 \qquad y'(0) = 0 \!$$

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

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

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

$$ \lambda_1 = -2 + 3i \qquad \lambda_2= -2 - 3i$$

We can now write the general solution to the problem as

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

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.

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

Using the second pair of initial conditions yields

$$y'(0) = 0 = -2 \color{Red} (1) \color{Black} e^{0} \cos(0) + 3\color{Red} C_2 \color{Black} e^{0} \ cos(0) \color{Red} \rightarrow C_2 = 2/3$$

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