Differential Equations w/ Eigenvectors

Question

Find the solution to the differential equation $\frac{d^2x}{dt^2} = -x$ with the initial conditions $x(0) = 5,$ $\frac{dx}{dt}(0) = 0$. The initial condition, $\vec{v} = \begin{bmatrix} 5 \\ 0 \end{bmatrix}$ as a linear combination of the eigenvectors $\begin{bmatrix} i \\ 1 \end{bmatrix}$ and $\begin{bmatrix} i \\ -1 \end{bmatrix}$.

So, we know that $\begin{bmatrix} x'(t) \\ y'(t) \end{bmatrix}$ = A$\begin{bmatrix} x(t) \\ y(t) \end{bmatrix}$, thus:

$$\begin{bmatrix} 5 \\ 0 \end{bmatrix} = c_1 e^\text{?} \begin{bmatrix} i \\ 1 \end{bmatrix} + c_1 e^\text{?} \begin{bmatrix} i \\ -1 \end{bmatrix}$$

I understand this much, but am confused about how to go about finding the eigenvectors and then the role $\frac{d^2x}{dt^2} = -x$ plays in it?

Answer 1

Let $$u_1=x$$ and $$u_2=x'$$

Your equation $$x''+x=0$$ transform to a system of $$ u'=Au$$ where $$A= \begin {bmatrix}0&1\\-1&0\end {bmatrix}$$

This matrix $A$ has the eigenvalues $i$ and $-i$ and eigenvectors as mentioned in your problem.

You may find $C_1$ and $C_2$ by your initial conditions.

Answer 2

The simpler approach is to note that the equation $x''(t)=-x(t)$ has the obvious solutions $ x(t)=\sin t$ and $x(t)=\cos t$ and, since it is a second order equation, the general solution has the form $$ x=C_1 \sin t + C_2 \cos t $$

and, from the initial conditions we have:

$x'(0)=0 \Rightarrow C_1\cos(0) - C_2\sin(0)=0 \Rightarrow C_1=0$

$ x(0)=5 \Rightarrow C_1\sin(0)+C_2 \cos(0)=5 \Rightarrow C_2=5$