Solving a system of ordinary differential equations

Question

Consider the simultaneous system of differential equations: $$ \begin{equation} x'(t)=y(t) -x(t)/2\\ y'(t)=x(t)/4-y(t)/2 \end{equation} $$ If $ x(0)=2 $ and $ y(0)=3 $, then what is $ \lim_{t\to\infty}((x(t)+y(t)) $?

Here is what I do:

$$ \frac{dy}{dx}=\frac{\frac{1}{4}x-\frac{1}{2}y}{-\frac{1}{2}x+y}=-\frac{1}{2} $$ So $$ y=-\frac{1}{2}x+4 $$ and $$ x(t)+y(t)=\frac{1}{2}x(t)+4 .$$ Now solve for $ x(t) $, we have $$ x(t)=4-\frac{2}{e^t} .$$ Hence $ \lim_{t\to\infty}((x(t)+y(t))=2+4=6 $.

However, there should be another method involving using matrices in the standard way. How to do it via matrices?

The question is from:(14) of https://math.uchicago.edu/~min/GRE/files/week2.pdf

Answer 1

The associated matrix system is $$\begin{bmatrix}-\frac12&1\\\frac14&-\frac12\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x'\\y'\end{bmatrix}$$ The square matrix has eigenvalues 0 and $-1$, with corresponding eigenvectors $(2,1)^T$ and $(2,-1)^T$. Therefore $$\begin{bmatrix}x\\y\end{bmatrix}=a\begin{bmatrix}2\\1\end{bmatrix}+be^{-t}\begin{bmatrix}2\\-1\end{bmatrix}$$ Solving for the initial conditions yields $2a+2b=2$ and $a-b=3$ or $a=2$ and $b=-1$. In the limit $t\to\infty$, the $b$ term tends to zero, leaving $x(t)+y(t)=2(2+1)=6$.

Answer 2

The matrix of coefficients is $$A=\begin {bmatrix}-1/2&1\\1/4&-1/2\end {bmatrix}$$ with eigenvalues of $-1$ and $0$

The eigenvectors are $\begin {pmatrix}2\\-1\end {pmatrix}$ and $\begin {pmatrix}2\\1\end {pmatrix}$

Thus your sollutions are $$\begin {pmatrix}x\\y\end {pmatrix}= c_1e^{-t} \begin {pmatrix}2\\-1\end {pmatrix} +c_2\begin {pmatrix}2\\1\end {pmatrix}$$

You find the constants from initial values.