Solution to Asymptotic ODE System for Small ε: x and y Expressions

[member 428835]

Homework Statement

Find through $O(\epsilon)$, for $\epsilon \ll 1$, the solution to the system $$\epsilon x'(t) = -x+y\\
\epsilon y'(t) = -(\epsilon+1)y+x$$

Homework Equations

$x = \sum x_n\epsilon^n$ and $y = \sum y_n\epsilon^n$

The Attempt at a Solution

Substituting the $\epsilon$ expansions of $x$ and $y$ into the governing ODE's yields the same equation for $x$ and $y$ equations:
$$O(1): x_0 = y_0.$$
The next two ODE's are
$$O(\epsilon)\: x: x_0'(t) + x_1(t) - y_1(t) = 0\\
O(\epsilon)\: y: y_0'(t) - x_1 + y_0 + y_1 = 0$$
Summing these two ODE's yields
$$O(\epsilon): x_0'+y_0' + y_0 = 0\implies\\
2y_0'+y_0=0\implies \boxed{x_0=y_0=A\exp(-t/2)}.$$
The next two ODE's are
$$O(\epsilon^2)\: x: x_1' + x_2 - y_2 = 0\\
O(\epsilon^2)\: y: y_1' - x_2 + y_1 + y_2 = 0.$$
Summing these two ODE's yields
$$O(\epsilon^2): y_1'+x_1' + y_1 = 0$$

From $O(1)$ equation $x$ we see that $x_0'(t) + x_1(t) - y_1(t) = 0 \implies x_1'=y_1'-x_0''$. Substituting this into the summed equation for $O(\epsilon^2)$ yields
$$O(\epsilon^2):2 y_1'+ y_1 = x_0''\implies\\
\boxed{y_1 = \frac{1}{8} A t \exp(-t/2) + B\exp(-t/2)}.$$
The $O(\epsilon)$ equation in $x$ states $x_0' + x_1 - y_1 = 0 \implies x_1=y_1-x_0'$ which implies
$$\boxed{x_1 = \frac{1}{8} \exp(-t/2) (8 B + A (4 + t))}.$$

How does this look so far?

 

[DrDu]

Note that the original system of equations is easily solved exactly. You could compare the approximation to the exact solution.