1. Straight-line solutions
SECTION 3.2

2. Summary of the handout Consider a two-variable system dY/dt = AY.
Suppose we can find two linearly independent solutions Y1(t) and Y2(t). Then the general solution of the system is Y(t) = k1Y1(t) + k2Y2(t).
If the system has two families of straight-line solutions, we can use these to find Y1 and Y2.
If we can find a vector V such that V is proportional to AV with proportionality constant , then Y(t) = etV is a straight-line solution to dY/dt = AY. Here’s why:
dY/dt = (d/dt)(etV) = etV = et(V) = etAV
AY = A (etV) = etAV, so the DE is satisfied.
Therefore, if A has two eigenvalues, we can use this technique to find the general solution to dY/dt = AY.

3. Exercises
p. 271: 1, 3, 5, 7