y" + p(x)y' + q(x) = 0 (*)
Fix x0, y0, y0'. Then there is a unique solution y to (*) such that y(x0) = y0 and y'(x0) = y0' . I drew this out.
For example for y" + y = 0 we have linearly independent solutions cos(x) and sin(x), so the general solution is c1 cos(x) + c2 sin(x). With x0 = 0, this leads to c1 = y0, c2 = y0' : very nice. A pair of solutions to (*) like this is called "normalized at x0."
Then consider y" - y = 0 with linearly independent solutions e^x and e^{-x}. They are not normalized at 0, and I'll find that
cosh(x) = (e^x + e^{-x})/2 , sinh(x) = (e^x - e^{-x})/2
is a normalized pair. These are "hyperbolic functions," and they satisfy many properties analogous to properties of the "circular functions" cos, sin. For example cosh' = sinh, sinh' = cosh. Also, cosh^2 - sinh^2 = 1, which means that they parametrize the right branch of the hyperbola x^2 - y^2 = 1 by x = cosh(t), y = sinh(t). The parameter t is twice the area of the region above the x-axis, to the left of the hyperbola, and below the line joining the origin to (x,y) (just as in the circular case).
Then I'll point out that the following are equivalent:
w = y1 y2' - y1' y2
must is nonzero.
Thus: w is either never zero (in which case y1 and y2 are linearly independent) or always zero (in which case y1 and y2 are linearly dependent). EP represents this as a good way to check linear independence. This is silly, but w is still very useful. One application occurs in PS3. Here's another.
You notice that the zeros of cos and sin alternate. This is not accidental. Suppose y1, y2 is ANY pair of linearly independent solutions to ANY homogeneous linear ODE (*) and suppose y1(a) = 0. Then y1'(a) is nonzero, since if y1'(a) = 0 then y1 and 0 would satisfy the same initial condition at a and so would coincide.
Assume y1'(a) is greater than 0. Let b be the next zero of y1, so that y1'(b) is less than 0.
w(a) = y1(a) y2'(a) - y1'(a) y2(a) = - y1'(a) y2(a)
so signs of y2(a) and w(a) are opposite.
w(b) = y1(b) y2'(b) - y1'(b) y2(b) = - y1'(b) y2(b)
so signs of y2(b) and w(b) are the same.
But w is never zero, so w(a) and w(b) must have the same sign. Therefore y2(a) and y2(b) have opposite signs, and so y2 must be zero somewhere between a and b.
The same argument works if instead y1'(a) is less than 0, and if you apply it with y1 and y2 reversed you see that y2 will have no more zeros between a and b: the zeros of y1 and y2 alternate.