Mathematical Tools for Physics

4—Differential Equations 95

This series looks suspiciously like the series for the sine function, but is has some of thex’s or some of the
factorials in the wrong place. You can fix that if you multiply the series in brackets byx. You then have

y(x) =a 0 x−^1 /^2

[

x−

x^3

3!

+

x^5

5!

−...

]

=a 0

sinx

x^1 /^2

(18)

I’ll leave it to problem 15 to find the other solution.

4.4 Trigonometry via ODE’s
The differential equationu′′=−uhas two independent solutions. The point of this exercise is to derive all (or at
least some) of the standard relationships for sines and cosinesstrictly from the differential equation.The reasons
for spending some time on this are twofold. First, it’s neat. Second, you have to get used to manipulating a
differential equation in order to find properties of its solutions. This is essential in the study of Fourier series later.
Two solutions can be defined when you specify boundary conditions. Call the functionsc(x)ands(x), and
specify their respective boundary conditions to be

c(0) = 1, c′(0) = 0, and s(0) = 0, s′(0) = 1 (19)

What iss′(x)? First observe thats′satisfies the same differential equation assandc:

u′′=−u =⇒ (u′)′′= (u′′)′=−u′, and that shows the desired result.

This in turn implies thats′is a linear combination ofsandc, as that is the most general solution to the original
differential equation.
s′(x) =Ac(x) +Bs(x)

Use the boundary conditions:
s′(0) = 1 =Ac(0) +Bs(0) =A

From the differential equation you also have

s′′(0) =−s(0) = 0 =Ac′(0) +Bs′(0) =B

Put these together and you have

s′(x) =c(x) And a similar calculation shows c′(x) =−s(x) (20)