4—Differential Equations 80As a check for the plausibility of this result, look at the special case of small times. Use the power
series expansion of the cosine, keeping a couple of terms, to get
x(t)≈
F 0
mω^20
[
1 −
(
1 −(ω 0 t)^2 / 2
)]
=
F 0
mω 02
ω 02 t^2
2
=
F 0
m
t^2
2
and this is just the result you’d get for constant accelerationF 0 /m. In this short time, the position
hasn’t changed much from zero, so the spring hasn’t had a chance to stretch very far, so it can’t apply
much force, and you have nearly constant acceleration.
This is a sufficiently important subject that it will be repeated elsewhere in this text. A completely
different approach to Green’s functions will appear is in section15.5, and chapter 17 is largely devoted
to the subject.
4.7 Separation of Variables
If you have a first order differential equation — I’ll be more specific for an example, in terms ofxand
t— and if you are able to move the variables around until everything involvingxanddxis on one
side of the equation and everything involvingtanddtis on the other side, then you have “separated
variables.” Now all you have to do is integrate.
For example, the total energy in the undamped harmonic oscillator isE=mv^2 /2 +kx^2 / 2.
Solve fordx/dtand
dx
dt
=
√
2
m
(
E−kx^2 / 2
)
(4.36)
To separate variables, multiply bydtand divide by the right-hand side.
dx
√
2
m(
E−kx^2 / 2
)=dt
Now it’s just manipulation to put this into a convenient form to integrate.
√m
k
dx
√
(2E/k)−x^2
=dt, or
∫
dx
√
(2E/k)−x^2
=
∫ √
k
m
dt
Make the substitutionx=asinθand you see that ifa^2 = 2E/kthen the integral on the left simplifies.
∫
acosθdθ
a
√
1 −sin^2 θ
=
∫ √
k
m
dt so θ= sin−^1
x
a
=ω 0 t+C
or x(t) =asin(ω 0 t+C) where ω 0 =
√
k/m
An electric circuit with an inductor, a resistor, and a battery has a differential equation for the
current flow:
L
dI
dt
+IR=V 0 (4.37)
Manipulate this into