4First-order differential equations
IA Differential Equations
4.6 Solution curves (trajectories)
Example. Consider the first-order equation
dy
dt
= t(1 − y
2
).
We can solve it to obtain
dy
1 − y
2
= t dt
1
2
ln
1 + y
1 − y
=
1
2
t
2
+ C
1 + y
1 − y
= Ae
t
2
y =
A − e
−t
2
A + e
−t
2
We can plot the solution for different values of
A
and obtain the following graph:
t
y
−1
1
But can we understand the nature of the family of solutions without solving
the equation? Can we sketch the graph without solving it?
We can spot that
y
=
±
1 are two constant solutions and we can plot them
first. We also note (and mark on our graph) that y
0
= 0 at t = 0 for any y.
Then notice that for t > 0, y
0
> 0 if −1 < y < 1. Otherwise, y
0
< 0.
Now we can find isoclines, which are curves along which
dy
dt
(i.e.
f
) is constant:
t
(1
− y
2
) =
D
for some constant
D
. Then
y
2
= 1
− D/t
. After marking a few
isoclines, can sketch the approximate form of our solution:
t
y
−1
1
In general, we sketch the graph of a differential equation
dy
dt
= f(t, y)
by locating constant solutions (and determining stability: see below) and iso-
clines.