San José State University
Math 133A, Fall 2005
Solutions to graded Homework 2 problems
Ex. 1.3, #14. Denote the values of y for which f (y) = 0 by y0 , y1 , and y2 , where y0 < 0 < y1 < y2 .
These are the equilibria of the differential equation dy/dt = f (y). That is, for each i = 0, 1, 2,
the function y∗ (t) ≡ yi is an equilibrium solution. Observe that f (y) > 0 for y0 < y < y1 and
y > y2 . Therefore, whenever a solution y(t) is in any of these intervals, it must increase (since
then dy/dt > 0). If y < y0 or y1 < y < y2 , we have f (y) < 0, so if a solution y(t) is any of
these intervals, it must decrease. Interpreting this in terms of slope fields, we get the following
sketch.
Ex. 1.3, #16. (a) We can sketch the whole slope field. The reason is that f on the right-hand side
of the equation does not depend on y, which means that the slope field doesn’t change if we
move in the vertical (y-) direction. So if we shift the given solution up or down by any amount,
we will obtain another solution.
(b) Based on (a), to get the solution with y(0) = 2, we just need to shift the given solution
(which satisfies y(0) = 1) by one unit up.
Ex. 1.3, #20. If 0 ≤ t < 3, then V (t) = K, so the equation we need to solve is
dvc
K − vc
=
.
dt
RC
Separation of variables gives
− log |K − vc | =
t
+ A,
RC
for some constant A. Solving for vc , we obtain
vc (t) = K + Be−t/RC ,
for some constant B. Since vc (0) must equal 1, we can solve for B and obtain B = 1 − K.
Therefore,
vc (t) = K + (1 − K)e−t/RC .
If t > 3, then the equation is dvc /dt = −vc /RC. Its solution (again by separation of variables)
is
vc (t) = De−t/RC ,
for some constant D. Remember that we are modeling an RC circuit, so our solution vc (t)
should be continuous for all t. In other words, the value of the first solution at t = 3 should
equal the value of the second solution at t = 3. This gives us the equation
K + (1 − K)e−3/RC = De−3/RC .
Therefore, D = K(e3/RC − 1) + 1. To summarize,
(
K + (1 − K)e−t/RC
vc (t) =
[K(e3/RC − 1) + 1]e−t/RC
Ex. 1.4, #2. Using EulersMethod, we get the following result:
if 0 ≤ t ≤ 3,
if t ≥ 3.