MATH 2280-002
Lecture Notes: 01/14/2013
Math 2280 Section 002 [SPRING 2013]
1
Review
We’ve already seen some important techniques for solving certain types of differential equations.
What DE can we solve so far?
• We can solve nth -order DE of the form
dn y
= f (x)
dxn
by integrating n times.
• We can also solve separable DE
dy
= g(y)f (x)
dx
using separation of variables.
Also, slope fields give us a technique for understanding the graphs of solutions to a first-order DEQ.
2
Models involving Separable DE’s
Separable differntial equations appear frequently in science or engineering when modeling various
real-world situations. Here are a few examples:
dy
dx
= ky
dP
= kP
– Population Change.
dt
(P = population, k = birth rate - death rate)
dA
– Compound Interest.
= rA
dt
(A = account balance, r = annual interest rate)
dN
– Radioactive Decay.
= −kN
dt
(N = number of atoms of radioactive isotope, k = decay constant)
• Equations of the form
dy
= k(ay − b)
• Equations of the form dx
dI
1
– RL Circuit.
= (V − IR)
dt
L
(I = current, L = inductance, V = voltage, R = resistance)
dT
= −k(T − A)
– Newton’s Law of Cooling.
dt
(T = temperature of an object, A = ambient temperature)
dy
– National Income.
= k(1 − L)(y − N )
dt
(y = national income, L = income spent on luxuries,
N = income spent on necessities)
1
MATH 2280-002
Lecture Notes: 01/14/2013
• Logistic equations, dP
dt = (b − aP )P , often provide a more realistic model than exponential
equations because they take into account the carrying capacity of an ecological environment.
Example. Solve the initial value problem
1
dI
= (V − IR),
dt
L
I(0) = 0,
when V = 100 volts, L = 2 H, R = 20 Ω. When we plug in these numbers and simplify, our
differential equation becomes
dI
= 50 − 10I.
dt
1. Separate the variables.
1
dy = dx
50 − 10I
2. Integrate both sides.
Z
Z
1
dy = dx
50 − 10I
1
− ln |50 − 10I| = x + C1
10
ln |50 − 10I| = −10x + C2 (C2 = −10C1 )
50 − 10I = C3 e−10x (C3 = eC2 )
We argued that we can drop the absolute value because when we looked at the slope field, we
observed that since I(0) = 0, I(t) is a stictly increasing function which takes on values in the
interval [0, 5). Using the initial condition I(0) = 0, we get
50 − 0 = C3 e0 .
That is, C3 = 50, and 50 − 10I = 50e−10x . Solve for I to get our solution
I(t) = 5 − 5e−10t .
Notice that this expression agrees with our observations about the slope field. This is a good way
to check that our answer makes sense.
3
Linear Differential Equations
Definition. A first-order differential equation is called linear if it can be written in the form
dy
+ P (x)y = Q(x).
dx
We usually think of linear as meaning constant coefficient, and notice that Q(x) and P (x) are
functions of x that are constant with respect to y.
We’ll learn many different techniques for solving linear differential equations, and the first technique
we’ll learn involves integrating factors. As usual, we’ll begin by looking at the theory behind the
technique.
2
MATH 2280-002
THEORY:
Lecture Notes: 01/14/2013
We want to make the left-hand side of the DEQ “look
R like” the product rule. To do
this, multiply both sides by the integrating factor e P (x)dx to get
R
R
dy R P (x)dx
e
+ P (x)e P (x)dx y = Q(x)e P (x)dx .
dx
By the product rule,
R
R
d y(x)e P (x)dx = Q(x)e P (x)dx .
dx
R
R
d
We have that dx
y(x)e P (x)dx = Q(x)e P (x)dx . This means that
R
y(x)e
R
Dividing by e
P (x)dx
P (x)dx
=
Z
R
Q(x)e
P (x)dx
We’ll
dx + C
gives
y(x) = e
−
R
P (x)dx
Z
R
Q(x)e
P (x)dx
dx + C .
talk about what this technique looks like in practice tomorrow. I do want to point out that, just
like separation of variables is essentially just the chain rule, the integrating factor technique is just a
clever application of the product rule.
3