MATH 2280-002
Lecture Notes: 01/15/2013
Math 2280 Section 002 [SPRING 2013]
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Linear Differential Equations (Continued)
Last time, we talked about linear differential equations and discussed a way to solve DE’s of the form
y 0 + P (x)y = Q(x).
Today we will write down explicitly the steps involved in solving a linear DE of this form using the
integrating factor method.
TECHNIQUE:
Step 1. Compute the integrating factor u(x) = e
R
P (x)dx .
Step 2. Multiply both sides by the integrating factor.
Step 3. Rewrite the resulting DE as
R
R
d y(x)e P (x)dx = Q(x)e P (x)dx .
dx
Step 4. Integrate both sides.
Step 5. Solve for y.
Example. Solve the differential equation
(x2 + 1)y 0 + 3xy = 6x.
To begin with, we must rewrite our equation in the form y 0 + P (x)y = Q(x) so that we can use
our integrating factors method. Our differential equation becomes
y0 +
In this case, P (x) =
3x
x2 +1
and Q(x) =
3x
6x
y= 2
.
+1
x +1
x2
6x
.
x2 +1
1. Compute the integrating factor u(x).
R
u(x) = e
3x
dx
x2 +1
3
= e 2 ln |x
2 +1|
.
Notice that when we integrated, we didn’t worry about a constant. That is because our method
only requires some anti-derivative of P (x) = x23x+1 . Since any antiderivative will work, we just
choose to set the constant term equal to 0. (We could let the constant term be any real number
we want, but this is the simplest choice.) We want to simplify the rightmost expression further,
so we observe that we can drop the absolute value (x2 + 1 is always positive).
3
u(x) = e 2 ln(x
2 +1)
= eln[(x
3
So our integrating factor is just (x2 + 1) 2 .
1
3
2 +1) 2 ]
3
= (x2 + 1) 2 .
MATH 2280-002
Lecture Notes: 01/15/2013
2. Multiply both sides by u(x).
3
3
3x
6x
(x2 + 1) 2 y = 2
(x2 + 1) 2
+1
x +1
3
y 0 (x2 + 1) 2 +
x2
3
1
1
y 0 (x2 + 1) 2 + 3x(x2 + 1) 2 y = 6x(x2 + 1) 2
R
R
d
3. Rewrite the resulting DE as dx
y(x)e P (x)dx = Q(x)e P (x)dx .
3
1
d 2
y(x + 1) 2 = 6x(x2 + 1) 2
dx
4. Integrate both sides.
Z
3
d 2
y(x + 1) 2 dx =
dx
2
3
2
y(x + 1) dx =
Z
6x(x2 + 1) 2 dx
Z
6x(x2 + 1) 2 dx
1
1
(We didn’t write down an integrating constant on the left-hand side because we knew we would
get one on the right-hand side.)
If we’ve done our work correctly, we will always be able to write down an equation that looks
like
Z
R
R
y(x)e P (x)dx = Q(x)e P (x)dx dx,
so once you get comfortable using this method, you’ll probably start skipping steps 2 and 3.
Z
3
1
3
2
2
y(x + 1) dx = 6x(x2 + 1) 2 dx = 2(x2 + 1) 2 + C.
5. Solve for y. Divide by u(x) to get
3
y(x) = 2 + C(x2 + 1)− 2 ,
our final answer.
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Existence and Uniqueness
What we notice is that, as long as Q(x) and P (x) are “nice enough,” we can always our method to
get a unique solution. In other words, we’ve shown the following existence and uniqueness theorem.
Theorem. If the functions P (x) and Q(x) are continuous on an open interval I containing the point
x0 , then the IVP
dy
+ P (x)y = Q(x), y(x0 ) = y0
dx
has a unique solution y(x) on I, given by
Z
R
R
− P (x)dx
P (x)dx
y(x) = e
Q(x)e
dx + C .
Even though this theorem only applies to first-order, linear DE’s, this theorem is very useful because
it tells us exactly what the unique solution has to be. Also, note that here were guaranteed a unique
solution on the entire interval I, not just on some possibly smaller interval like we had for the theorem
from section 1.3. Linear differential equations are nice that way.
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MATH 2280-002
3
Lecture Notes: 01/15/2013
Mixing Tank Problems
The primary application we’ll consider for this section is mixing tank problems. Although our focus
in lectures will be tanks containing brine (salt-water), the same set-up works more generally (for
example, when modeling lakes containing pollution).
Imagine a tank with a mixture of solute (salt) and solvent (water). There’s inflow and outflow, and
we’re interested in the amount of salt in the tank at time t.
•
•
•
•
•
•
x(t) = amount of salt in tank (grams)
V (t) = volume of mixture in tank (liters)
ri = flow rate into tank (liters/second)
ro = flow rate out of tank (liters/second)
ci = concentration into tank (grams/liter)
co = concentration out of tank (grams/liter)
For us, ri and ro will just be constant. Observe that the concentration of the tank co =
x
.
V
The volume V (t) changes at the rate
dV
= ri − ro .
dt
Integrate
dV
dt
= ri − ro to get
V (t) = (ri − ro )t + V0 ,
where V0 = V (0). When the rate of inflow is the same as the rate of outflow, the volume of the tank
will remain constant (as expected). Assuming the mixture is uniform, the amount of salt changes at
the rate
dx
= rate of salt entering - rate of salt leaving
dt
= ri ci − ro co .
This gives us the linear differential equation
dx
x(t)
= ri ci − ro
dt
V (t)
= ri ci − ro
x(t)
.
(ri − ro )t + V0
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MATH 2280-002
Lecture Notes: 01/15/2013
Let’s start an example that you’ll be asked to finish in your homework. This example looks more
complicated than the situation described above because it deals with two tanks instead of one.
Example. Consider the cascade of two tanks where V1 = 100 (gal) and V2 = 200 (gal) are the
volumes of brine in the two tanks. Each tank initially contains 50 lb of salt. The three flow rates
are each r = 5 gal/min, with pure water flowing into Tank 1. Find the amount of salt in Tank 2
at time t.
If the amount of salt in Tank 1 is denoted x, then
dx
x(t)
1
=0−
r = − x(t).
dt
V1
20
Since only pure water is flowing into Tank 1, the concentration of the inflow is 0. We know how to
solve differentials equations that look like this. We use separation of variables. Earlier we showed
that the DE x0 = kx has the solution x(t) = Aekt , so we know the solution here is
t
x(t) = Ae− 20 .
Using the initial condition x(0) = 50, we get that A = 50.
If the amount of salt in Tank 2 is denoted y, then
t
x(t)
1
1
1
dy
y(t)
1
=
· 50e− 20 − y(t).
r−
r = x(t) − y(t) =
dt
V1
V2
20
40
20
40
I’ll leave the rest to you. Note that our differential equations for x and y together form a system
of differential equations. We’ll talk about these in greater detail later in the semester.
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