Math 222
Name:
Quiz 4 Answer
3 Nov 2013
TA: Jiyuan Han
Problem 1 (15 points) The following first-order non-linear differential equation is a type
of “Riccati Equation”:
dy
= y 2 + 3xy
dx
Given the initial value: y(0) = 0, solve the following three questions:
a Find one solution that satisfies this initial value problem.
(You are not trying to solve this equation “generally”. Try to “guess” a simple solution,
and then check that the solution you find sastifies this Riccati differential equation.
b Using the solution you find in (a) to find the exact value of y(1). Then use Euler’s
method to find an approximate value of y(1), with step length h = 12 .
c Again, use Euler’s method to find an approximate value of y(1), with step length h = 31 .
(The result’s of Euler’s method helps to demonstrate your “guessing solution” is correct!)
[ 1]Solution: a. You would probably get an intuitative guessing solution: y(x) = 0. (If you
can not get an intuitative solution, try to solve b, c first. When you use Euler’s method, you
would find at each point, the derivative is 0. So you would probably think y(x) doesn’t change
at all. So y = 0 is a possible solution)
Check y(x) = 0 is a solution to the initial-value problem above.
dy
= 0 anywhere. And also y 2 + 3xy = 0 anywhere. Thus
1. Since y(x) = 0 is constant, then dx
dy
y(x) = 0 satisfies dx
= y 2 + 3xy.
2. When x = 0, y = 0. So y(x) = 0 satisfies the initial condition.
Thus y(x) = 0 is a solution to this initial-value problem.
b. Since y(x) = 0, the exact value of y(1) is 1.
Now use Euler’s method to approximate y(1).
Choose h = 12 .
Then:
y( 21 ) = y(0) + y 0 (0)h = 0 + 0 · h = 0;
y(1) = y( 12 ) + y 0 ( 12 )h = 0 + 0 · h = 0;
So By Euler’s method with h = 21 , y(1) = 0.
c. Now use Euler’s method to approximate y(1) with step length h = 31 .
Choose h = 12 .
Then:
y( 31 ) = y(0) + y 0 (0)h = 0 + 0 · h = 0;
y( 23 ) = y( 13 ) + y 0 ( 13 )h = 0 + 0 · h = 0;
y(1) = y( 32 ) + y 0 ( 23 )h = 0 + 0 · h = 0;
So By Euler’s method with h = 31 , y(1) = 0.
Problem 2 (15 points): Mixture problem: A 200-gal tank is half full of distilled water.
At time t = 0, a solution containing 0.5 lb/gal of concentrate enters the tank at the rate of 5
gal/min, and the well-stirred mixture is withdrawn at the rate of 3 gal/min.
a At what time will the tank be fulled?
b At the time the tank is full, how many pounds of concentrate will it contain?
[ 1]Solution: a. Since at the beginning, the 200-gal tank is half full, the time to fill this tank is:
T =
100
= 50
5−3
b. Since we can use the mass of concentrate m(t) as our variable, we can also use the density
of the concentrate ρ(t) as our variable, there are two ways to build the differential euqation.
(Actually they are the same)
Method 1: use m(t) as variable.
First we build the differential equation:
Let’s use min as the unit of time.
At time t, the volume is:V (t) = 100 + 2t.
Let m(t) denote the mass of concentrate in the tank at time t. Then we have:
m(t + 1) − m(t) = {5 · 0.5 − 3 ·
m(t)
}
V (t)
is the density of the concentrate at time t. And V (t) = 100 + 2t as we got above.
where m(t)
V (t)
Then we have:
dm
m(t)
= {2.5 − 3 ·
}
dt
100 + 2t
Simplify it, we have:
dm
m(t)
+3·
= 2.5
dt
100 + 2t
We solve this first-order linear differential equation by integrating factor:u(t) = e
3
3
e 2 ln(100+2t) = (100 + 2t) 2
Then we have:
Z
3
3
m(t) · (100 + 2t) 2 = 2.5 (100 + 2t) 2 dt
which leads to:
−3
1
· (100 + 2t) + C · (100 + 2t) 2
2
C
Use unitial condition: m(0) = 0, we have: 0 = 50 + 1000
, C = −50000.
Then:
3
m(t) = 50 + t + −50000(100 + 2t)− 2
m(t) =
Method 2: Use ρ(t) as variable.
R
3
100+2t
=
First we build the differential equation:
Let’s use min as the unit of time.
At time t, the volume is:V (t) = 100 + 2t.
Let ρ(t) denote the concentration of solution in the tank at time t. Then we have:
ρ(t + 1) − ρ(t) =
Then we have:
{5 · 0.5 − 3ρ(t)}
V (t)
dρ
2.5 − 3ρ(t)
=
dt
100 + 2t
Simplify this equation, we get:
dρ
3
2.5
+
ρ=
dt 100 + 2t
100 + 2t
3
Use integrating factor to solve this first-order linear differential equation, with m(t) = e 2 ln(100+2t) =
3
(100 + 2t) 2
Then we have:
Z
3
1
2
(100 + 2t) ρ = 2.5 (100 + 2t) 2 dt
Then we have:
3
5
+ C(100 + 2t)− 2
6
With our initial condition: t = 0, ρ(0) = 0, we can fix C = 2500
3
Then we have:
−3
5 2500
ρ= −
(100 + 2t) 2
6
3
Then m(t) = ρ(t) · V (t)
ρ(t) =