ID:
Please Name:
UCLA circle your section:
Yin/Tuesday (3A)
Marshak/Tuesday (3C)
Baron/Tuesday (3E)
Yin/Thursday (3B)
Marshak/Thursday (3D)
Baron/Thursday (3F)
Instructions:
1. Solve all problems completely. Give complete reasoning unless it is stated
that you dont need to.
2. Show all your work. Partial credit will be given where it is possible to
see that you have a partial answer.
3. No calculators are allowed, or necessary.
4. No notes or references (including but not limited to the textbook) are
allowed.
5. You must answer the questions entirely yourself.
6. If you need more space to work, please continue each problem on the back
of the same sheet.
7. Look at all the problems before starting! You may nd that some seem
easier, or parts of them; do those rst.
The actual exam will have 5 questions
Question
Points
1
12
2
12
3
12
4
12
5
12
6
12
Total:
72
Score
Math 3A (Ryan Reich) Midterm 1
Page 1 of 7
1. (12 points) Mark each statement as true or false; one point each, and work will not
be taken into account.
T
T
T
T
T
T
T
T
T
T
T
T
F Every separable dierential equation is linear.
F The dierential equation 4x(1 + 2y 2 )y + 5y = 0 is separable.
F The dierential equation 4x(1 + 2y 2 )y + 5y = 0 has an integrating factor
u(x).
F The dierential equation 4x(1 + 2y 2 ) dy + 5y dx = 0 is exact.
F Any dierential equation in normal form y = f (x, y ) can be put in dierential
form P (x, y ) dx + Q(x, y ) dy , and any equation in dierential form can be put
in normal form.
F Every dierential equation can be put in dierential form P (x, y ) dx +
Q(x, y ) dy = 0.
F The solution to the initial value problem y = 1 + y 2 with y (0) = 0 has an
interval of existence smaller than R = (, ).
F All solutions to an initial value problem have the same interval of existence.
F If two solution curves of a rst-order dierential equation intersect at any
point, then they are equal.
F The dierential equation (x2 + y 2 ) dx + (x2 y 2 ) dy = 0 is homogeneous.
F Every autonomous equation is separable.
F Every equilibrium point of the equation y = (x 2)2 (x 3)2 is stable.
Math 3A (Ryan Reich) Midterm 1
Page 2 of 7
2. (12 points) Find the general solution to
(x2 + 2y ) + (2x 3y )
dy
= 0.
dx
Express the solution for y (x) explicitly if possible; otherwise, implicitly is ne.
Solution: Its not linear and its not separable, so we will have to treat it as a
dierential-form equation:
(x2 + 2y ) dx + (2x 3y ) dy = 0.
Fortunately, this equation is exact:
Dy (x2 + 2y ) = 2
Dx (2x 3y ) = 2
so we can just integrate it term-by-term. First the dx term:
F (x, y ) =
1
(x2 + 2y ) dx = x3 + 2xy + f (y ),
3
where F (x, y ) is the implicit solution we want and f (y ) is a constant of integration.
For compatibility we need
3
?
Dy F (x, y ) = 2x + f (y ) = 2x 3y = f (y ) = 3y = f (y ) = y 2 + C.
2
Therefore, the solution is
3
1
F (x, y ) = x3 + 2xy y 2 = C.
3
2
Math 3A (Ryan Reich) Midterm 1
Page 3 of 7
3. (12 points) Use the method of variation of parameters to nd the general solution to
y + y = t.
(Note: if you cannot evaluate an integral, then leave the integral in your solution.)
Solution: First we have to solve the homogeneous equation
yh + yh = 0.
That is, yh = yh , which we recognize as a form of the equation y = ky , whose
general solution is y = Cekt . Therefore yh = Cet . Varying the constant, we guess
y = v (t)et .
Now we enforce that it solves the original equation:
?
y + y = v et vet + vet = t = v et = t.
This is separable, and we can just integrate directly:
v=
tet dt = tet
et dt = tet et + C.
So the general solution is
y = (tet et + C )et = t 1 + Cet .
Math 3A (Ryan Reich) Midterm 1
Page 4 of 7
4. (12 points) Does a solution exist for the following initial value problem? If so, is it
unique?
1
dy
y (3) = 0, t > 0.
t = y t2 ,
dt
3
Solution: is This linear, so we could solve it directly, but it is better to use the
existence theorem. First, we put it in nomal form:
y=
t
y
,
t
3
y (3) = 0,
t > 0.
The right-hand side does have discontinuities whenever t = 0, but we are instructed
only to nd a solution for t > 0, so this is not a problem. Otherwise, it is continuous,
in particular at t = 3, so according to the existence theorem, there is a solution
passing through the initial value (3, 0).
As for uniqueness, we need to check that the partial derivative is also continuous:
y
1
=,
y
t
which it is, as long as t = 0. Therefore the solution through (3, 0) is unique.
Math 3A (Ryan Reich) Midterm 1
Page 5 of 7
5. The Shell oil company wants to add nitrogen to regular gasoline. They claim that
nitrogen enriched gasoline is better because it cleans your engine. Of course, any
scientist who has studied combustion knows that nitrogen is an inert substance, and
therefore this is the equivalent of watering down the gasoline to increase prots.
You have been put in charge of the gasoline processing plant. A 1,600 gal tank initially
contains 1,000 gal of pure gasoline. At time t = 0, an incoming source of nitrogen-enriched
gasoline enters the tank at a rate of 10 gal/min and contains 4 lb/gal of nitrogen. At the
same time, an outow of nitrogen enriched gasoline begins to exit the tank at a rate of
8 gal/min. Assume perfect mixing.
(a) (4 points) Write a dierential equation that describes the nitrogen content x(t) (lbs)
in the tank.
Solution: We need, as usual, the in and the out rates. Gas coming in has
a constant concentration and a constant rate, so xin is just their product. Gas
going out has a constant rate but the concentration depends on the volume.
That gives:
xin = (10 gal/min)(4 lb/gal) = 40 (in lb/min)
x
8x
xout =
(8 gal/min) =
(in lb/min).
volume
1000 + 2t
So the dierential equation is:
x = xin xout = 40
8x
.
1000 + 2t
(b) (4 points) Find the particular solution to the dierential equation from part (a).
Solution: In addition to the equation, we also have the initial condition x(0) = 0,
when there is no nitrogen. The dierential equation is linear but not separable,
so well use an integrating factor to nd the general solution rst. Here the
function a(t) = 8/(1000 + 2t), so the integrating factor is
u = exp
8
dt = exp
1000 + 2t
1
8 ln(1000
2
+ 2t) = (1000 + 2t)4 .
The function f (t) = 40, so the solution after multiplying by the integrating
factor is
ux =
40(1000 + 2t)4 dt = 4(1000 + 2t)5 + C,
x = 4(1000 + 2t) +
C
.
(1000 + 2t)4
or
Math 3A (Ryan Reich) Midterm 1
Page 6 of 7
Imposing the condition x(0) = 0, we nd that
0 = 4(1000) +
Therefore,
C
C = 4(1000)5 .
4=
(1000)
4(1000)5
x(t) = 4(1000 + 2t)
.
(1000 + 2t)4
(c) (4 points) What is the nitrogen content in the tank at the time in which the tank
becomes full? (You may leave the answer as an expression for x without evaluating
it.)
Solution: When the tank is full, it has 1,600 gal of material in it. Since it gains
2 gal/min and has 600 gal to gain, it lls up in t = 300 min. Putting that in the
answer:
x(300) = 4(1600)
4(1000)5
4
5,789
4 = 6,400 4,000 1.6
(1600)
Math 3A (Ryan Reich) Midterm 1
Page 7 of 7
6. Consider the following direction eld plot:
5
4
3
2
1
0
1
2
3
4
5
2
1
0
1
2
3
4
5
6
7
8
9
10
(a) (6 points) Sketch several solution curves on the direction eld plot above.
Solution: (See the plot.)
(b) (6 points) Is the dierential equation autonomous (give justication)? If so, draw a
phase line diagram (you may approximate the locations of any equilibrium points).
Solution: It is autonomous, since (as you can see from the direction eld) it
has horizontal symmetry: the directions are all the same on any horizontal line,
so are independent of the time coordinate. The phase line diagram is:
3?
0
3?