Download MAT B44 Midterm Exam , Wednesday, November 2, 2011 5 pm – 7

MAT B44 Midterm Exam , Wednesday, November 2, 2011
5 pm – 7 pm
No books or calculators are allowed.
SOLUTION
1. (20 points) Suppose the birth rate (births per unit time per unit
population) is proportional to the population, while the death rate (deaths
per unit time per unit population) is independent of the population.
(a) Find a differential equation satisfied by the population P (t) as a
function of time t.
Solution: The birth rate is
B = bP
while the death rate is
D=d
where b and d are positive constants. Then the rate of change of the population is
dP/dt = (bP − d)P =
(b) Solve this differential equation. Solution:
1
dP = dt
P (bP − d)
B
A(bP − d) + BP
1
A
+
=
=
P
bP − d
P (bP − d)
P (bP − d)
so Ab + B = 1 and −dA = 1 so A = −1/d and B = 1 − b/d. Integrating,
A ln(P ) + B/b ln(bP − d) = t + C
or
P A (bP − d)B/b = Ket
where K = eC is a nonnegative constant.
2. (13 points) Find the solution of the difference equation
yn+1 =
n
yn
n+1
1
for integers n ≥ 1. (with y1 = 1 Give a proof by induction. If you are unable
to give a proof by induction, write the formula for yn for n = 2, 3, 4, 5 and
write your guess for general n.
Solution:
y2 = 1/2y1 = 1/2
y3 = 2/3y2 = (2/3)(1/2) = 1/3
y4 = 3/4y3 = 3/41/3 = 1/4
y5 = 4/5y4 = 4/51/4 = 1/5
Guess formula:
yn = 1/n
for n ≥ 2.
Proof by induction:
Step 1: True for n = 2 (obvious)
Step 2: Suppose true for n − 1 in other words suppose yn−1 = 1/(n − 1).
Then yn = (n−1)/n·yn−1 (recursion relation) = (n−1)/n·1/(n−1) (because
it’s assumed true for n − 1) = 1/n as we wanted to prove.
3. (24 pts) (a) Show that the equation
ydx + (2x − yey ))dy = 0
is not exact but becomes exact after multiplication by the integration factor
y k for some real number k. Find a value of k for which the equation is exact
after multiplying by y k .
Solution:
We want k for which
y k (ydx + (2x − yey ))dy = 0)
is exact. Write
M = y k+1,
N = y k (2x − yey )
This is exact iff My = Nx , in other words
(k + 1)y k = 2y k
so k = 1 is the solution.
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(b) Then find the general solution of the equation.
We want to find F (x, y) such that
y2 =
∂F
∂x
and
∂F
.
∂y
y(2x − yey ) =
Integrating the first equation gives
F (x, y) = xy 2 + h(y).
So
∂F
= 2xy + dh/dy
∂y
and this is equal to
2xy − y 2ey
by the second equation. Hence
dh/dy = −y 2 ey
Integrate by parts:
Z
y 2 ey dy =
Z
udv = uv −
Z
vdu
where u = y 2 and dv = ey dy So this equals
2 y
y e −
Z
= uv −
Z
y
2yey dy
ye dy =
(u = y, v = ey dy)
This means
Z
Z
Z
udv
vdu = yey − ey
y 2ey dy = y 2ey − 2(yey − ey )
= y 2 ey − 2yey + 2ey .
3
So
h = −(y 2 ey − 2yey + 2ey )
and
F = xy 2 − y 2 ey + 2yey − 2ey .
The solution is
F (x, y) = C
4. (23 pts) Find the general solution of the problem
y ′′ − 2y ′ + y = 2ex + 4
Solution:
Undetermined coefficients guess for
y ′′ − 2y ′ + y = 4
is y = 4.
Repeated roots: characteristic equation (r − 1)2 = 0 so 2 linearly independent solutions to the homogeneous equation are ex , xex . So guess for
y ′′ − 2y ′ + y = 2ex
is
y = Ax2 ex .
Substitute:
y ′ = (2Ax)ex + Ax2 ex
y ′′ = A(x2 + 4x + 2)ex .
So
A(x2 + 4x + 2 − 4x − 2x2 + x2 )ex = 2ex .
This gives A = 1.
5. (20 pts) Use the method of reduction of order to solve the following
equation
(x − 1)y ′′ − xy ′ + y = 0, x > 1
given that one solution is y1 (x) = ex .
4
Solution: Look for a solution y2 = y1 v. Then
y2′ = y1′ v + y1 v ′
and y2′′ = y1 v ′′ + 2y1′ v ′ + y1′′ v Let t = x − 1, then d/dt = d/dx. Our equation
becomes
ty ′′ − (t + 1)y ′ + y = 0
y ′′ − (1 + 1/t)y ′ + 1/ty = 0
By a formula taught in class,
y2 = vy1
where (let dv/dt be denoted by w)
w=
e−
R
P dt
y12
where P is the coefficient of y ′ in the preceding equation; in this case P =
−(1 + 1/t).
(see text
section 3.4 (30))
R
R
So − P dt = (1 + 1/t)dt = t + ln t
e−
v=
Z
e−
R
R
P dt
P dt
= tet
/y12 = te−t
te−t dt = C − te−t − e−t
So the solution is
y2 = y1 v = et v = Cet − t − 1 = Cet − x
(or equivalently x).
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