108-3: Engineering Mathematics (I)
Exam-1
Time: 2020-07-11 Tue. 8:10AM-10:40AM (No extension)
Contact
Name
E-mail
Location
Extension
Lecturer
Wenson Chang
wenson@ee.ncku.edu.tw
92A31
62392
TA1
TA2
程浩偉
king295705@gmail.com
92912
62400-1612
李庚道
philiplovemaster@gmail.com
92912
62400-1612
Problem 1(25%): An first order differential equation has the form
M (x, y) + N (x, y)y 0 = 0 .
(1)
(a)(10%) Show that there is a potential function φ(x, y) which satisfies
∂φ
= M (x, y) ,
∂x
∂φ
= N (x, y) .
∂y
(2)
(3)
And φ(x, y) = C is the general solution of (1), where C is a constant.
Hint: If f (x(t), y(t)), then by chain rule, we can have
∂f dx ∂f dy
df
=
+
.
dt
∂x dt
∂y dt
(4)
∂M
∂N
=
.
∂y
∂x
(5)
(b)(5%) The exactness property says if
Then a potential function (φ(x, y)) can exist. But, if the equality of (5) doesn’t exist, what should you do to
make it become exact?
(c)(10%) Find the general solution for
2y(1 + x2 ) + xy 0 = 0 .
(6)
Hint: this is one of the selected problems for practice. You may need to find a integrating factor different from
xa y b . Think about the exponential function!
Sol:
(a) If we have (2) and (3), then (1) can be rewritten as
∂φ
∂x
∂φ dx
∂x dx
+
+
∂φ dy
=0 ,
∂y dx
∂φ dy
=0 .
∂y dx
(7)
(8)
Referring to (4), (8) means
dφ
∂φ dx
=
dx
∂x dx
+
1
∂φ dy
=0 .
∂y dx
(9)
Therefore, φ = c is a general solution of (1), where c is a constant.
(b)Check Definition 1.5 on page 33 of the textbook(6ed).
2
(c)(1) you may use the integrating factor of µ = xa ebx as suggested by the textbook(6ed); or (2) you can use
1
by observation or calculating as following.
u = xy
∂M
∂y
−
∂N
∂x
= −
yN − xM
µ(xy)
= e
R
1
xy
1
− xy
d(xy)
=
1
.
xy
(10)
2
Then you can find your solutions written as (1) x2 yex = c; or (2) ln |x2 y| + x2 = c.
Problem 2(15%): Find the general solution for
xy 0 = x cos(y/x) + y .
(11)
Sol:
The differential equation is homogeneous, and y = xu yields the general solution defined by
y
y
sec ( ) + tan ( ) = cx
x
x
(12)
(For the procedure of solving the homogeneous equation, please refer to pages 26-27.)
Problem 3(20%): Find the general solution for
y0 =
2x − 5y − 9
.
−4x + y + 9
(13)
Sol:
Using the result of Problem 15 of Ch1.4 with x = X + 2 and y = Y − 1, we obtain a homogeneous equation.
Solving the homogeneous equation using the steps showed in Ch1.4.1 on pages 26-27, you can get the solution
as what follows.
(2x + y − 3)2 = K(y − x + 3) .
Problem 4(20%):
Prove the properties of the Wronskian test which is listed as follows.
Theorm 2.3 Wronskian Test
Let y1 and y2 be solutions of y 00 + p(x)y 0 + q(x)y = 0 on an open interval I. Then,
1. (10%) Either W (x) = 0 for all x in I, or W (x) 6= 0 for all x in I.
2. (10%) y1 and y2 are linearly independent on I if and only if W (x) 6= 0 on I. Sol:
For the 1st conclusion, begin to write
y100 + py10 + qy1
=
0
y200 + py20 + qy2
=
0.
Multiply the first equation by y2 and the second by −y1 and add the resulting equations to obtain
y100 y2 − y200 y1 + p(y10 y2 − y20 y1 ) = 0 .
2
(14)
Since W = y1 y20 − y2 y10 , then W 0 = y1 y200 − y2 y100 , so
−(W 0 + pW ) = y100 y2 − y200 y1 + p(y10 y2 − y20 y1 ) = 0 .
Therefore,
the Wronskian satisfies the linear differential equation W 0 + pW = 0. This has integrating factor
R
p(x)dx
e
and can be written
0
R
W e p(x)dx
= 0
W
= ce−
R
p(x)dx
.
If c = 0, then this Wronskian is zero fro all x in I. If c 6= 0, then W 6= 0 for x in I because the exponential
function does not vanish for any x.
Next, we prove the 2nd conclusion.
• →: If y1 and y2 are linearly dependent on I, then y1 (x) = Ky2 (x) for all x ∈ I, where K is a real constant.
Then, it is straightforward to get W (x) = 0.
• ←: Assume W (x) = 0 for all x ∈ I and that for some x0 ∈ I, we have y2 (x) 6= 0. That means
y2 (x) 6= 0 ∀ x ∈ J ⊂ I. Now
y10 y2 − y20 y1
d y1 (x)
=
dx y2 (x)
y22
W (x)
= −
2 =0 ,
[y2 (x)]
which means y1 (x) = Cy2 (x) , ∀ x ∈ J. Now consider the initial value problem y 00 + py 0 + qy = 0,
y(x0 ) = Cy2 (x0 ), y 0 (x0 ) = Cy20 (x), which by Thm. ?? has a unique solution ∀ x ∈ I. Clearly y = Cy2 is
a solution and y = y1 is also a solution. Although y1 (x) = Cy2 (x) , ∀ x ∈ J ⊂ I, uniqueness can lead
to y1 (x) = Cy2 (x) , ∀ x ∈ I. Problem 5(20%):
Consider
y0 +
1
y = 3x2 y 3 .
x
1. (5%) Is this a Homogeneous, Reccati or Bernoulli equation?
2. (15%) Find the general solution using the variable transformation you specified in the first question.
Sol:
1. It is a Bernoulli equation.
2. It is a Bernoulli equation with P (x) = 1/x, R(x) = 3x2 , and α = 3. Let v = y −2 and y = v −1/2 , then
y 0 (x)
1
1
− v −3/2 v 0 (x) + v −1/2
2
x
2
v0 − v
x
1
= − v −3/2 v 0 (x)
2
=
3x2 v −3/2
= −6x2 (obtained via multiplying by −2v 3/2 ) .
3
(15)
Now, following to the procedure in Ch1.3, as integrating factor can be obtained as e−
Therefore, multiplying (15) by x−2 gives
x−2 v 0 − 2x−3 v
= −6
(x−2 v)0
= −6
x
−2
v
= −6x + C
v
= −6x3 + Cx2 .
The general solution of the Bernoulli equation is
1
1
=√
. y(x) = p
2
Cx − 6x3
v(x)
4
R
(2/x)dx
= x−2 .