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California State University Northridge
MATH 351: Differential Equations
Midterm Exam 1 Solutions
March 5, 2014. Duration: 75 Minutes. Instructor: Jing Li
Student Name: Student number:
Take your time to read the entire paper before you begin to write, and read each question carefully. Remember that certain questions are worth more points than others. Make a note of the questions that you feel confident you can do, and then do those first: you do not have to proceed through the paper in the order given.
- You have 75 minutes to complete this exam.
- This is a closed book exam, and no notes of any kind are allowed. The use of cell phones, pagers or any text storage or communication device is not permitted.
- Only the Faculty approved TI-30 calculator is allowed.
- The correct answer requires justification written legibly and logically: you must convince me that you know why your solution is correct. Answer these questions in the space pro- vided. Use the backs of pages if necessary.
- Where it is possible to check your work, do so.
- Good Luck!
Problem 1 2 3 4 5 6 7 8 9 Total
Points 12 12 16 12 14 20 32 18 24 160
Your Marks
Question 1. [12 points] Match the differential equation with the appropriate direction field:
(a) [3 points] For d yd x = y + x , the subfigure C in the above figure is the direction field.
(b) [3 points] For d yd x = y − x , the subfigure D in the above figure is the direction field.
(c) [3 points] For d yd x = y (1 − y ), the subfigure B in the above figure is the direction field.
(d) [3 points] For d yd x = x (1 − x ), the subfigure A in the above figure is the direction field.
Question 3. [16 points] (a)[8 points] Without solving the differential equation, find the largest interval where the initial value problem is guaranteed to have a unique solution.
( x^2 − 4) y ′^ + x y = ex 2 , y (0) = π.
Solution: The standard form of the given equation is
y ′^ + x x^2 − 4 y = ex 2 x^2 − 4
with
p ( x ) = x x^2 − 4 and f ( x ) = ex 2 x^2 − 4 Since this equation is a 1st-order linear equation, the problems points are the points at which either p ( x ) or f ( x ) are discontinuous. That is,
x^2 − 4 = 0 =⇒ ( x − 2)( x + 2) = 0 =⇒ x = 2 or x = −2.
Then since − 2 < 0 < 2, the largest interval is
(−2, 2).
(b)[8 points] Consider the following differential equation d y d x
p x y
determine a region of the x y -plane for which the given differential equation would have a unique solution whose graph passes through a point ( x 0 , y 0 ) in the region. Solution: Let f ( x , y ) = p x y , then
- f ( x , y ) = p x y is continuous when x ≥ 0, y ≥ 0 or x ≤ 0, y ≤ 0;
- ∂ f ∂y
( x y ) (^12) − 1 · x =
x p x y Therefore, by the Fundamental Theorem of Existence & Uniqueness, in the region of
R = {( x , y )| x > 0, y > 0, or x < 0, y < 0}
would have a unique solution whose graph passes through a point ( x 0 , y 0 ) in the region.
Question 4. [12 points] Solve the initial value problem
x d y d x
x + 2
y = cos( x ) e − x^ , with y ( π ) = 0.
On what interval around x = π is the solution defined? Solution: Divid the above equation by x , we have
d y d x
x + 2 x y = cos( x ) e − x x
This is a 1st-order linear equation with p ( x ) = 1 + (^2) x and q ( x ) = cos( x ) e
− x x. The integrating factor is
μ ( x ) = e
∫ (^) p ( x ) d x = e
∫ (^) (1+ 2 x ) d x^ = ex +2 ln^ x^ = x^2 ex^.
Multiplying through by this function, we obtain
d d x ( x^2 ex^ y ) = cos( x ) e − x x · x^2 ex^ = x cos( x ).
To integrate the right side, we need to use integration by parts, ∫ x cos xd x =
xd (sin x ) = x sin x −
sin xd x = x sin x − (− cos x ) + c.
Therefore, x^2 ex^ y = x sin x + cos x + c ⇒ y = e − x^ ( sin x^ x + cos x 2 x + (^) xc 2 ). This gives the general solution of the DE. To determine the constant c , we impose the initial condition y ( π ) = 0 ⇒ 0 = e − π ( sin π^ π + cos π 2 π + (^) πc 2 ) ⇒ 0 = e − π (0 − (^) π^12 + (^) πc 2 ) ⇒ c = 1. Thus, the solution of this IVP is
y ( x ) = e − x^ ( sin x^ x + cos x 2 x + (^) x^12 ).
Note that the solution grows unbounded as x approaches 0. So, the interval around x = π over which the solution is well-defined (or valid) is
x ∈ (0, ∞).
Question 6. [20 points] Solve the following differetial equations by using an appropriate sub- stitution :
(a) [10 points] t^2 y ′^ + 2 t y − y^3 = 0, t > 0 [ Hint: This is a Bernoulli equation.] Solution: Rewriting the above equation to
y ′^ +
t y =
t^2 y^3
y ′^ =
t^2 y^3 −
t y
This is a Bernoulli’s equation with n = 3, so using substitution u = y^1 − n^ = y^1 −^3 = y −^2 , we have du d t
= − 2 y −^2 −^1 d y d t
= − 2 y −^3
t^2
y^3 −
t
y
t^2
t
y −^2
i.e., du d t
t^2
t u =⇒ du d t
t u = −
t^2
With p ( t ) = − (^4) t , we have the integrating factor μ ( t ) = e
∫ (^) − 4 t d t^ = e −4 ln^ t^ = t −^4. Hence, d d t ( t −^4 u ) = −
t^2 · t −^4 =⇒ t −^4 u =
t^2 · t −^4 d t + C =⇒ t −^4 u =
t^6 d t + C =
t −^5 + C
Then u =
t −^1 + C t^4
Thus,
y −^2 =
t −^1 + C t^4 =⇒
y = ±
2 + 5 C t^5 5 t
, i.e., y = ±
5 t 2 + 5 C t^5
(b) [10 points] d y d x
x^2 + 3 y^2 2 x y
[ Hint: This is a homogeneous equation.]
Solution: Note that
d y d x
x^2 + 3 y^2 2 x y
( (^) y x
2 yx Using substitution of v = (^) xy , i.e., y = v x =⇒ d yd x = v + x d vd x , we have
v + x d v d x
1 + 3 v^2 2 v =⇒ x d v d x
1 + 3 v^2 2 v − v = 1 + v^2 2 v
2 v 1 + v^2 d v =
x d x =⇒
∫ (^2) v 1 + v^2 d v =
x d x
Then ln( v^2 + 1) = ln | x | + C =⇒ ln
[( (^) y x
]
= ln | x | + C =⇒
( (^) y x
y^2 + x^2 = C x^3.
Question 7. [32 points] Consider the first order autonomous differential equation d y d t = y^2 (16 − y^2 )
Answer the following questions without solvoing the equation.
(a) [6 points] Denote f ( y ) = y^2 (16 − y^2 ), sketch the graph of f ( y ) versus y ; Solution: The graph of f ( y ) versus y is given as follow
−100−10 −8 −6 −4 −2 0 2 4 6 8 10
−
−
−
−
0
20
40
60
80
100
y
f(y)
Figure 3: The graph of f ( y ) versus y
(b) [4 points] Determine the equilibrium solutions of the given differential equation; Solution: Solving f ( y ) = 0, i.e., y^2 (16 − y^2 ) = 0, we get the equilirium solutions: y = 0, 4, −4.
(c) [7 points] Draw the phase line for the given differential equation; Solution: The phase line diagram is y
−
0
4
Figure 4: The phase line diagram
Question 8. [18 points] A 120-gallon tank initially contains 90 lb of salt dissolved in 90 gal of water. Water containing 2 lb/gal of salt enters the tank at a rate of 4 gal/min. The well-stirred mixture flows out of the tank at a rate of 3 gal/min.
(a) [5 points] Set up an initial value problem (give both a differential equation and an initial condition) modeling this process (prior to the instant when the solution begins to overflow). Solutions: Let Q ( t ) denote the amount of salt in the tank at time t , then
- Rate in = 2 · 4 lb/min;
- Rate out (^90) + Q (4( t −^ )3) t · 2 lb/min. (Here we note that inflow and outflow are at different rates, therefore the water volume in the tank is changing according to 90 + (4 − 3) t .) We also note that the solution will begin to overflow when
90 + (4 − 3) t = 120 =⇒ t = 30mins
There fore the initial value problem is
dQ ( t ) d t
Q ( t ) 90 + (4 − 3) t
dQ d t =^8 −^
3 90 + t · Q , ( t^ <^ 30), with I.C.^ Q (0)^ =^ 90.
(b) [9 points] Solve the initial value problem in part (a) to find the amount of salt in the tank at any time prior to the instant when the solution begins to overflow. Solutions: We need to solve dQ d t
90 + t · Q = 8, t < 30
Subject to Q (0) = 90. The differential equation above is a 1st-order linear equation with p ( t ) = 3 90 + t , thus the integrating factor is μ ( t ) = e
∫ (^) p ( t ) d t = e
∫ (^3) 90 + t d t^ = e 3 ln( t^ +90)^ = ( t + 90)^3
Hence,
d d t
[
( t + 90)^3 Q
]
= 8 · ( t + 90)^3 =⇒ ( t + 90)^3 Q =
8 · ( t + 90)^3 d t + C =⇒ ( t + 90)^3 Q = 2( t + 90)^4 + C
=⇒ Q = 2( t + 90) + C ( t + 90)−^3
Applying I.C. Q (0) = 90, we have
90 = 2 · 90 + C (0 + 90)−^3 =⇒ C = − 904.
Therefore, the amount of salt in the tank at time t prior to the instant when the solution begins to overflow is given by
Q ( t ) = 2 · ( t + 90) − 90 4 ( t +90)^3 ,^ t^ <^ 30.
(c) [4 points] How much salt does the tank contain when it begins to overflow? Solutions: The amount of salt in the tank when it is on the point of overflowing t = 30 is given by
Q (30) = 2 · (30 + 90) − 90 4 (30+90)^3 =^ 202.03125 lbs.
μ ( x ) = e
∫ ∂ ∂My −^ ∂ ∂Nx N ( x , y ) d x^ = e
∫ (^) x + 1 x d x^ = ex +ln^ | x |^ = xex^.
(c) [8 points] y + (2 x − ye y^ ) y ′^ = 0 Solution: We note that M ( x , y ) = y , and N ( x , y ) = 2 x − e y
Then we have ∂M ∂y = 1 and
∂N
∂x
Thus, ∂ ∂My 6 = ∂ ∂Nx. Hence, the given equation is not exact. However, we note that
∂N ∂x −^
∂M ∂y M ( x , y )
y
y
which is only dependent on y. Therefore there exist an integrating factor in the format of μ ( y ), which can be determined as follow:
μ ( y ) = e
∫ ∂ ∂Nx −^ ∂ ∂My M ( x , y ) d y^ = e
∫ (^1) y d y^ = e ln^ | y |^ = y.
