Download Math 3120-001: Test One

Math 3120-001: Test One
June 02, 2016
Your Name: Solutions
Your Name: Solutions
Math 3120-001 ( ): Test One
Problem 1
Problem 1
(20 points ) Classify each of the following as an Ordinary Differential Equation or a partial differential
equation, linear or nonlinear, autonomous or non-autonomous give the order, and indicate the dependent
and independent variables.
(a)
dy
x sin(y)
=
.
dx
x+4
(Solution) It is first order non-autonomous nonlinear ordinary differential equation. The dependent variable is y
and the independent variable is x.
(b)
d5 P
= P cos(P ).
dt5
(Solution) It is fifth order autonomous nonlinear ordinary differential equation. The dependent variable is P and
the independent variable is t.
(c)
∂u
∂4u 1
=
+
∂t
∂x4
x
∂u
∂x
5
+ 13u.
(Solution) It is fourth order non-autonomous nonlinear partial differential equation. The dependent variable is u
and the independent variables are t and x.
(d)
d3 y
y sin(y)
+ cot(y)y =
.
dx3
y+4
(Solution) It is third order autonomous nonlinear ordinary differential equation. The dependent variable is y and
the independent variable is x.
Page 1 of 6
Your Name: Solutions
Math 3120-001 ( ): Test One
Problem 2
Problem 2
(20 points)Verify that
y = tan(x3 + C)
satisfies the differential equation
dy
= 3x2 (y 2 + 1).
dx
Find the values of C so that y(0) = 1.
Solution:
Left hand side of the equation =
d dy
=
tan(x3 + C) = sec2 (x3 + C)3x2 = 3x2 sec2 (x3 + C)
dx
dx
Right hand side the equation = 3x2 (y 2 + 1) = 3x2 (tan2 (x3 + C) + 1) = 3x2 sec2 (x3 + C),
since 1 + tan2 (θ) = sec2 (θ). Therefore y = tan2 (x3 + C) is a solution to the above equation.
Next, we find C. Using the initial condition we get:
y(0) = tan(02 + C)
1 = tan(C).
Since the solution is unique, C =
π
4
Page 2 of 6
Your Name: Solutions
Math 3120-001 ( ): Test One
Problem 3
Problem 3
(20 points)Determine whether the function
y = cos(4x) + 2x
is a solution to the equation
d2 y
+ 16y = 32x.
dx2
Solution:
y = cos(4x) + 2x
dy
= −4 sin(4x) + 2
dx
d2 y
= −16 cos(4x)
dx2
Left hand side of the equation =
d2 y
+16y = −16 cos(4x)+16(cos(4x)+2x) = −16 cos(4x)+16 cos(4x)+32x = 32x)
dx2
Right hand side the equation = 32x
. Therefore y = cos(4x) + 2x is a solution to the above equation.
Page 3 of 6
Your Name: Solutions
Math 3120-001 ( ): Test One
Problem 4
Problem 4
(20 points)Determine whether the given relation
x2 + y 2 = 100
is an implicit solution to the
x
dy
=− .
dx
y
Solution:
d 2
x + y 2 = 100
dx
d 2
d 2
d
x +
y =
[100]
dx
dx
dx
dy
=0
2x + 2y
dx
dy
=0
x+y
dx
dy
x
=− .
dx
y
Therefore the above relation is a solution to the above equation.
Page 4 of 6
Your Name: Solutions
Math 3120-001 ( ): Test One
Problem 5
Problem 5
(10 points)Consider the differential equation
dP
= P (P − 4)(P − 6)2
dt
for the population P (in thousand) of a certain species at time t.
(a) If the initial population is 5999 what can you say about the limiting population limt→∞ P (t)?
(Solution) Note that
(i) if P > 6 then
dP
dt
= p(p − 4)(p − 6)2 > 0.
(ii) If 4 < P < 6 then
(iii) if 0 < P < 4 then
dP
dt
dP
dt
= p(p − 4)(p − 6)2 > 0.
= p(p − 4)(p − 6)2 < 0.
Since P is in 1000 unit scale then P = 5.999. Then by (ii) limt→∞ P (t) = 6
(b) Can a population of 1400 ever decrease to 1000?
(Solution) P = 1.4 decreases to P = 1 since the function is decreasing by part (iii)
Page 5 of 6
Your Name: Solutions
Math 3120-001 ( ): Test One
Problem 6
Problem 6
(10 points)Determine whether Existence Theorem of uniqueness implies that the given initial value problem
has a unique solution.
(a)
dy
= y 2 sin(x), y(0) = 4
dx
(Solution)
f (x, y) = y 2 sin(x))
∂f (x, y)
= 2y sin(x).
∂y
Therefore for any rectangle containing the point (0, 4) f and
Therefore the Theorem applies.
∂f
∂y
are continuous on the rectangle.
(b)
dy
y
+4
= 0, y(0) = 10
dx
3x
(Solution)
f (x, y) =
−4y
3x
∂f (x, y)
−4
=
.
∂y
3x
Therefore for any rectangle containing the point (0, 10) f and
Therefore the Theorem does not apply.
∂f
∂y
are not continuous at point (0,10).
Page 6 of 6