AMS361 Recitation 03/04 Notes 1
1
Differential Equations
• Ordinary Differential Equations (ODE) or Partial Differential Equations (PDE)
• Order of DE
• Linear or Non-linear
• Homogenous or Non-homogenous
• Autonomous or Non-autonomous
1.1
Example Problem 1.1.2
Question: Verify by substitution that the given function is a solution of the given DE.
Primes denote derivatives WRT x.
y 0 + 2y = 0,
y = 3e−2x
Solution: “Verify”=⇒“Substitution”. We want to check LHS = RHS
y = 3e−2x
y 0 = −6e−2x
LHS = −6e−2x + 2 ∗ 3e−2x = 0 = RHS
So, y = 3e−2x is a solution of the given DE.
1.2
Example Problem 1.1.18
Question: Verify that y(x) satisfies the given DE and then determine a value of the constant
C so that y(x) satisfies the given initial condition.
y 0 = 2y,
y(x) = Ce2x ,
y(0) = 2
Solution: “Verify”=⇒“Substitution”. We want to check LHS = RHS
y 0 = 2Ce2x
1
LHS = 2Ce2x = 2 ∗ Ce2x = RHS
So, y(x) = Ce2x satisfies the given DE.
Then, by plugging in y(0) = 2, we have y(0) = C ∗ e2∗0 . This gives C = 2.
Thus, y = 2e2x is a solution of the given DE and the given initial condition.
2
Mathematical Models
• Newton’s Law of Cooling
dT (t)
= k(A − T (t))
dt
• Newton’s law of Motion
d2 x(t)
= F (t)
dt2
dv
dx
a=
, v=
dt
dt
m
• Torricelli’s Law for Draining
A(y(t))dy(t)
√
= −k y
dt
• Population Model
dP (t)
= kP (t)
dt
• A Swimmer’s Problem
V0
dy(x)
x2
=
(1 − 2 )
dx
Vs
a
2.1
Example Torricelli’s Law for Draining
According to Torricelli’s Law, the time rate of change of the volume V of water or other
liquid in a draining sink is proportional to the square root of the depth y of water in the
tank. Denote v(t) as the volume of water in the sink at time t and y(t) as the height of the
water surface.
dv
√
= −k y
dt
Let A(y) be the cross section area of the container at height y. Then, dv = A(y)dy. Now,
we can write the DE as
A(y)dy
√
= −k y.
dt
2
2.2
Example A Swimmer’s Problem
Suppose that the velocity VR at which the water flows increases as one approaches the
center of the river and is given in terms of distance x (x ∈ [−a, a]) from the center by
VR = V0 (1 −
x2
)
a2
where VR (x) is the water speed at location x, V0 is the water speed at the center of the rive
and VS is the swimmer speed.
Note that the formula VR given is a well-known result in mechanics of fluids under the
no-slip condition.
The swimmer’s velocity vector relative to the ground had horizontal component VS and
vertical component VR . hence the swimmer’s instantaneous directional angle α is given by
dy
VR
V0
x2
= y 0 = tan(α) =
=
(1 − 2 )
dx
VS
VS
a
where y(x) represents the trajectory of the swimmer.
3
Integrals as General and Particular Solutions
For a simple form of first-order equation
dy
= f (x),
dx
the solution can be obtained by direct integral. The general solution is given by
Z
y(x) = f (x)dx + C,
where C is an arbitrary constant.
3.1
Problem 1.2.8
Question: find a function y = f (x) satisfying the given differential equation and the prescribed initial condition.
dy
= cos(2x), y(0) = 1
dx
Solution: According to the direct integral, the general solution is
Z
Z
y(x) = f (x)dx + C = cos(2x) + C,
1
= sin(2x) + C.
2
Use the initial condition y(0) = C = 1. Therefore, the particular solution is y(x) =
1
2 sin(2x) + 1.
3
4
Slope Field and Solution Curves
• Slope: first derivative dy/dx
• Curvature: second derivative d2 y/dx2
Theorem: Existence and Uniqueness of Solutions:
Suppose that both the function f (x, y) and its partial derivative Dy f (x, y) are continuous
on some rectangle R in the xy-plane that contains the point (a, b) in its interior. Then, for
some open interval I containing the point a, the initial value problem
dy
= f (x, y), y(a) = b
dx
has one and only solution that is defined on the interval I.
4.1
Problem 1.3.11
Question:Determinant the existence and uniqueness of the solution.
dy
= 2x2 y 2 , y(1) = −1.
dx
Solution: In this problem,
f (x, y) = 2x2 y 2 , Dy f (x, y) = 4x2 y.
Both functions are continuous everywhere and the region contains the initial point (1, −1).
So, the solution exists and is unique.
4