Download Assignment 2 Solutions - Differential Equations | MATH 441 and more Assignments Differential Equations in PDF only on Docsity!
Differential Equations, Math 441 Spring 2006:
Solutions for Homework 2
Problem 2.1: 1 Solve the ODE
y′^ + 3y = t + e−^2 t
Solution: I only do part c), the explicit solution. The above ODE is a linear first order ODE, so we do the integrating factor trick: The integrating factor must be a solution of μ′^ = 3μ, so we can take μ(t) = e^3 t. With this we have
(e^3 ty(t))′^ = e^3 ty′(t) + 3e^3 ty(t) = e^3 t(y′^ + 3y) = e^3 t(t + e−^2 t) = te^3 t^ + et.
Integrating both sides, we see that
e^3 ty(t) − y(0) =
∫ (^) t
0
se^3 s^ ds +
∫ (^) t
0
es^ ds.
The second integral is easy, ∫ (^) t
0
esds = et^ − 1 ,
the first is done with the help of integration by parts, ∫ (^) t
0
se^3 s^ ds = [s
e^3 s 3
]t 0 −
∫ (^) t
0
e^3 s^ ds
t 3
e^3 t^ −
(e^3 t^ − 1).
Thus
e^3 ty(t) = y 0 +
t 3
e^3 t^ −
e^3 t^ + et^ −
So the solution with initial condition y 0 at time zero is
y(t) = y 0 e−^3 t^ +
t 3
e−^3 t^ + e−^2 t.
So we that y(t) − (
t 3
) = y 0 e−^3 t^ −
e−^3 t^ + e−^2 t
which goes to zero. So for large t,
y(t) ∼
t 3
Problem 2.1: 33 Show that if a, λ > 0, and b ∈ R, then every solution of
y′^ + ay = be−λt
goes to zero as t → ∞. 1
Solution: Here the integrating factor must obey μ′^ = aμ, so μ(t) = eat must work. We see that
(eaty)′^ = eaty′^ + aeaty = eat(y′^ + ay) = eatbe−λt^ = be(a−λ)t.
Consider the case a = λ first. In this case we see that
(eaty)′^ = b
which integrate immediately to
eaty(t) − y 0 = bt
or y(t) = y 0 e−at^ + bte−at.
Since limt→∞ te−at^ = 0, we see that limt→∞ y(t) = 0 for any y 0 ∈ R. In the case a 6 = λ, we have (eaty)′^ = be(a−λ)t
which we again integrate to
eaty(t) − y 0 =
b a − λ
(e(a−λ)t^ − 1).
Hence
y(t) = y 0 e−at^ +
b a − λ
(e−λt^ − e−at).
again, the right hand side goes to zero as t → ∞, hence y(t) → 0 as t → ∞ irrespective of the initial condition.
Problem 2.1: 42 Use variation of parameters to solve the ODE
2 y′^ + y = 3t^2
Solution: We bring this equation in the form
y′^ +
y 2
3 t^2 2
Then we know that y 1 = e−t/^2 solves
y′ 1 +
y 2
We make the ansatz y = vy 1 for some function v. Calculating
3 t^2 2
= y′^ +
y 2
= (vy 1 )′^ +
y 2
= v′y 1 + vy 1 ′ +
y 2
= v′y 1 − v
y 1 2
y 2
= v′y 1
we see that
v′e−t/^2 =
3 t^2 2
Problem 2.2: 26 Solve the ODE
y′^ = 2(1 + x)(1 + y^2 ) with y(0) = 0
and determine where the solutions attains it minimum value.
Solution : This is a separable equation which we can rewrite as
y′ 1 + y^2
= 2 + 2x.
Integrating both sides, we see that ∫ 1 1 + y^2
dy =
dy/dx 1 + y^2
dx =
(2 + 2x) dx + c == 2x + x^2 + c.
Noticing that
(1 + y)−^1 dy = arctan(y), we get arctan(y) = 2x + x^2 + c
Using the initial condition, we find 0 = arctan(0) = c, so c = 0. Thus
y(x) = tan(2x + x^2 ).
Note that this function is defined on any open interval I which contains 0 and for which −π 2 < x^2 + 2x < π 2. (Since tan(α) is only defined for α 6 = π 2 + kπ with k an integer). Notice x^2 + 2x = (x + 1)^2 − 1 ≥ −1 for all x. Also notice that Now, to find the minimum, we differentiate w.r.t. x and look for a zero of the derivative. But HALT. Don’t differentiate. We know from the differential equation, that y′^ = 2(1 + x)(1 + y^2 ), so if y′(x) = 0, we must have
0 = y′(x) = 2(1 + x)(1 + y^2 )
and hence x = −1. This shows that at x = −1, there is an extremum, to see that it is a minimum, you can compute the second derivative from the explicit formula for y(x), or you notice that the largest open interval containing zero on which the tan is defined is the interval (−π/ 2 , π/2). So the solution y(x)is defined for x such that x^2 + 2x ∈ (−π/ 2 , π/2). Since x√+ 2 x = (x + 1)^2 − 1, we must have (x + 1)^2 − 1 < π/2. Hence |x + 1| < π/2 + 1, that is −
π/2 + 1 < x + 1 <
π/2 + 1 − 1, or
− 1 −
π 2
π/2 + 1 − 1.
Thus the solution y is defined only on the interval (− 1 −
√π
√^2 + 1,^ −1 + π 2 + 1). If x converges to one of the endpoints of the above interval one has that tan(x^2 + 2x) diverges to ∞. Since y(x) = tan(x^2 + 2x) has only one extremum in the interval (− 1 −
√π 2 + 1,^ −1 +^
√π 2 + 1), namely at^ x^ =^ −1,
and diverges to plus infinity at the endpoints of the interval, x = −1 must be the minimum! Problem 2.3: 12 (radiocarbon dating)
a) If Q satisfies Q′^ = −rQ, find the decay constant for Carbon, given that its half-life time is 5730 years. b) Find an expression for Q(t) at any time if Q(0) = Q 0. c) Suppose that remains are found with residual amount of carbon-14 is 22% of the original value. How old are the remains?
Solution: a) We know that Q must obey the equation
Q(t) = Q 0 e−rt
(Since the solution is unique and the above is certainly a solution of the equation.) At the half-time we have only half of what we started with left. So with y = year,
Q(5730y) =
Q 0
= Q 0 e−r^5730 y.
So er^5730 y^ = 2 or
r =
ln(2) 5730 y
= 0. 00012097 y−^1
b) According to part a) we have
Q(t) = Q 0 e−^0.^00012097 t/years.
c) In this case, we are given that Q(t) = 0. 2 Q 0. So
- 2 Q 0 = Q(t) = Q 0 e−rt.
Thus ert^ = 5
or
t =
ln(5) r
ln(5) − 0. 00012097
years = 13, 300years.
