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Homework 1, due September 1, 2004.
Posted: Wed Aug 25, 2004 12:17 pm
by goodwine
Problem 1: 1.2, number 11 hint
Posted: Mon Aug 30, 2004 2:43 pm
by goodwine
You may find the following identity useful to get your answer in the same form as the back of the book. Whether or not you will need it will depend on the integration method or table that you use.
arctanh x = (1/2) ln [(1 + x)/(1 - x)]
Unless otherwise specified, using an integration table is fine.
tech mouse icons
Posted: Mon Aug 30, 2004 7:37 pm
by nopatience
For the problems with the little blue mouse icons, are we just supposed to plug those into matlab etc?
Problem 11 1.2
Posted: Mon Aug 30, 2004 8:42 pm
by sshomber
Hey does anyone have an idea about problem 11 in section 1.2? I've gone on and haven't had much trouble after that but I still can't get started on that one.
Re: tech mouse icons
Posted: Mon Aug 30, 2004 9:07 pm
by goodwine
nopatience wrote:For the problems with the little blue mouse icons, are we just supposed to plug those into matlab etc?
It just means that you
may plot them with a computer. Do whatever is easiest. You can plot them by hand if you want.
Re: Problem 11 1.2
Posted: Mon Aug 30, 2004 9:10 pm
by goodwine
sshomber wrote:Hey does anyone have an idea about problem 11 in section 1.2? I've gone on and haven't had much trouble after that but I still can't get started on that one.
That one is separable, but the integral is tricky. You may use a table if you want, but may have to then use the hyperbolic trigonometric identity I provided above.
Problem 32 2.4
Posted: Mon Aug 30, 2004 11:49 pm
by lisaturtle
How do you find the solution for the interval t > 1, when the equation is homogeneous? I thought I could separate the equation but apparently that doesn't work...
Re: Problem 32 2.4
Posted: Tue Aug 31, 2004 6:59 am
by goodwine
lisaturtle wrote:How do you find the solution for the interval t > 1, when the equation is homogeneous? I thought I could separate the equation but apparently that doesn't work...
You can use an integrating factor if the equation is linear. The integral with mu(s)g(s) will be zero since g(t)=0, but 1/mu times the constant is still there.
Also, if you want you could assume an exponential solution and simply substitute to find lambda. The former is reflected in the solution method decision tree since it's the approach that the book gives. We'll add the latter at some later point.
Plots
Posted: Tue Aug 31, 2004 2:10 pm
by NDChevy07
Do we have to plot every problem that has a blue mouse even if the problem doesn't call for it? (Ch. 2.2 Problem 25 and Ch. 3.1 Problem 26 specifically)
Re: Plots
Posted: Tue Aug 31, 2004 2:35 pm
by goodwine
NDChevy07 wrote:Do we have to plot every problem that has a blue mouse even if the problem doesn't call for it? (Ch. 2.2 Problem 25 and Ch. 3.1 Problem 26 specifically)
No. You only have to do what the problem says. If there is a blue mouse and a computer can help you with the problem, then use the computer, if you choose, in any manner that you want.
2.4,#32
Posted: Tue Aug 31, 2004 5:47 pm
by norris
I don't know what I'm missing, but by using the integrating factor and a y0=1 I'm not getting anything close to the book. Is the constant y0=1 wrong?
Re: 2.4,#32
Posted: Tue Aug 31, 2004 7:17 pm
by goodwine
norris wrote:I don't know what I'm missing, but by using the integrating factor and a y0=1 I'm not getting anything close to the book. Is the constant y0=1 wrong?
y(0)=0 as is specified in the book. What you want to do is solve the problem for g(t)=1. Then find the
general solution for g(t)=0 (with an unknown constant in it). Then use the value of your first solution at t=1 to find the constant in the second solution.
Section 1.2 number 11
Posted: Tue Aug 31, 2004 7:21 pm
by goodwine
I told about 5 people in my office hours the hard way to do part (a), which was to solve the equation with an unknown gamma and compute the limiting value to determine what gamma needs to be. In other words, I said to solve parts (b), (c), etc. to solve part (a).
The easy way, as pointed out to me by a student, is to recognize that if there is a steady state value, then dv/dt=0. So, all you have to do to find gamma is to take the original differential equation, set dv/dt=0, plug the limiting value in for v, the parameter values for m and g and then solve for gamma.