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Differential Equations exam help

  • Thread starter Gwilim
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  • #1
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First a word of explanation about my circumstances, skip the italics if you're not interested: I have a (vitally important) exam on differential equations coming up in two weeks, and I have no institutional support (lectures, tutors, etc). It's not a subject I've studied in any depth before but I'm a quick learner and I have to try. I'm sure given enough time I could find the information I need without having to ask, but I don't have a lot of time, and with other exams also coming up I need to concentrate on those too. Forgive me if the questions are insultingly easy, I simply the lack the knowledge to answer them myself

Some questions from a past exam paper, I don't need a full set of workings and answers, just enough so that I can guess the rest. Or just giving me links to pages which the explanations I need would be fantastic.

1. Solve each of the following equations given the initial conditions:
i) y' - 3y = 0, with y(0) = 2
ii) y'' + y' -6y = 0, with y(0) = 0 and y'(0) = 1
iii) y'' + 2y' +y = 0, with y(0) = 2 and y'(0) = 0
iv) y'' - 2y' +2y = 0, with y(0) = 0 and y'(0) = 1

I'm not even sure what this is asking. Substituting the values into the equations, clearly in the case of i) y'(0) = 6 and in the others would give a value for y''(0) but is this a solution?

2.
i) Consider the following differential equation for y(x):

y'' + y = 1 - x

a) Find the complementary function for this equation
b) find the particular integral for this equation
I don't know what these terms mean
c) Write down the solution to this equation, given the initial conditions y(0) = 1 and y'(0) = 1
ii) Consider the following differential equation for y(x):

y'' + y' = e^-x
a) b) and c) as above.

There are three more questions on the paper, but I think I suffer from the same basic lack of knowledge for all of them, and help with these could shed some light on the others. Sorry for not using the template or making a serious attempt at an answer, I will make one if someone can point me in the right direction.
 

Answers and Replies

  • #3
Defennder
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I have a (vitally important) exam on differential equations coming up in two weeks, and I have no institutional support (lectures, tutors, etc). It's not a subject I've studied in any depth before but I'm a quick learner and I have to try. I'm sure given enough time I could find the information I need without having to ask, but I don't have a lot of time, and with other exams also coming up I need to concentrate on those too.
This is exceedingly odd. You're sitting for an exam for which you are not taught anything about. Do you have notes or textbooks which you can study? If not, then your circumstances are certainly quite unique.

Gwilim said:
1. Solve each of the following equations given the initial conditions:
i) y' - 3y = 0, with y(0) = 2
ii) y'' + y' -6y = 0, with y(0) = 0 and y'(0) = 1
iii) y'' + 2y' +y = 0, with y(0) = 2 and y'(0) = 0
iv) y'' - 2y' +2y = 0, with y(0) = 0 and y'(0) = 1

I'm not even sure what this is asking. Substituting the values into the equations, clearly in the case of i) y'(0) = 6 and in the others would give a value for y''(0) but is this a solution?
These are known as homogenous linear ordinary differential equations. A homogenous linear ODE is expressed in the form y'' + ay' + by' = 0, where a,b are constants. It would be extremely lengthy to tell you what you need to know in order to solve the above. Try this:
http://tutorial.math.lamar.edu/Classes/DE/IntroSecondOrder.aspx


2.
i) Consider the following differential equation for y(x):

y'' + y = 1 - x

a) Find the complementary function for this equation
b) find the particular integral for this equation
I don't know what these terms mean
I don't know what 'complementary function' and 'particular integral' mean either. I'm guessing the latter actually refers to a particular solution to the non-homogenous DE.

I don't think anyone is able to help you here unless you understand the basics and techniques of solving such ODEs. Try the above link I posted. Read through them and then attempt the questions. Post your attempts and here and I'm sure plenty of us are willing to help.
 
  • #4
tiny-tim
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Welcome to PF!

Hi Gwilim! Welcome to PF! :smile:
I'm not even sure what this is asking. Substituting the values into the equations, clearly in the case of i) y'(0) = 6 and in the others would give a value for y''(0) but is this a solution?
No … if ' means d/dx, then "solution" means expressing y as a function of x.

In the case of i), rewrite it as dy/dx - 3y = 0, and solve it. That will give you y = f(x) + a constant.

To find the constant, you need one further piece of information … that is why they have given you y(0) = 2. (There is nothing special about 0 … they could have given you y(1) or even y'(1).)

The others have y'', so there will be two constants, which is why they have given you two extra pieces of information. :smile:
a) Find the complementary function for this equation
b) find the particular integral for this equation
I don't know what these terms mean
The case where f(x) = 0 is called a homogeneous equation and its solutions are called complementary functions. It is particularly important to the solution of the general case, since any complementary function can be added to a solution of the inhomogeneous equation to give another solution (by a method traditionally called particular integral and complementary function).
This is from http://en.wikipedia.org/wiki/Linear_differential_equation

Have a look at the whole of the article. :smile:

This is a really complicated topic … I suggest you set aside some extra time for practising it.

Let us know how you're getting on! :smile:
 
  • #5
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This is exceedingly odd. You're sitting for an exam for which you are not taught anything about. Do you have notes or textbooks which you can study? If not, then your circumstances are certainly quite unique.
Sadly this is the case, a full explanation would probably distract from what I'm actually asking about here though.

Hi Gwilim! Welcome to PF! :smile:
Hi, glad I could make it here.
No … if ' means d/dx, then "solution" means expressing y as a function of x.

In the case of i), rewrite it as dy/dx - 3y = 0, and solve it. That will give you y = f(x) + a constant.
You mean integrate to eliminate the dy/dx? I'm not sure if that notation works here, being as there's no x to integrate with respect to, and I'm fairly sure y represents a function not a variable. I'm 100% certain I said something stupid in that paragraph.

This is a really complicated topic … I suggest you set aside some extra time for practising it.

Let us know how you're getting on! :smile:
Sadly I don't have any 'extra' time, and I made a mistake in my first paragraph, the exam is not in two weeks it's in one. Like I say though, I have to try my best no matter how hopeless it might seem, it's not just important, it's personal. Thanks for the encouragement.

Defennder, I tried that link you gave me, and I've come up with the following solutions based on the method it gave me:

i) y' - 3y = 0, with y(0) = 2
y(t) = ke^rt
r - 3 = 0, r = 3.
putting in the intial value, y(t) = 2e^3t, y'(t) = 6e^3t

ii) y'' + y' - 6y = 0, with y(0) = 0 and y'(0) = 1
y(t) = ke^rt
r^2 + r - 6 = 0, r = 2 or -3
taking either value of r and any k does produce an equation like the one in the question, but the initial condition states y(0) = 0, and ke^0 = k. I could add a constant to y(t), for example y(t) = e^2t - 1, and this would satisfy the initial condition while the constant would dissapear when differentiated, but then it would no longer be true that y'' +y' -6y = 0.

Similarly for iii) I have r = -1 and for iv) I have r = 1 +/- i, but the zeros in the initial conditions give the same problem when using exponents as answers.
 
  • #6
tiny-tim
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… y represents a function not a variable.
y is a both a function of t and a variable. :smile:

(Your t is my x, of course.)
ii) y'' + y' - 6y = 0, with y(0) = 0 and y'(0) = 1
y(t) = ke^rt
r^2 + r - 6 = 0, r = 2 or -3
taking either value of r and any k does produce an equation like the one in the question, but the initial condition states y(0) = 0, and ke^0 = k. I could add a constant to y(t), for example y(t) = e^2t - 1, and this would satisfy the initial condition while the constant would dissapear when differentiated, but then it would no longer be true that y'' +y' -6y = 0.
Hi Gwilim! :smile:

ok … you've obviously picked up some of the technique, but not all of it.

You know how to find the roots, but not what to do with them.

If the two roots (I'll call them r and s) are different, then the solutions are Ae^rt + Be^st, for any constants A and B.

There are no other solutions.

If the two roots (I'll call them both r) are the same, then the solutions are (A + Bt)e^rt, for any constants A and B.

There are no other solutions.

If the equation has y''', and there are three equal roots, then the solutions are (A + Bt + Ct^2)e^rt, for any constants A and B and C.

And so on … :smile:
 
  • #7
If you can't even solve linear homogeneous de's and your test is in two weeks, I wouldn't even bother. What you should have done was buy a textbook, study it, work the exercises and if you did that for an hour or two a day in a few months you would have been ready for the exam.

As it stands you can't even do the most simplest problems, and I doubt that you'll master the material in two weeks. I'm sorry I don't mean to sound like a jerk, but you need blunt honesty more than politeness right now.
 
  • #8
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If you can't even solve linear homogeneous de's and your test is in two weeks, I wouldn't even bother. What you should have done was buy a textbook, study it, work the exercises and if you did that for an hour or two a day in a few months you would have been ready for the exam.

As it stands you can't even do the most simplest problems, and I doubt that you'll master the material in two weeks. I'm sorry I don't mean to sound like a jerk, but you need blunt honesty more than politeness right now.
If one did nothing else for two weeks, but did examples from a book, one might pass.

I did this once in upper level physics class. Didn't do the homework for five weeks because I was swamped in other classes. Did some 50 multipart homework questions in one week and passed the test. It was like 60 hours of hell catching up.
 
  • #9
tiny-tim
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hmm … DavidWhitbeck has a point.

You said you have other exams also coming up.

You should seriously consider how you're goind to divide your time.

You say this exam is vitally important. If the others aren't, then I'm afraid you've got to give this one priority. If the others are, then you may be better off deciding to ditch one and retake later, so as to get good grades in the others.

If you want advice on this, start a new thread in the "Academic & Career Guidance" forum.

Sorry to be depressing … :redface:
 
  • #10
If one did nothing else for two weeks, but did examples from a book, one might pass.

I did this once in upper level physics class. Didn't do the homework for five weeks because I was swamped in other classes. Did some 50 multipart homework questions in one week and passed the test. It was like 60 hours of hell catching up.
Well you only (thought that is alot) caught up on homework for a month of the class. And you at least had only the homework to catch up on, you did go to lecture didn't you? the OP has no lecture, no homework, no book, no tutoring nothing. Cold turkey learning a class in two weeks is way harder than catching up on a few weeks missed homework imho.
 
  • #11
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If you can't even solve linear homogeneous de's and your test is in two weeks, I wouldn't even bother. What you should have done was buy a textbook, study it, work the exercises and if you did that for an hour or two a day in a few months you would have been ready for the exam.

As it stands you can't even do the most simplest problems, and I doubt that you'll master the material in two weeks. I'm sorry I don't mean to sound like a jerk, but you need blunt honesty more than politeness right now.
You're personifying the attitude that got me into my current situation in the first place. The fact is that I couldn't do linear homogenous DEs, but in another hour or so I will be able to, unless I give up now. I could not bother, but that would be too easy, and like I say, it's personal now. As long as I do as much as I can between now and the exam, I can't fail even if I do, if you get me.

y is a both a function of t and a variable. :smile:

(Your t is my x, of course.)


Hi Gwilim! :smile:

ok … you've obviously picked up some of the technique, but not all of it.

You know how to find the roots, but not what to do with them.

If the two roots (I'll call them r and s) are different, then the solutions are Ae^rt + Be^st, for any constants A and B.

There are no other solutions.

If the two roots (I'll call them both r) are the same, then the solutions are (A + Bt)e^rt, for any constants A and B.

There are no other solutions.

If the equation has y''', and there are three equal roots, then the solutions are (A + Bt + Ct^2)e^rt, for any constants A and B and C.

And so on … :smile:
Alright thanks! So with that new information, the answers to the remaining three subquestions are as follows;
ii) y(t) = e^-3t - e^2t
iii) oh damn I still can't do this. The repeated root of -1 gives y(t) = Ae^-t + Bte^-t and
y'(t) = -Ae^-t - Bte^-t. Substituting in y(0) = 2 gives A = 2, but putting that value into y'(t) contradicts the condition that y'(0) = 0 (since t=0, y'(0) = -Ae^-t, and A =/= 0)
iv) y(t) = [e^(1+i)t]/2i - [e^(1-i)t]/2i

If someone could check those and/or provide me help with question ii) that would be great
 
  • #12
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hmm … DavidWhitbeck has a point.

You said you have other exams also coming up.

You should seriously consider how you're goind to divide your time.

You say this exam is vitally important. If the others aren't, then I'm afraid you've got to give this one priority. If the others are, then you may be better off deciding to ditch one and retake later, so as to get good grades in the others.

If you want advice on this, start a new thread in the "Academic & Career Guidance" forum.

Sorry to be depressing … :redface:
I might just do that. But to give a brief account here, I have 6 exams coming up, 5 of which are core modules. This subject is the last I've started work on, but also the first exam, which is silly. I think I can still retake though.
 
  • #13
I might just do that. But to give a brief account here, I have 6 exams coming up, 5 of which are core modules. This subject is the last I've started work on, but also the first exam, which is silly. I think I can still retake though.
If you can retake it, then just try your best now and hope that it's enough. Just remember the best way to do any kind of long term project is to set a schedule, and work on it regularly for some part of the day, everyday, instead of waiting and cramming at the last minute. That advise is for the future.
 
  • #14
tiny-tim
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ii) y(t) = e^-3t - e^2t

iii) oh damn I still can't do this. The repeated root of -1 gives y(t) = Ae^-t + Bte^-t and
y'(t) = -Ae^-t - Bte^-t. Substituting in y(0) = 2 gives A = 2, but putting that value into y'(t) contradicts the condition that y'(0) = 0 (since t=0, y'(0) = -Ae^-t, and A =/= 0

iv) y(t) = [e^(1+i)t]/2i - [e^(1-i)t]/2i
Hi Gwilim! :smile:

ii) looks fine.

In iii), you should have had y'(t) = -Ae^-t - Bte^-t + Be^-t.

Lesson: it is not a good idea to split (A + Bt)e^-t … there's no advantage in it, it makes it longer, and it can lead to error (which it did!) :rolleyes:
iv) is correct, but you would be expected to write it as a combination of e^t cost and sint. :smile:

(on the other matter, you really should check whether you can retake, and also whether you can only retake if you fail. if you want to discuss this further, you really should do it on the other forum … that's not a rule … it's just that far more people with relevant experience read the other forum :smile:)
 
  • #15
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Thanks to everyone who helped, I have this type of problem pretty much figured now, though I will be back tomorrow with more questions on more topics.

As for the 'other matter', I've tried to explain it as best I can in this thread: physicsforums.com/showthread.php?p=1727616 (apparently I'm not allowed to post URLs yet) on the specified subforum
 

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