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Homework Help: Linear Differential Equation with a substitution.

  1. Sep 30, 2008 #1
    Wow I am so very bad at differential equations.. :(

    The problem

    Here is the exact problem I'm given: http://img229.imageshack.us/img229/9025/10640260bs5.jpg [Broken]

    attempt at a solution

    I'm guessing that I need to differentiate y(x) that I am given and substitute that into the left hand side and then put the y function [sinx + 1/u(x)] as y in the right hand side (y^2).
    Then re-arrange for u(x) and differentiate to get du/dx and hopefully it will be the third given equation.

    After that I think I just have to solve that ODE and get it back in terms of y.

    Assuming that is what I have to do, sadly I cannot even get that far..

    differentiating y(x) = sinx + 1/u(x)
    must this be done implicitly? I get confused as u(x) is a function and not a variable.

    I can see that the third equation must be solved linearly such that dy/dx + utan(x) = -1/2 sec(x)

    And I (think) that I solved the integrating factor to be 1/cos(x).

    But I am so bad at differential equations and the sources on the internet seem to be so confusing that I'm quite stuck from here but this is what I have:

    (1/cos(x))*du/dx + u*(1/cos(x))*tan(x) = -(1/2)*(1/cos(x))*sec(x)=d/dx(u/cos(x)) <- multiplying by integrating factor and copying form of examples from the net

    u/cos(x) = integral of -(1/2)*(1/cos(x))*sec(x)

    and I know I'm way off here so I think I might aswell just stop...

    Also it would be great if someone could point me in the direction of a site that shows how to type all those math symbols such as the integral sign.

    Thanks in advance. A lot.
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Sep 30, 2008 #2

    CompuChip

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    Homework Helper

    Good guess :smile:

    I don't know what you call "implicitly" but I don't think so. All you need is the chain rule:
    [tex]\frac{d}{dx} \frac{1}{u(x)} = \frac{d (1/u)}{du} \frac{du}{dx}[/tex]
     
  4. Sep 30, 2008 #3
    Thanks CompuChip.

    For dy/dx I got cos(x) - 1/u^2

    so that's on the left side, for the right side I subbed in sin(x) + 1/u

    but it just gets really messy, and I can't get u on its own.. This is probably one of the most simple things to do but I just can't do it.

    Maybe I should drop out of engineering.....
     
  5. Sep 30, 2008 #4

    CompuChip

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    You still forgot the chain rule:

    dy/dx = cos(x) - u' / u^2,
    where u' denotes differentation of u w.r.t. x.
    And you don't want u on its own, but du/dx = u'.
     
  6. Sep 30, 2008 #5
    Thanks tons!
    I can't show how grateful I am for being so patient with me. I am so slow with this kind of thing..

    Now I'm 99% sure I have to do a linear differential equation to get the solution, but doing that will probably result in colossal failure on my part.
    As I said earlier I'm sure the integrating factor is 1/cos(x), and I multiplied it through my differential equation such that I had it in the form of I(x)dy/dx + I(x)P(x) = Q(x)
    but I don't really know what to do from here.
    Sites that I've looked at seem to use terminology and symbols that are far beyond me..
     
  7. Sep 30, 2008 #6
    Okay, linear differential equations confuse the hell out of me, but by any chance is the answer:

    [tex]u = -\frac{1}{2} \int sec(x)sec(x)dx = -\frac{1}{2} tan(x) + C [/tex]
    [tex] y(x) = sin(x) + \frac{1}{u(x)} [/tex]
    [tex]y(x) = sin(x) - \frac{2}{tan(x)+C}[/tex]

    Did I do this right?
    Cheers
     
  8. Sep 30, 2008 #7
    Somehow I don't think that this is right, because I have dy/dx, subbed in u, took the derivative of that. Did the differential equation, but shouldn't I need to do another differential equation to reach y(x)? It confuses me because I'm given dy/dx aswell as y(x) in terms of u(x).
    Thanks
     
  9. Oct 1, 2008 #8
    Okay, I think that I have done something wrong in that it's not:
    [tex]
    u = -\frac{1}{2} \int sec(x)sec(x)dx
    [/tex]

    but rather [tex]
    I(x)u = -\frac{1}{2} \int sec(x)sec(x)dx = -\frac{1}{2} tan(x) + C
    [/tex]

    where [tex] I(x) = sec(x) [/tex]
    so

    [tex]
    u = -\frac{1}{2} tan(x)cos(x) + C
    [/tex]

    Then [tex]
    y(x) = sin(x) - \frac{2}{tan(x)cos(x)+C} = sin(x) - 2cot(x)sec(x)
    [/tex]

    If anyone could confirm this, it would be most greatly appreciated.
    Cheers.
     
  10. Oct 2, 2008 #9
    Bump thx.
     
  11. Oct 2, 2008 #10
    Yeap, you've got it right. Just one thing, where did your constant go for the final answer? Btw, we are on the same course doing the same subject. I found the assignment really difficult esp question 3.
     
  12. Oct 2, 2008 #11
    Cheers jeffreylze,
    well my constant of integration is [tex] \frac{1}{cos(x)} [/tex]

    which is the [tex] sec(x) [/tex]

    in the final answer:

    [tex] sin(x) - 2cot(x)sec(x) [/tex]

    And yes a very difficult assignment, luckily we got over the break for it!
     
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