1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Initial Value Problem (with y(0) ?)

  1. Mar 25, 2006 #1
    Initial Value Problem

    I (kind of) understand how to do initial value problems. I know that if the problem is [tex] y' = x, y(4) = 3[/tex] then you just differentiate it, solve for y(4) and then replace the constant C (in the equation I differentiated) with this answer.

    But what if the initial value is y(0) = 6??

    The problem I am trying to do is:

    [tex] y' = 36 + y^2 [/tex]

    [tex]y(0) = 6 [/tex]

    [tex] 0 < c < \frac{\pi}{2} [/tex]

    Differentiating, I get [tex] y = 36y + \frac{1}{3}y^3 + C[/tex].

    But if I solve for y(0) = 6, then I get [tex] y = 36(0) + \frac{1}{3}(0)^3 + C[/tex].

    But the answer can't be [tex] y = 36y + \frac{1}{3}y^3 + 6[/tex] - can it?!

    And what is the significance of [tex] 0 < c < \frac{\pi}{2} [/tex] - why is this needed?

    I've been looking all over my calculus book, but I just don't understand how to do this. I am taking Engineering Mathematics (Kreyzig Book) and I can't help but feel that there is this large gap between this book and my calculus skills.. everything just looks so different.. *sigh*

    Can anyone help me out?
     
    Last edited: Mar 25, 2006
  2. jcsd
  3. Mar 25, 2006 #2

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Do you know what an independent variable is, and what it means to integrate with respect to that?
     
  4. Mar 25, 2006 #3
    No, I'm afraid not. Does it have anything to do with implicit differentiation?
     
  5. Mar 25, 2006 #4

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    x is your independent variable, y is your dependent.

    Go back to your fallacious "integration" of the diff. eq.

    On your left hand side, you have found the anti-derivative of y'(x) correctly, that is simply y(x).

    But which variable have you integrated with respect to on your right hand side?
    The dependent or the independent variable?
     
  6. Mar 25, 2006 #5
    Thank you for telling me about the independent and dependent variable. It now rings a bell.

    I just realised that I said differentiation instead of integration.

    As you can see, I am hopelessly confused..

    I have only encountered equations where (for dy/dx) there are x's on the right hand side. I have never encountered equations with y's and x's and, as in this case, only y's.

    I.. don't really know where to start. What should I be thinking of doing?
     
    Last edited: Mar 25, 2006
  7. Mar 25, 2006 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You are right about the discrepancy between what you are trying to do and your calculus skills! Plan to spend a lot of time reviewing calculus.

    In the first place, you aren't "differentiating", you are "integrating". And the integral of y with respect to x is not (1/2)y2: y is some function of x and without knowing what function you don't know how to integrate it- that's the whole point of differential equations!

    Actually, the part you say you are stuck on- that you are given y(0) rather than y(4) is trivial- just use x= 0 instead of x= 4! However, the part that you think you can do is completely wrong.

    Your equation is y'= 36+ y2. You might see it better if you wrote that as [tex]\frac{dy}{dx}= 36+ y^2[/tex]. Separating into "differentials" [tex]dy= (36+ y^2)dx[/tex]. Now can you see that the right hand side has "dx" rather than "dy"? You can't just treat that as an integral. In this simple example, a "separable" equation in which x and y can be completely separated, however, you can write it as
    [tex]\frac{dy}{y^2+ 36}= dx[/tex].
    Can you integrate that (think "arctan")?

    Once you've done that integration and have y as a function of x, with the constant of integration, set x= 0 and y= 6 to determine the constant.
     
    Last edited: Mar 25, 2006
  8. Mar 25, 2006 #7
    My god. You are a GOD. Thank you SO much. I feel as a mortal would feel when a god demonstrates his powers by parting the sea or something. I am finally liberated from hours of trying and failing to understand my textbook.

    I thank you. From the bottom of my heart. You don't know how happy I feel right now!

    :D


    So to continue, it becomes

    [tex] \int \frac{dy}{y^2+6^2} = \int dx [/tex]

    [tex] \frac{1}{6}arctan\frac{y}{6} + C = x[/tex]

    Then

    [tex] C = x - \frac{1}{6}arctan\frac{6}{6} = 0 - \frac{1}{6} . \frac{\pi}{4} = - \frac{\pi}{24} [/tex]




    But then.. if I do this, I come up with a different answer:

    [tex] \frac{1}{6}arctan1 = x - C [/tex]

    [tex] arctan1 = 6x - 6C [/tex]

    [tex] arctan1 = 6(0) - C [/tex]

    [tex] C = -arctan1 = - \frac{\pi}{4}[/tex]


    What am I doing wrong?!

    Also, if you cannot separate the variables, what do you do? Do equataions that are not separable have a separate name?
     
    Last edited: Mar 25, 2006
  9. Mar 25, 2006 #8

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Separable differential equations constitute a subset of differential equations.
    Differential equations that are not separable are called..non-separable.
     
  10. Mar 25, 2006 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Woah! My sarcasm detector is going off big time!
     
  11. Mar 25, 2006 #10
    No, no, no!! Though I am fairly sarcastic at times, I assure you I was NOT sarcastic this time!! I realise that I did overdo it a little bit, but I really was overwhelmed with joy! You don't know how happy I was to be liberated from hours of frustration... and to be able to solve a question in this awful book for once! I also recognise you from helping me out a few days ago.. And most probably, I will need your help again in the future! (I have a very long and hard term ahead of me..) I don't think I'd be sarcastic, even if i wanted to - and this book has a way of humbling people like me. But seriously, I was, and still am very grateful! :)

    As for the 6C.. I thought that anything multiplied with a constant, remains a constant. I've seen this in my calculus book.. Does it not apply in this situation? Though, yes, I now understand that it is that 6C that is the difference between the two versions..
     
    Last edited: Mar 26, 2006
  12. Mar 26, 2006 #11
    I also encountered something new and I'm not sure what to do:

    The problem is:

    [tex] y' = \frac{1-7y-49x}{1+y+7x} [/tex]

    with [tex] \inline y^2(1)=7 [/tex] using [tex] \inline (y + 7x = v) [/tex]

    I've managed this so far:

    [tex] v'-7 = \frac{1-7v}{1+v} [/tex]

    [tex] v' = \frac{-7(1+v)}{1+v} + \frac{8}{1+v} + 7 [/tex]

    [tex] \frac{dv}{dx} = \frac{8}{1+v} [/tex]

    [tex] \int (1 + v) dv = \int 8 dx [/tex]

    [tex] v + \frac{1}{2}v^2 = 8x + C [/tex]

    But how do I plug that into [tex] (y = v - 7x) [/tex] to get the constant?!

    How do I reduce [tex] v + \frac{1}{2}v^2 = 8x + C [/tex] for a single v?
     
  13. Mar 26, 2006 #12

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Assuming you have calculated correctly,
    solve for v(x) using the quadratic formula.
     
  14. Mar 26, 2006 #13
    Oh! But... doesn't a quadratic equation have to be in the form ax^2 + bx + c = 0; doesn't it have to equal zero?

    How do I do it if it equals 8x?

    Could you tell me about the 6C = C bit from the previous equation? I'm running into a lot of equations where I have to multiply a constant, and it is very important for me to know whether 6C = C or 6C = 6C (so that if it is 6C = 2, then C = 2/6)..
     
    Last edited: Mar 26, 2006
  15. Mar 26, 2006 #14

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    As for the "C"-thing:
    Different C's are for y-functions with different initial conditions than yours, but fulfilling the same differential equation.

    C is fixed for a particular solution of the diff. eq.

    As for the quadratic formula:
    Write this as:

    [tex]\frac{1}{2}v^{2}+v+(B-8x)=0 (B=-C)[/tex]

    That is:
    [tex]v=-1\pm\sqrt{1-2(B-8x)}[/tex]
    The constant B is determined from initial conditions.
     
  16. Mar 26, 2006 #15

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    I'm sorry, I've suddenly seen that you've made a total mess of evaluating your initial condition earlier. I'll post a reply later.
     
  17. Mar 26, 2006 #16

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    What ARE you doing here? x equals zero here!
    So, you've got:
    [tex]\frac{1}{6}arctan(1)=-C\to{C}=-\frac{\pi}{24}[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Initial Value Problem (with y(0) ?)
  1. Initial value problem (Replies: 7)

  2. Initial-value problem (Replies: 2)

Loading...