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Differential equations - cannot solve one

  1. Feb 16, 2013 #1
    Greetings everyone.
    Can you help me please with solving this differential equation?

    [tex]\large \frac{dv}{dt} = a_g + \alpha v^n [/tex]

    where [tex]\Large a_g \alpha n[/tex] are constants. Artelnatively with specific n as n = 1, 2.

    I have no idea what to do...
    Thank you very much
     
  2. jcsd
  3. Feb 16, 2013 #2
    It seems quite simple at the first glance, as the equation is separable.

    The first thing I'd do is to separate the two variables to get [itex]\displaystyle \frac{dv}{a_g+\alpha v^n}=dt[/itex]. From there, you integrate both sides.

    The "trick" here is that the left hand side doesn't seem integrable. In fact, it isn't very easy and pleasant to integrate for most n's, but it can be done. I can't find a general algorithm though.
     
    Last edited: Feb 16, 2013
  4. Feb 16, 2013 #3
    Okay. Let's say n = 1. Then it should be resolvable. And what if n = 2 ? For n = 0 it is trivial....
     
  5. Feb 16, 2013 #4

    Dick

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    For n=1 a simple substitution will give you log. For n=2 a trig substitution will get you an arctan.
     
  6. Feb 16, 2013 #5
    Can you show me please ?
     
  7. Feb 16, 2013 #6

    SammyS

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    How does it give you arctan ?

    What is [itex]\displaystyle \ \ \int\frac{dx}{1+x^2}\ ?[/itex]
     
  8. Feb 16, 2013 #7

    HallsofIvy

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    Have you never taken a Calucus class? If n= 0, The integral is just
    [tex]\int \frac{dv}{a_g+ \alpha}= \frac{1}{a_g+ \alpha}\int dv[/tex]

    If n= 1 it is
    [tex]\int\frac{dv}{a_g+ \alpha v}[/tex]

    If n= 2 it is
    [tex]\int\frac{dv}{a_g+ \alpha v^2}[/tex]

    Can you integrate those?
     
  9. Feb 16, 2013 #8
    No, I haven't. I do not know how to substitute. The first one I can. The next two I cannot.
     
  10. Feb 16, 2013 #9

    SammyS

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    What is the derivative of arctan(x) ?
     
  11. Feb 16, 2013 #10
    For integrals of the type

    [tex]\int \frac{dx}{a^2 +x^2}[/tex], try the following substitution :

    [tex]x=a \tan \theta[/tex]

    With a bit of algebra, you'll be able to put your integral at this form.
     
  12. Feb 16, 2013 #11
    [tex]\int \frac{dx}{a^2 +x^2}[/tex]
    [tex]x=a \tan \theta[/tex]
    [tex]dx=\frac {a} {\cos^2 \theta} d \theta[/tex] - is it like this ?

    so we get:
    [tex]\frac {1}{a^2}\int \frac{\frac{a}{\cos^2}d \theta}{1 +\tan^2 \theta} = \frac {1}{a}\int \frac{d\theta}{\cos^2\theta +\sin^2 \theta} = \frac{\theta}{a} = \frac {\arctan \frac{x}{a}}{a}[/tex]
    Okay and what next ?
     
  13. Feb 16, 2013 #12
    Fine. Now, compare the two integrals and identify your "a".
     
  14. Feb 16, 2013 #13
    Well
    [tex]a = \sqrt{a_g} \\
    x = \sqrt{\alpha}v[/tex]

    So I got
    [tex]\frac {\arctan({\sqrt{\frac{\alpha}{a_g}} v})}{\sqrt{a_g}} = t[/tex]
    didn't I?
    So
    [tex]v = \frac {\tan(\sqrt{a_g}t)}{\sqrt{\frac{\alpha}{a_g}}}[/tex]
    IS IT RIGHT ?
     
  15. Feb 16, 2013 #14
    Actually, do this:

    [tex]\frac{1}{\alpha} \int \frac{dv}{\frac{a_g}{\alpha}+v^2}[/tex]

    Then, use the substitution to integrate.
     
  16. Feb 16, 2013 #15
    I would get the same, wouldn't I ? One thing I'd like you to tell me. How do you know what subtitution you have to use?
     
  17. Feb 16, 2013 #16
    Not exactly. Notice there's an [tex]\sqrt \alpha[/tex] missing on the denominator.

    Btw, I know the substitution simply by having done similar integrals many times over. Also, you can get the idea if you remember the usual trigonometric identities.
     
  18. Feb 16, 2013 #17
    Got it. So you say it's all about practise, like in normal algebra, right?
     
  19. Feb 16, 2013 #18
    Exactly. After some time, you'll just get used to it.
     
  20. Feb 17, 2013 #19
    I found a general algorithm to integrate these kinds of functions.

    You do a partial fraction decomposition using complex numbers. You'll need the nth roots of unity for this. An example is [itex]\dfrac{1}{1+x^4}[/itex], which can be decomposed using the equality [itex]1+x^4=(x-\sqrt{i})(x+\sqrt{i})(x-i\sqrt{i})(x+i\sqrt{i})[/itex] where [itex]\sqrt{i}=\dfrac{\sqrt{2}}{2}(1+i)[/itex] from Euler's identity. This leaves you with 4 fractions with denominators of degree one. In general, this yields an expression with the hypergeometric function involved, but it can be simplified for integer n to an expression involving only the complex logarithm (or real logarithms + inverse tangent.)
     
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