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Math Help - Nonlinear ODE

  1. #1
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    Nonlinear ODE

    1. The problem statement, all variables and given/known data

    Solve the nonlinear ODE

    du/dx=(u+x√(x^2+u^2 ))/(x-u√(x^2+u^2 ))

    by changing variables to x=rcos(theta), u=rsin(theta) and converting the equation to one for d(theta)/dr.

    3. The attempt at a solution

    Not sure if i'm going in the right direction.

    du/dx = du/d(theta) x d(theta)/dx

    u = rsin(theta), du/d(theta) = rcos(theta)
    x = rcos(theta), dx/d(theta) = -rsin(theta),
    i.e. d(theta)/dx = -1/rsin(theta)

    so du/dx = rcos(theta)/-1rsin(theta) = - cos(theta)/sin(theta)

    then i sub x=rcos(theta), u=rsin(theta) in the main equation

    - cos(theta)/sin(theta) = rsin(theta)+rcos(theta)√(r^2sin^2(theta) etc etc..

    so i gather the r^2, and make the sin^2+cos^2 both to one, then the √r^2 goes to just 'r'.

    then i take divide the whole equation by r

    du/dx = -cos(theta)/sin(theta) = [sin(theta)+rcos(theta)]/[rcos(theta)-rsin(theta)

    so i times the sin(theta) over the right side and the other to the right..
    and i get 1=0...

    can someone put me in the right direction on what i did wrong?

    thankyou.
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  2. #2
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    Hi

    I would say

    u = r\:\sin \theta
    x = r\:\cos \theta

    du = \sin \theta \:dr + r\:\cos \theta\:d\theta
    dx = \cos \theta \:dr - r\:\sin \theta\:d\theta

    du = \frac{u+x\sqrt{x^2+u^2}}{x-u\sqrt{x^2+u^2}} \:dx

    \cos \theta \:dr - r\:\sin \theta\:d\theta)" alt="\sin \theta \:dr + r\:\cos \theta\:d\theta = \frac{r\:\sin \theta+r^2\:\cos \theta}{r\:\cos \theta-r^2\:\sin \theta} \\cos \theta \:dr - r\:\sin \theta\:d\theta)" />

    \sin \theta \:dr + r\:\cos \theta\:d\theta) = (r\:\sin \theta+r^2\:\cos \theta) \\cos \theta \:dr - r\:\sin \theta\:d\theta)" alt="(r\:\cos \theta-r^2\:\sin \theta)\\sin \theta \:dr + r\:\cos \theta\:d\theta) = (r\:\sin \theta+r^2\:\cos \theta) \\cos \theta \:dr - r\:\sin \theta\:d\theta)" />

    d\theta = dr

    Seems strange but ... why not ?
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Hi

    I would say

    u = r\:\sin \theta
    x = r\:\cos \theta

    du = \sin \theta \:dr + r\:\cos \theta\:d\theta
    dx = \cos \theta \:dr - r\:\sin \theta\:d\theta
    Hi Running thank you!,
    How did you get from
    u = r\:\sin \theta to du = \sin \theta \:dr + r\:\cos \theta\:d\theta

    is that a derivative rule? like partial derivative?
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  4. #4
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    u(r,\theta) = r\:\sin \theta

    du = \frac{\partial u}{\partial r}\:dr + \frac{\partial u}{\partial \theta}\:d\theta

    which leads to

    du = \sin \theta \:dr + r\:\cos \theta\:d\theta
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  5. #5
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    OHH i was putting 'r' as a constant, instead of a function of 'u', my bad!
    since d\theta = dr

    is the answer
    \theta = r
    or
    d\theta/dr = 1

    because the question says to convert the equation into d\theta/dr, but does it mean leave it in that form?
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  6. #6
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    \theta = r
    or
    d\theta/dr = 1

    so x=rcosr and u=rsinr

    I'm lost.
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  7. #7
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    Well, since the question says to convert the equation into \frac{d\theta}{dr}, I would say that \frac{d\theta}{dr} = 1 is the required answer

    Is there any additional question ?
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  8. #8
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    Quote Originally Posted by running-gag View Post
    Well, since the question says to convert the equation into \frac{d\theta}{dr}, I would say that \frac{d\theta}{dr} = 1 is the required answer

    Is there any additional question ?
    Actually the last part of the question is

    "The differential equation you find in these new variables should be the simplest possible equation
    you could ever have to solve!"

    Sorry i lefted that out.
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  9. #9
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    Quote Originally Posted by ynotidas View Post
    Actually the last part of the question is

    "The differential equation you find in these new variables should be the simplest possible equation
    you could ever have to solve!"

    Sorry i lefted that out.
    I think an easier one is \frac{d\theta}{dr} = 0 but that's just me
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  10. #10
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    Quote Originally Posted by danny arrigo View Post
    I think an easier one is \frac{d\theta}{dr} = 0 but that's just me
    For sure !
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  11. #11
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    the answer is d\theta/dr = 1 ?
    I don't need to substitute back to get it in terms of 'x' and 'u'?
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  12. #12
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    As I said before if the question is only to convert the equation into \frac{d\theta}{dr}, then \frac{d\theta}{dr} = 1 is the required answer.
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  13. #13
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    OH thank you very much!

    I'll ask my lecturer on monday if he wants me to do anything more for this question!

    thanks again!
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  14. #14
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    That is the best way to be sure !
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