# Math Help - Nonlinear ODE

1. ## 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.

2. Hi

I would say

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

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

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

$\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)" 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)" />

$(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)" 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 ?

3. Originally Posted by running-gag
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?

4. $u(r,\theta) = r\:\sin \theta$

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

$du = \sin \theta \:dr + r\:\cos \theta\:d\theta$

5. OHH i was putting 'r' as a constant, instead of a function of 'u', my bad!
since $d\theta = dr$

$\theta = r$
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?

6. $\theta = r$
or
$d\theta/dr = 1$

so $x=rcosr$ and $u=rsinr$

I'm lost.

7. 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 ?

8. Originally Posted by running-gag
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.

9. Originally Posted by ynotidas
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

10. Originally Posted by danny arrigo
I think an easier one is $\frac{d\theta}{dr} = 0$ but that's just me
For sure !

11. the answer is $d\theta/dr = 1$ ?
I don't need to substitute back to get it in terms of 'x' and 'u'?

12. 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.

13. 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!

14. That is the best way to be sure !