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First order nonlinear differential equation

by process91
Tags: differential, equation, nonlinear, order
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process91
#1
Sep26-11, 05:35 PM
P: 106
1. The problem statement, all variables and given/known data
Find the orthogonal trajectories of the given families of curves.
[tex]x^2 + y^2+2Cy=1[/tex]

2. Relevant equations
The book has covered homogeneous and separable methods.


3. The attempt at a solution
To find the orthogonal trajectories, we simply find the curves whose tangents are perpendicular to the tangents of any curves in the original family. Implicit differentiation of the original equation yields
[tex]2x+2y y' + 2C y' = 0[/tex]
Solving for C in the original equation:
[tex]C = \frac{1-x^2-y^2}{2y}[/tex]
Solving for y' and substituting C:
[tex]y' = \frac{-2xy}{1+y^2-x^2}[/tex]
Our solutions are all curves whose tangent lines are the negative reciprocal of this, however this is not a separable or homogeneous equation as far as I can tell, so neither is its reciprocal, and thus I am stuck.

The answer, as given by the book, is
Spoiler
[itex]x^2+y^2-2Cx+1=0[/itex]

which I have verified is correct by checking that y' is the negative reciprocal of the calculated y' above, but how can this be derived without knowing the answer first?
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dynamicsolo
#2
Sep26-11, 05:46 PM
HW Helper
P: 1,662
Quote Quote by process91 View Post

Find the orthogonal trajectories of the given families of curves.
[tex]x^2 + y^2+2Cy=1[/tex]


[tex]2x+2y y' + 2C y' = 0[/tex]
I think you should have been less in a hurry to rid yourself of the 'C'. Write y' for the first family of curves as you had it at first, then find y' for the orthogonal family.

Now solve the resulting separable equation, leaving the 'C' in place for the moment (you know its relationship to x and y , but save that for the end). Upon integration, you will have another "arbitrary constant", but this one really is. Now you could replace 'C'.

(Incidentally, the 'C' in the solution in your "spoiler" is not the same 'C' you had earlier. I would have written the new one as 'A', or something other than 'C'...)

EDIT: I'm not buying that solution anyway, for two reasons. First, if you differentiate it implicitly, you don't get the expression for y' which is the reciprocal of the result for y' of the original family of curves. Second, try plotting the "solution" for various values of "C". The original curves are a set of circles of varying radii and centers along the y-axis, all of which pass through ( 1 , 0 ); you can see this after "completing the square". The "solution" would be a set of circles with centers along the x-axis, but with r2 = -1 (!) . If you try changing the sign in the "solution" to -1 , you will now get circles centered along the x-axis, but they are pretty definitely not orthogonal curves.

(LATER: I take this back -- I should have pushed on a bit further and the graphing problem was also resolved... can't win 'em all...)
process91
#3
Sep26-11, 06:02 PM
P: 106
I tried that initially, but it did not seem to work.

Leaving C in place,
[tex]y' = \frac{-x}{y+C}[/tex]
So we are looking for curves with derivatives which meet the following criteria:
[tex]y' = \frac{y+C}{x}[/tex]

This is separable, and we get
[tex]\ln|y+C|=ln|x|+D[/tex] for some new constant D.
From the original equation, when x=1 we have that y=0 or y=-2C, both implying that D=ln|C| so
[tex]|y+C| = |Cx|[/tex]
I'm not sure where to go from here. If I can justify removing the absolute value signs, which I think I can, then I would substitute C from the original equation.

[tex]C=\frac{1-x^2-y^2}{2y}[/tex]
[tex]y^2+1-x^2 = 2Cxy[/tex]

But this isn't the answer... where did I go wrong?

process91
#4
Sep26-11, 06:08 PM
P: 106
First order nonlinear differential equation

It is strange to me that this doesn't work, I would think that if C is a constant I should be able to toss it around and integrate it just like any other constant, however I think perhaps because C is implicitly defined in the first equation this is not a valid assumption.
process91
#5
Sep26-11, 06:15 PM
P: 106
Quote Quote by dynamicsolo View Post
EDIT: I'm not buying that solution anyway, for two reasons. First, if you differentiate it implicitly, you don't get the expression for y' which is the reciprocal of the result for y' of the original family of curves. Second, try plotting the "solution" for various values of "C". The original curves are a set of circles of varying radii and centers along the y-axis, all of which pass through ( 1 , 0 ); you can see this after "completing the square". The "solution" would be a set of circles with centers along the x-axis, but with r2 = -1 (!) . If you try changing the sign in the "solution" to -1 , you will now get circles centered along the x-axis, but they are pretty definitely [b]not[/i] orthogonal curves.
Here's how I verified the solution:
Switching constants per your suggestion, the answer given in the book is
[tex]x^2+y^2 -2Ax+1=0[/tex]
Solving for A yields
[tex]A = \frac{x^2+y^2+1}{2x}[/tex]
Implicitly differentiating the answer:
[tex]2x+2yy'-2A=0[/tex]
Solving for y':
[tex]y' = \frac{A-x}{y}[/tex]
Substituting A:
[tex]y'=\frac{1+y^2-x^2}{2xy}[/tex]

which is the negative reciprocal of the original.
dynamicsolo
#6
Sep26-11, 06:42 PM
HW Helper
P: 1,662
Sorry, I should have followed through on the substitution of 'A' : that does work.

There is a restriction of the values of 'A' that should be noted (which is why I wasn't getting anything to plot). For the first family of curves, we complete the square to obtain

x2 + ( y + C )2 = 1 + C2 ;

since the right-hand side is always positive, this will produce a circle for any real value C . However, the orthogonal curves produce circles with

( x - A )2 + y2 = A2 - 1 ,

which will only produce circles for | A | > 1 . (And now I'm mystified as to what I was plotting before, but these circles are orthogonal to the first set.) Interestingly, this set of circles is non-intersecting.

My apologies for the misses on this problem. I'm still looking for a way to derive the solution curves.
process91
#7
Sep26-11, 07:01 PM
P: 106
Ah, I think it's a Bernoulli Equation:
I need to find all curves which meet this criteria:
[tex]y'=\frac{1+y^2-x^2}{2xy} = \frac{1-x^2}{2xy} + \frac{1}{2x}y[/tex]
Proceeding on this track now...
process91
#8
Sep26-11, 07:13 PM
P: 106
Yes, that worked. Thanks for your help!
dynamicsolo
#9
Sep27-11, 12:04 AM
HW Helper
P: 1,662
I'm glad you did succeed in completing the solution. I only had time to get as far as finding that the DE is not exact (misses by a minus sign), but had to wait until later to go on. It is a Bernoulli equation with n = -1 ; I was hoping to find a simpler way to arrive at the solution curve, but I think this is the best way there.
process91
#10
Sep27-11, 12:06 AM
P: 106
Sorry, what do you mean "misses by a minus sign", did I type it in wrong on here? I ended up getting the correct result treating it as a Bernoulli equation.

Any chance you could help me with this one:
http://www.physicsforums.com/showthread.php?t=534106

It's a follow up question, very similar, but looks even more complicated unfortunately.
dynamicsolo
#11
Sep27-11, 12:13 AM
HW Helper
P: 1,662
Quote Quote by process91 View Post
Sorry, what do you mean "misses by a minus sign", did I type it in wrong on here? I ended up getting the correct result treating it as a Bernoulli equation.
I re-wrote the y' = (y + C)/x equation, with C substituted in, in the form M dx + N dy = 0 . It turns out that My = -Nx, so the DE isn't "exact" (which would have made it easier to solve). The method you arrived at seems to be the simpler approach to use.

I'll take a look at the other problem, but I won't promise much right now, since it's after midnight here and my reliability starts to plummet...


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