ODE substitution question

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  • #1
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I would appreciate any advice on the following ODE substitution question:
xy' = y + (x^2 + y^2)^.5
Dividing thru by x and using the usual y/x substitution, I get:
y' = v + (1 + v^2)^.5 but I don't know if that is right or how to integrate the left side. The book has the answer of y + (x^2 + y^2)^.5 = Cx^2
Am I not handling the radical correctly?
 

Answers and Replies

  • #2
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It looks like you didn't differentiate implicitly correctly. You should have a step that looks like:

v+xdv/dx = v+sqrt(1+v^2)

then

x*dv/dx = sqrt(1+v^2)
 
  • #3
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You're right, I did leave that out but I think my problem is not knowing how to integrate that radical. I tried entering it in mathematica and it came up with a hyperbolic sine if I remember correctly, is that what you are seeing?
 
  • #4
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Whenever I see a constant + variable^2 underneath the radical, I think tangent substitution. See if that you takes you somewhere.
 
  • #5
dextercioby
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Yes, the solution is in terms of hyperbolic sine.
 
  • #6
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Great. We "skipped" those in my Calc 2 class. If you'd be willing to opine on one other one that I am not getting right that I know doesn't involve a wild integral, I'd appreciate it. I'll post it here in a sec in this thread.
 
  • #7
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The solution doesn't have to be in terms of sinh. It can be a log.
 
  • #8
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Did you cover trig subs in your Calc2 class? That'll get you there--you don't need hyperbolics for it, though they "clean up" the end result.
 
  • #9
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x(x+y)y' + y(3x+y) = 0
I did the following steps:
x(x+y)y' = -y(3x+y)
Divide thru...
y' = -v * [ (3x+y)/(x+y) ]
Then I divided what was in the brackets by x, obtaining:
v + xv' = -v * (3 + v)/(1+v)
Speeding up a little, I get:
(1+v)/(3-v^2) dv = dx / x
I'm under the impression I need to split the left side into:
1/ (3-v^2) + v/(3-v^2) but again, I'm not sure I know how to do the integral.
The solutions manual does the whole thing in just a few steps using a method I haven't seen in my class but makes it look more simple than it really is.
 
  • #10
dextercioby
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The first integral can be expressed in terms of argtanh x, while the second one is very easy to integrate by a simple substitution.
 

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