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1st order nonlinear differential equation

  • Thread starter magnifik
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  • #1
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this is the problem:
x2y' = (2y2 - x2)

here's what i have done so far:

dy = (2y2/x2 - 1)dx
(2y2/x2 - 1)dx - dy = 0
i used the substitution y = xv then found an integrating factor and got
dx/x - dv/(2v2 - 2v - 1) = 0
but i am stuck at this point..
i know ln(x) + C is the first part, but is there an easy way to find the second part? or am i totally off to begin with?

this is the solution:
[PLAIN]http://www4c.wolframalpha.com/Calculate/MSP/MSP50419e2227898a2f2f0000026hh297a4ebd6b3h?MSPStoreType=image/gif&s=33&w=154&h=47 [Broken]

i am at a loss for how they got to this point
 
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Answers and Replies

  • #2
tiny-tim
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hi magnifik! :smile:
dy = (2y2/x2 - 1)dx
(2y2/x2 - 1)dx - dy = 0

dx/x - dv/(2v2 - 2v - 1) = 0
isn't it (2v2 - v - 1) ?
 
  • #3
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i have tried it several times and seem to always get 2v2 - 2v - 1. do you get the correct solution using 2v2 - v - 1??
 
  • #4
tiny-tim
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dy = vdx + xdv :wink:
 
  • #5
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ok, i see how you got 2v2 - v - 1

now i have
dx/x - dv/2v2 - v - 1 = 0
...
[ln(x) + C] - [1/3(ln(1-v) - ln(2v+1))+C] = 0
multiply everything by 3
3ln(x) + 3C1 - ln(1-v) + ln(2v+1) + 3C2

stuck here :\
 
  • #6
Char. Limit
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ok, i see how you got 2v2 - v - 1

now i have
dx/x - dv/2v2 - v - 1 = 0
...
[ln(x) + C] - [1/3(ln(1-v) - ln(2v+1))+C] = 0
multiply everything by 3
3ln(x) + 3C1 - ln(1-v) + ln(2v+1) + 3C2

stuck here :\
Define C3=3C1+3C2. Remember that the RHS is 0 or a constant (let's make it 0, merging that constant with C3 as well). Then rearrange to solve for v.
 
  • #7
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combining the constants, i have
3ln(x) - [ln(1-v) + ln(2v+1)] = C
using some properties of natural log,
ln(x3) - [ln((1-v)(2v+1))] = C
which simplifies to
ln(x3/(-2v2 + 3v - 1) = C
(x3/(-2v2 + 3v - 1) = eC
trying to solve for v i get,
x3/eC = -2v2 + 3v - 1
not sure if this step is correct:
0 = -2v2 + 3v - (1 - x3/eC)

i'm assuming i do the quadratic formula to find the answer then use v = y/x to get y?? is this correct?
 
  • #8
Char. Limit
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combining the constants, i have
3ln(x) - [ln(1-v) + ln(2v+1)] = C
using some properties of natural log,
ln(x3) - [ln((1-v)(2v+1))] = C
which simplifies to
ln(x3/(-2v2 + 3v - 1) = C
(x3/(-2v2 + 3v - 1) = eC
trying to solve for v i get,
x3/eC = -2v2 + 3v - 1
not sure if this step is correct:
0 = -2v2 + 3v - (1 + x3/eC)

i'm assuming i do the quadratic formula to find the answer then use v = y/x to get y?? is this correct?
Fixed that for you. And yes, that's what you gotta do.
 
  • #9
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thanks for the help. much appreciated!
 
  • #10
Char. Limit
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It's not a problem. I actually enjoy solving these problems.
 
  • #11
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I thought y = ux, not vx.
 
  • #12
Char. Limit
Gold Member
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The only real difference is using a u instead of a v. It's similar to using t or x in your equations. It doesn't really matter, as long as it's uniquely defined.
 
  • #13
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ok...i tried to solve the quadratic equation but am still not getting the correct answer

x = -b + sqrt(b2 - 4ac)/2a

0 = -2v2 + 3v - (1 + x3/eC)
v = -2 + sqrt(9 - 4(-2)(-1-x3/eC))/-4
v = y/x
y = x[-2 + sqrt(9 - 4(-2)(-1-x3/eC))/-4]

i'm not sure how this would simplify to the correct answer, which is
[PLAIN]https://www.physicsforums.com/latex_images/30/3082166-0.png [Broken]
 
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  • #14
tiny-tim
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hi magnifik! :smile:

(just got up :zzz: …)
ok, i see how you got 2v2 - v - 1
i didn't "get" it :rolleyes:it was already there!! :rofl:
dx/x - dv/2v2 - v - 1 = 0
...
[ln(x) + C] - [1/3(ln(1-v) - ln(2v+1))+C] = 0
in the exams, you must write out everything in full

it'll stop you making mistakes, and even if you do make a mistake, you'll get more marks for following the right method …

in this case, your partial fractions are wrong: one of the ln should have a "2" in front of it :wink:
combining the constants, i have
3ln(x) - [ln(1-v) + ln(2v+1)] = C
noooo :redface: … you've copied your own equation wrong … that + should be a - :cry:
 
  • #15
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tiny tim,

i believe the integral is correct - there should not be a 2 in front of anything.
and the + should not be a -, but i did, however, group them mistakenly after i already distributed a -1
 

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