1st order nonlinear differential equation

[magnifik]

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

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

 

[tiny-tim]

hi magnifik!

dy = (2y²/x² - 1)dx
(2y²/x² - 1)dx - dy = 0
…
dx/x - dv/(2v² - 2v - 1) = 0

isn't it (2v² - v - 1) ?

 

[magnifik]

i have tried it several times and seem to always get 2v² - 2v - 1. do you get the correct solution using 2v² - v - 1??

 

[tiny-tim]

dy = vdx + xdv

 

[magnifik]

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 :\

 

[Char. Limit]

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.

 

[magnifik]

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) = e^C
trying to solve for v i get,
x³/e^C = -2v² + 3v - 1
not sure if this step is correct:
0 = -2v² + 3v - (1 - x³/e^C)

i'm assuming i do the quadratic formula to find the answer then use v = y/x to get y?? is this correct?

 

[Char. Limit]

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) = e^C
trying to solve for v i get,
x³/e^C = -2v² + 3v - 1
not sure if this step is correct:
0 = -2v² + 3v - (1 + x³/e^C)

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 got to do.

 

[magnifik]

thanks for the help. much appreciated!

 

[Char. Limit]

It's not a problem. I actually enjoy solving these problems.

 

[pr0blumz]

I thought y = ux, not vx.

 

[Char. Limit]

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.

 

[magnifik]

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/e^C))/-4
v = y/x
y = x[-2 + sqrt(9 - 4(-2)(-1-x³/e^C))/-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

 

[tiny-tim]

hi magnifik!

(just got up :zzz: …)

ok, i see how you got 2v² - v - 1

i didn't "get" it … it was already there!

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

combining the constants, i have
3ln(x) - [ln(1-v) + ln(2v+1)] = C

noooo … you've copied your own equation wrong … that + should be a -

 

[magnifik]

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