1st order nonlinear differential equation

In summary: C1 - ln(1-v) + ln(2v+1) + 3C2dx/x - dv/2v2 - v = 03ln(x) + 3C1 - ln(1-v) + ln(2v+1) = 0In summary, the problem is that the problem is that the integrating factor for dy is not found
  • #1
360
0
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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  • #2
hi magnifik! :smile:
magnifik said:
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
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
dy = vdx + xdv :wink:
 
  • #5
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
magnifik said:
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
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
magnifik said:
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 got to do.
 
  • #9
thanks for the help. much appreciated!
 
  • #10
It's not a problem. I actually enjoy solving these problems.
 
  • #11
I thought y = ux, not vx.
 
  • #12
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
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
hi magnifik! :smile:

(just got up :zzz: …)
magnifik said:
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:
magnifik said:
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
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
 

1. What is a 1st order nonlinear differential equation?

A 1st order nonlinear differential equation is an equation that involves a function and its derivative, with the function being raised to a power or multiplied by itself. It is called nonlinear because the function and its derivative cannot be expressed as a linear combination of each other.

2. How do you solve a 1st order nonlinear differential equation?

There is no universal method for solving 1st order nonlinear differential equations. However, some common techniques include separation of variables, substitution, and transforming the equation into a linear equation. It is also important to check for exact solutions and use numerical methods if necessary.

3. What are the applications of 1st order nonlinear differential equations?

1st order nonlinear differential equations have many applications in various fields such as physics, engineering, biology, and economics. They are used to model real-world phenomena that involve nonlinear relationships, such as population growth, chemical reactions, and fluid dynamics.

4. How do you verify the solution of a 1st order nonlinear differential equation?

To verify the solution of a 1st order nonlinear differential equation, you can substitute the proposed solution into the original equation and check if it satisfies the equation. Another way is to take the derivative of the proposed solution and see if it matches the derivative in the original equation.

5. Can 1st order nonlinear differential equations have more than one solution?

Yes, 1st order nonlinear differential equations can have multiple solutions. This is because they are not unique, unlike linear differential equations, which have a single unique solution. The number of solutions can also depend on the initial conditions given in the problem.

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