Cant find the general solution

  • Thread starter Laney5
  • Start date
5
0

Main Question or Discussion Point

Im looking for a general solution to the following equation but i cant seem to get an answer.
3xdy = (ln(y[tex]^{6}[/tex]) - 6lnx)ydx

I've got this far anyway..

[tex]\frac{3x}{y}[/tex] dy = (ln(y[tex]^{6}[/tex])-ln(x[tex]^{6}[/tex])) dx

[tex]\frac{dy}{dx}[/tex] = [tex]\frac{y}{3x}[/tex] . 6ln[tex]\frac{y}{x}[/tex]

[tex]\frac{dy}{dx}[/tex] = [tex]\frac{y}{x}[/tex] . 2ln[tex]\frac{y}{x}[/tex]

Let z = [tex]\frac{y}{x}[/tex]

[tex]\frac{dz}{dx}[/tex] = (x[tex]\frac{dy}{dx}[/tex] - y) / x[tex]^{2}[/tex]

[tex]\frac{dz}{dx}[/tex] = [tex]\frac{1}{x}[/tex]([tex]\frac{y}{x}[/tex] . 2ln[tex]\frac{y}{x}[/tex]) - [tex]\frac{y}{x^2}[/tex]

z = [tex]\frac{y}{x}[/tex] gives

[tex]\frac{dz}{dx}[/tex] = [tex]\frac{1}{x}[/tex](2zlnz) - [tex]\frac{z}{x}[/tex]

[tex]\frac{dz}{dx}[/tex] = [tex]\frac{1}{x}[/tex](2zlnz-z)

[tex]\frac{dz}{2zlnz - z}[/tex] = [tex]\frac{1}{x}[/tex]dx

now im stuck...??
 

Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
41,734
893
I presume it is the
[tex]\frac{dz}{2zlnz- z}[/tex]
that is giving you the problem!

Write it as
[tex]\frac{1}{2ln z- 1}\frac{dz}{z}[/tex]
and let u= 2ln z- 1.
 
5
0
Ya, its that part alright!

Do i write it like [tex]\frac{1}{u}[/tex] [tex]\int[/tex] [tex]\frac{dz}{z}[/tex]?

[tex]\frac{1}{u}[/tex] [tex]\int[/tex] [tex]\frac{dz}{z}[/tex] = Integral([tex]\frac{dx}{x}[/tex])



[tex]\frac{1}{u}[/tex] (lnz) = lnx + C

[tex]\frac{1}{2lnz - 1}[/tex] (lnz) = lnx + C

Is this right?
 
Last edited:
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,598
6
Ya, its that part alright!

Do i write it like [tex]\frac{1}{u}[/tex] [tex]\int[/tex] [tex]\frac{dz}{z}[/tex]?

[tex]\frac{1}{u}[/tex] [tex]\int[/tex] [tex]\frac{dz}{z}[/tex] = Integral([tex]\frac{dx}{x}[/tex])



[tex]\frac{1}{u}[/tex] (lnz) = lnx + C

[tex]\frac{1}{2lnz - 1}[/tex] (lnz) = lnx + C

Is this right?
No. Notice that u is a function of z and is therefore not constant under integration with respect to z. Therefore you need to make a change of variable from [itex]dz\mapsto du[/itex].
 
5
0
u=2lnz-1
du=2([tex]\frac{1}{z}[/tex])dz
([tex]\frac{1}{2}[/tex])du=([tex]\frac{1}{z}[/tex])dx

([tex]\frac{1}{u}[/tex]).([tex]\frac{1}{2}[/tex])du=([tex]\frac{1}{x}[/tex])dx

[tex]\frac{1}{2}\int[/tex][tex]\frac{1}{u}[/tex]du=[tex]\int\frac{1}{x}[/tex]dx

([tex]\frac{1}{2}[/tex])lnu=lnx + C

ln(u)[tex]^{1/2}[/tex] - lnx = C

ln[[tex]\frac{(u)^{1/2}}{x}[/tex]] = C

u=2lnz-1

ln[[tex]\frac{(2lnz-1)^{1/2}}{x}[/tex]] = C

z=[tex]\frac{y}{x}[/tex]

ln[[tex]\frac{(2ln\frac{y}{x}-1)^{1/2}}{x}[/tex]] = C

[tex]e^{ln[\frac{(2ln\frac{y}{x}-1)^{1/2}}{x}] }[/tex] = [tex]e^C[/tex] , Let [tex]e^C[/tex] = B

[tex]\frac{(2ln\frac{y}{x}-1)^{1/2}}{x}[/tex] = B

Am i doing it correctly?
 
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,598
6
Looks okay to me :smile:
 

Related Threads for: Cant find the general solution

  • Last Post
Replies
1
Views
1K
Replies
1
Views
631
Replies
4
Views
896
Replies
1
Views
2K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
2
Views
2K
Replies
2
Views
4K
  • Last Post
Replies
3
Views
2K
Top