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Cant find the general solution

  1. Apr 15, 2008 #1
    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...??
     
  2. jcsd
  3. Apr 15, 2008 #2

    HallsofIvy

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    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.
     
  4. Apr 15, 2008 #3
    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: Apr 15, 2008
  5. Apr 16, 2008 #4

    Hootenanny

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    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].
     
  6. Apr 16, 2008 #5
    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?
     
  7. Apr 17, 2008 #6

    Hootenanny

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    Looks okay to me :smile:
     
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