Results 1 to 9 of 9

Thread: Solve y'=y(1 + xy^4)

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    Solve y'=y(1 + xy^4)

    Last edited by mr fantastic; Oct 21st 2010 at 04:49 AM. Reason: Re-titled.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    Awards
    2
    It's Bernoulli, with standard form

    $\displaystyle y'-y=xy^{5}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    ok
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    bernulli solution problematic points

    $\displaystyle y'-y=5xy^5\\$
    $\displaystyle \frac{y'}{y^5}-\frac{1}{y^4}=5x \, \, \, \, \, \,z=\frac{1}{y^4} \, \, \, \, \, \,\, \,dz=\frac{1}{y^5}dy$
    how does from $\displaystyle dz=\frac{1}{y^5}dy $ we get z' and y' we only have dz and dy here???
    in order to get y' we need to have dy/dx we dont have it here.

    then
    $\displaystyle z'-z=5x \, \, \, \, \, \, a=-1 \, \, \, \, \, A=-z \, \, \, \, \, \ e^{-z}(z'-z)=e^{-z} 5z \, \, \, \, \, $$\displaystyle (e^{-z}z)'=e^{-z} 5z \, \, \, \, \, e^{-z}z=\int ^z e^{-t} 5t +c $ after solving by parts i get
    $\displaystyle e^{-z}z=-5ze^{-z}+5e^{-z} +c$

    $\displaystyle c=e^{-z}z+5ze^{-z}-5e^{-z} $
    what to do now?
    Last edited by transgalactic; Oct 20th 2010 at 11:28 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    First off, if you let $\displaystyle z = y^{-4}$ then $\displaystyle \displaystyle \frac{dz}{dy} = -4y^{-5}$.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    ok i missed that -4 in the defferentiation
    what about my questions
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Well, you need to transform the DE which is in terms of $\displaystyle \displaystyle \frac{dy}{dx}$ into a DE in terms of $\displaystyle \displaystyle \frac{dz}{dx}$.

    You have $\displaystyle z = y^{-4}$.


    By the chain rule, $\displaystyle \displaystyle \frac{dz}{dx} = \frac{dz}{dy}\,\frac{dy}{dx}$.

    So $\displaystyle \displaystyle \frac{dz}{dx} = -4y^{-5}\,\frac{dy}{dx}$ and thus $\displaystyle \displaystyle \frac{dy}{dx} = -\frac{1}{4}y^5\,\frac{dz}{dx}$.


    Substituting everything into the DE

    $\displaystyle \displaystyle \frac{dy}{dx} - y = 5x\,y^5$

    $\displaystyle \displaystyle -\frac{1}{4}y^5\,\frac{dz}{dx} - y = 5x\,y^5$

    $\displaystyle \displaystyle -\frac{1}{4}\,\frac{dz}{dx} - y^{-4} = 5x$

    $\displaystyle \displaystyle -\frac{1}{4}\,\frac{dz}{dx} - z = 5x$

    $\displaystyle \displaystyle \frac{dz}{dx} + 4z = -20x$.


    This is now first order linear, which you can solve using an integrating factor.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    i got a similiar lenear differential equation
    and i solved it by multiplying in exponet etc
    and it doesnt come out
    ??
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Your integrating factor is $\displaystyle e^{\int{4\,dx}} = e^{4x}$.

    So multiplying through by the integrating factor gives

    $\displaystyle \displaystyle e^{4x}\,\frac{dz}{dx} + 4e^{4x}z = -20x\,e^{4x}$

    $\displaystyle \displaystyle \frac{d}{dx}(e^{4x}z) = -20x\,e^{4x}$

    $\displaystyle \displaystyle e^{4x}z = \int{-20x\,e^{4x}\,dx}$

    $\displaystyle \displaystyle e^{4x}z = -20\int{x\,e^{4x}\,dx}$

    $\displaystyle \displaystyle e^{4x}z = -20\left(\frac{1}{4}x\,e^{4x} - \int{\frac{1}{4}e^{4x}\,dx}\right)$

    $\displaystyle \displaystyle e^{4x}z = -5x\,e^{4x} + 5\int{e^{4x}\,dx}$

    $\displaystyle \displaystyle e^{4x}z = -5x\,e^{4x} + \frac{5}{4}e^{4x} + C$

    $\displaystyle \displaystyle z = -5x + \frac{5}{4} + Ce^{-4x}$

    $\displaystyle \displaystyle y^{-4} = -5x + \frac{5}{4} + Ce^{-4x}$

    $\displaystyle \displaystyle y = \left(-5x + \frac{5}{4} + Ce^{-4x}\right)^{-\frac{1}{4}}$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. need to solve summation equation to solve sum(x2)
    Posted in the Statistics Forum
    Replies: 2
    Last Post: Jul 16th 2010, 10:29 PM
  2. Solve for x
    Posted in the Algebra Forum
    Replies: 6
    Last Post: Mar 3rd 2010, 06:34 PM
  3. Replies: 1
    Last Post: Jun 9th 2009, 10:37 PM
  4. how to solve ..
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Aug 2nd 2008, 08:17 AM
  5. please help me solve this
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Aug 1st 2008, 09:50 PM

Search Tags


/mathhelpforum @mathhelpforum