Results 1 to 4 of 4

Math Help - Differential Equations - Intergrating by factors

  1. #1
    Super Member
    Joined
    Dec 2008
    Posts
    509

    Differential Equations - Intergrating by factors

    Hi

    Need help on the following question:

    Find the general solution of the differential equation

    \frac{dy}{dx}-\frac{4y}{x}=e^{x^{-3}}

    p(x)=\frac{-4}{x}

    q(x)=e^{x^{-3}}

    I=e^{\int \frac{-4}{x} dx
    I=e^{ln|x^{-4}|}
    I=x^{-4}

    x^{-4}y=\int x^{-4}e^{x^{-3}} dx

    ok, i tried to integrate by parts however i never get the correct answer.

    this is what i did

    u = e^{x^{-3}}
    du = -3x^{x^{-3}}

    dv = x^{-5}
    v = \frac{x^{-4}}{-4}

    i have also tried the other way but still not correct.

    P.S
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,695
    Thanks
    1509
    \frac{dy}{dx} - \frac{4}{x}y = e^{x^{-3}}.

    The integrating factor is e^{\int{-\frac{4}{x}\,dx}} = e^{-4\ln{x}} = e^{\ln{x^{-4}}} = x^{-4}.


    Multiplying through by the integrating factor gives

    x^{-4}\frac{dy}{dx} - 4x^{-5}y = x^{-4}e^{x^{-3}}

    \frac{d}{dx}(x^{-4}y) = x^{-4}e^{x^{-3}}

    x^{-4}y = \int{x^{-4}e^{x^{-3}}\,dx}

    x^{-4}y = -\frac{1}{3}\int{e^{x^{-3}}(-3x^{-4})\,dx}.


    Now let u = x^{-3} so that \frac{du}{dx} = -3x^{-4} and the DE becomes

    x^{-4}y = -\frac{1}{3}\int{e^u\,\frac{du}{dx}\,dx}

    x^{-4}y = -\frac{1}{3}\int{e^u\,du}

    x^{-4}y = -\frac{1}{3}e^u + C

    x^{-4}y = -\frac{1}{3}e^{x^{-3}} + C

    y = -\frac{1}{3}x^4e^{x^{-3}} + Cx^4.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2008
    Posts
    509
    i don't understand how you got the following line?
    x^{-4}y = -\frac{1}{3}\int{e^{x^{-3}}(-3x^{-4})\,dx}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,695
    Thanks
    1509
    I knew I needed to "create" the derivative of x^{-3}, which is -3x^{-4}.

    I already had the x^{-4}, so all I needed to do was multiply by -3. But to keep the equation balanced, that means I had to multiply by -\frac{1}{3} as well.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: February 29th 2012, 08:28 AM
  2. Differential Equations Intergating Factors
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: November 8th 2010, 06:45 PM
  3. Differential Equations Intergating Factors
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: September 29th 2010, 05:09 AM
  4. Intergrating Factors
    Posted in the Differential Equations Forum
    Replies: 16
    Last Post: March 30th 2010, 08:12 AM
  5. Replies: 2
    Last Post: May 18th 2009, 03:49 AM

Search Tags


/mathhelpforum @mathhelpforum