Results 1 to 4 of 4

Math Help - 1st order linear differential equation

  1. #1
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381

    1st order linear differential equation

    find the general solution
    2(2xy+4y - 3)dx +  (x+2)^2dy  =  0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    find the general solution
    2(2xy+4y - 3)dx + (x+2)^2dy = 0

    \implies \frac{dy}{dx} + \frac{4}{x+2}y = \frac{6}{(x+2)^2}

    Theorem:
    the general solution of \frac{dy}{dx} + q(x)y = r(x) is given by y = cy_1 + y_2, where y_1 = \frac{1}{Q(x)} and y_2 = \frac{1}{Q(x)}\int_0^x r(\tau )Q(\tau ) d\tau and Q(x) = \exp \left[ \int_0^x q(\tau ) d\tau \right], and also y_1(0) = 1, y_2(0) = 0 so that y(0) = c, the initial condition.

    now, from the given, we have q(x) = \frac{4}{x+2} and r(x) = \frac{6}{(x+2)^2}

    so Q(x) = \exp \left[ \int_0^x \frac{4}{\tau +2} d\tau \right] = \exp \left[ 4 \left( {\ln (x+2) - \ln 2} \right) \right] = \frac{(x+2)^4}{16}

    therefore, y_1 = \frac{16}{(x+2)^4}

    y_2 = \frac{16}{(x+2)^4} \int_0^x \frac{6}{(\tau +2)^2} \cdot \frac{(\tau + 2)^4}{16} \, d\tau = \frac{6}{(x+2)^4} \int_0^x (\tau +2)^2\, d\tau

     = \frac{6}{(x+2)^4} \left( {\frac{x^3}{3} + 2x^2 + 2x} \right) = \frac{2}{(x+2)^4} \left( {x^3 + 6x^2 + 6x} \right) = \frac{2x^3 + 12x^2 + 12x}{(x+2)^4}

    therefore, the general solution is y = c\cdot \frac{16}{(x+2)^4} + \frac{2x^3 + 12x^2 + 12x}{(x+2)^4}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381
    thanks kuya
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    thanks kuya
    no problem kuya! (baka nga mas matanda ka pa sa akin.. )
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. First Order Linear Differential Equation
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: March 11th 2011, 04:38 AM
  2. 2nd order linear differential equation
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: January 16th 2011, 07:52 AM
  3. First Order Linear Differential Equation
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: August 11th 2010, 03:15 AM
  4. First-Order Linear Differential Equation
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: May 1st 2010, 10:58 AM
  5. Non linear 2nd order differential equation
    Posted in the Differential Equations Forum
    Replies: 8
    Last Post: July 27th 2009, 02:37 AM

Search Tags


/mathhelpforum @mathhelpforum