Results 1 to 5 of 5

Math Help - Second Order Linear Differential Equation

  1. #1
    Junior Member
    Joined
    Oct 2008
    Posts
    65

    Second Order Linear Differential Equation

    If anyone can guide me through this problem, it would be much helpful. Thank you.

    Let u be a nonzero solution of the second-order equation

    y'' + P(x)y' + Q(x)y = 0

    Show that the substitution y = uv converts the equation

    y'' + P(x)y' + Q(x)y = R(x)

    into a first-order linear equation for v'
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member kjchauhan's Avatar
    Joined
    Nov 2009
    Posts
    137
    Thanks
    2
    Quote Originally Posted by hashshashin715 View Post
    If anyone can guide me through this problem, it would be much helpful. Thank you.

    Let u be a nonzero solution of the second-order equation

    y'' + P(x)y' + Q(x)y = 0

    Show that the substitution y = uv converts the equation

    y'' + P(x)y' + Q(x)y = R(x)

    into a first-order linear equation for v'
    let

    y=uv
    then
    y'=u \frac{dv}{dx}+\frac{du}{dx}v
    and
    y''=u \frac{d^2v}{dx^2}+2\frac{du}{dx}\frac{dv}{dx}+\fra  c{d^2u}{dx^2}v
    Substitute these values in the given equation and determine u
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2008
    Posts
    65
    Ok, but if I do that, I will still have u' and u'' left over. What do I do with them? Aren't I only suppose to express the equation in terms of v's and their derivatives only?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member kjchauhan's Avatar
    Joined
    Nov 2009
    Posts
    137
    Thanks
    2
    Quote Originally Posted by hashshashin715 View Post
    Ok, but if I do that, I will still have u' and u'' left over. What do I do with them? Aren't I only suppose to express the equation in terms of v's and their derivatives only?

    After Substituting, we get,

    (u''v+2u'v'+uv'')+P(u'v+uv')+Quv=R

    \therefore uv''+(ru'+Pu)v'+(u''+Pu'+Qu)v=R

    Since uis a nonzero solution, then,

    u''+Pu'+Qu=0

    \therefore uv''+(2u'+Pu)v'=R

    \therefore v''+\left(\frac{2}{u}u'+P\right)v'=\frac{R}{u}

    Now put v'=p, so that, v''=\frac{dp}{dx}, then we have,

    \frac{dp}{dx}+\left(\frac{2}{u}u'+P\right)p=\frac{  R}{u}

    Which is a linear equation in p,

    \therefore I.F.= e^{\int \left(\frac{2}{u}u'+P\right)dx} =u^2e^{\int P dx}

    \therefore pu^2e^{\int P dx}=\int \left(\frac{R}{u}u^2e^{\int P dx}\right)dx+c_1
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2008
    Posts
    65
    Oh, I see. Thank you
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 2nd order linear differential equation
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: January 16th 2011, 07:52 AM
  2. First Order Non-Linear Differential Equation
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: October 3rd 2010, 10:46 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. Second order non linear differential equation.
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: February 20th 2010, 12:49 PM

Search Tags


/mathhelpforum @mathhelpforum