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Homework Help: Solve second order diff equation using substitution

  1. Jul 2, 2010 #1
    1. The problem statement, all variables and given/known data

    d2y/dx2-dy/dx+y*exp(2x) = x*exp(2x)-1

    substitute t=exp(x) and set z(t)=y(x) and rewrite hence find all solutions

    3. The attempt at a solution

    Rewriting gives:
    d2z/dt2-dz/dt+z*t^2=(ln(t) * t^2) - 1

    however I dont see how this in any way helps us...
     
  2. jcsd
  3. Jul 2, 2010 #2

    HallsofIvy

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    Science Advisor

    You haven't changed the derivatives properly. You cannot just replace "dy/dx" with "dy/dt". You have to use the chain rule.

    dy/dx= (dy/dt)(dt/dx) so if t= e^x, dy/dx= e^x(dy/dt)= t(dy/dt).

    d^2y/dx^2= (d/dx)(dy/dx)= d/dx(t dy/dt)= t(d/dt(t(dy/dt)). By the product rule, that is t^2 d^2y/dt^2+ t dy/dt.
    The equation becomes (t^2 d^2y/dt^2+ t dy/dt)- t dy/dt+ t^2 y= t^2(d^2y/dt^2+ y)= t^2ln(t)- 1 so that d^2y/dt^2+ y= ln(t)- t^{-2}.

    The related homogeneous equation is easy to solve and I would recommend "variation of parameters" to find a specific solution to the entire equation.
     
  4. Jul 2, 2010 #3
    first of all:
    "d2y/dx2" does not equals "d2z/dt2" and "dy/dx" does not equals "dz/dt".

    but you right in saying that exp(2x) = t^2 and x = ln(t)

    you need to use the "chain rule" in differentiation !

    t=e^x

    z(t)=y(x)

    dz(t)/dx = dy(x)/dx => (dz(t)/dt)*(dt/dx) = y'(x)

    *try to go from here :)
    *don't forget the differentiate correctly when you do d^2z(t)/dx^2 = d^2y(x)/dx^2, it has the form (u*v)' = u'*v + u*v'

    Hope it will lead you to the right answer ! :)
     
  5. Jul 2, 2010 #4
    HallsofIvy beat me to it :)

    also I love Lagrange's variation of parameters !

    Here is how you do it:

    (( in your case y=z, x=t ))

    y1(x), y2(x) : are two linearly independent homogeneous solutions.
    g(x) : the particular expression, in your case = ln(t)- t^{-2}.

    u'1(x)*y1(x) + u'2(x)*y2(x) = 0
    u'1'(x)*y'1(x) + u'2(x)*y'2(x) = g(x)

    solve the system of equations and find u1(x) and u2(x) by integration.
    the particular part of the answer will be:

    Y(x) = u1(x)*y1(x) + u2(x)*y2(x).

    the complete answer for the ODE (A,B are constants):

    y(x) = [A*y1(x)] + [B*y1(x)] + [u1(x)*y1(x) + u2(x)*y2(x)].

    *Lagrange was a genius !
     
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