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Homework Help: Differential Equation

  1. Feb 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Solve the inital value first order linear differential equation

    y'=y+x y(0) = 2

    2. Relevant equations

    3. The attempt at a solution


    That's as far as I got. I'm not sure how to approach this. I've looked through my notes and book, and I don't have any examples that are similar to this. I looked online and it said something about raising the coefficient of y to e, but I'm not sure what to do after.
  2. jcsd
  3. Feb 8, 2009 #2


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    Gold Member

    You should multiply by an exponential factor.
    Usually the exponential factor is of the form:
    if y'+f(x)y=g(x)
    then the exponential factor is:
    u=exp([tex]\int f(x)dx[/tex])
    Take the integral of xexp(-x) by parts, and then your'e done.
  4. Feb 8, 2009 #3
    Hi; I tried using an integration factor, and using the initial value provided (y(0)=2), I arrived at a particular solution of [tex]y=x-1+3e^{-x}[/tex].

    I just saw loop's reply above. The integration factor I obtained was [tex]e^x[/tex].
  5. Feb 9, 2009 #4


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    Science Advisor

    For the linear equation y'+ f(x)y= g(x), and integrating factor is a function m(x) such that multiplying by it, m(x)y'+ m(x)f(x)y= f(x)g(x), reduces the left side to a single derivative: (m(x)y)'. Since, by the product rule, (m(x)y)'= my'+ m'y= my'+ mfy, we must have m'= m(x)f(x) which is, itself, a separable differential equation: dm/m= f(x)dx so [itex]ln(m)= \int f(x)dx[/itex] and so [itex]m(x)= e^{\int f(x)dx}[/itex]. For this particular problem f(x) is the constant -1 so your integrating factor is [itex]e^{-x}[/itex], not [itex]e^x[/itex]. Multiplying the equation by [itex]e^{-x}[/itex], we have [itex]e^{-x}y'- e^{-x}y= (e^{-x}y)'= xe^{-x}[/itex]. Integrating both sides of that (the left side by parts, as loop quantum gravity said) you get [itex]e^{-x}y= -(x+1)e^{-x}+ C[/itex] or [itex]y= -x- 1+ Ce^x[/itex] and, since y(0)= 2, 2= -1+ C and C= 3.
    [itex]y= -x-1+ 3e^x[/itex]

    Notice that if [itex]y= x- 1+ 3e^{-x}[/itex] then [itex]y'= 1- 3e^{-x}[/itex] while [itex]y+ x= x- 1+ 3e^{-x}+ x= 2x- 1+ 3e^{-x}[/itex] so your y does NOT satisfy the differential equation.
    Last edited by a moderator: Feb 9, 2009
  6. Feb 9, 2009 #5
    Thanks for pointing out that mistake Ivy...
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