1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integration by substitution diff. eq.

  1. Oct 29, 2012 #1
    1. The problem statement, all variables and given/known data
    From Larson, 9th Edition: Section 4.5. Solve the differential equation

    [tex]\frac{\operatorname{d}y}{\operatorname{d}x}=4x+ \frac{4x}{\sqrt{16-x^2}}[/tex]


    2. Relevant equations



    3. The attempt at a solution
    Well, I can get my book's answer, but not through doing things in the prescribed way, because I get a different answer following the section's guidelines and examples. Using those, I choose a variable [itex]u=16-x^2[/itex]. Taking the derivative, we obtain [itex]du=2x \ \operatorname{d}x[/itex]. Rearranging this to get a [itex]4x[/itex] factor, we have [tex]-2 \operatorname{d}u = 4x \operatorname{d}x[/tex].

    Plugging this back into the differential equation and rearranging the d[itex]x[/itex] gives us:
    [tex]
    \int \operatorname{d}y = -2 \int \operatorname{d}u + (-2)\int u^\frac{-1}{2} \operatorname{d}u[/tex]

    Now I wind up with [tex]y=-2u - 4\sqrt{u} + C[/tex]
    Which, using substitution becomes [tex]2x^2-4\sqrt{16-x^2}-32+C[/tex], but the book's answer is [tex]2x^2 -4\sqrt{16-x^2} + C[/tex]

    Now I can get this answer if I just take the integral of the first [itex]4x[/itex] and only use substitution for the second term, but I am not sure why my answers should be different. Unless it really doesn't matter and the 32 from my answer just becomes part of the constant of integration [itex]C[/itex]
     
    Last edited: Oct 29, 2012
  2. jcsd
  3. Oct 29, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It's exactly what you think. The -32 can be absorbed into the C. Indefinite integrals can always have a constant added and they work just as well.
     
  4. Oct 29, 2012 #3
    Thanks Dick. Just out of curiosity, what would happen if this were a definite integral? The same, I am assuming, since we would have to subtract -32 from -32?
     
  5. Oct 29, 2012 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You wouldn't see the -32 in a definite integral, because again, as you say, it would cancel. That's why it's not important.
     
  6. Oct 30, 2012 #5

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    It's not necessary to use u-substitution to evaluate both integrals simultaneously.
    The integral of 4x is readily evaluated using the standard formula. The u-substitution
    can be used to evaluate the integral with the radical.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Integration by substitution diff. eq.
Loading...