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Homework Help: 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:
    \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


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    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


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    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


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    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.
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