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Homework Help: Seperation of variables (easy)

  1. Jul 23, 2008 #1
    1. The problem statement, all variables and given/known data

    show the solution to dy/dx = [1+(1/y)]^1/2 is given by ∫[y^1/2]/[1+y]^1/2 dy = x + c

    3. The attempt at a solution

    I know this will be solved by seperation of variables...
    If i take the whole rhs over to the lhs I get...

    dy/[1+(1/y)]^1/2 = dx

    RHS no problems when I integrate, however cannot solve the lhs.
    I've tried substitution with u = 1+1/y, but didn't seem to help

    Which way should I go about this?
  2. jcsd
  3. Jul 23, 2008 #2


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    First of all, show that
    [tex]\frac{1}{(1 + 1/y)^{1/2}} = \frac{y^{1/2}}{(1+y)^{1/2}}[/tex]
    which is not hard (hint, multiply by 1 = a/a, where you choose a conveniently)
    Then you are done, the question only asks to show you that the integral of that over dy is equal to the integral of 1 over dx, which is just x + c. It doesn't ask you to actually do the integration, does it?
  4. Jul 23, 2008 #3
    Lets look at this problem:

    dz/dx = f(z) show that integral of dz/f(z) = x + c
    dz/f(z) = dx

    integral of left side = integral of right side get:

    integral of dz/f(z) = integral of dx

    integral of dz/f(z) = x + c

    I have shown what is asked of me!

    Unless you are asking the question wrong, it is exactly this easy. You're just a little algebra away from the answer.
  5. Jul 23, 2008 #4
    The rhs is straighforward, but your wording for the lhs has me confused. Is [tex]\frac{1}{(1 + 1/y)^{1/2}}[/tex] multiplied by root(y)/root(y) correct? Pardon my lack of surd expansion knowledge
  6. Jul 23, 2008 #5


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    Yep, that's a good try. Now rewrite it a bit and you're done. You will need rules like
    [tex]\frac{a}{b} \cdot \frac{c}{d} = \frac{a c}{b d} = a \frac{c}{b d} = \frac{a c}{b} \cdot \frac{1}{d} \cdots[/tex]
    [tex]\sqrt{a}\sqrt{b} = \sqrt{a b}[/tex]
    (a, b >= 0).
    But they are essential for anything involving algeba, so you will want to learn them anyway, if you don't know them yet.
  7. Jul 23, 2008 #6


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    Yes, that is correct. You can multiply both numerator and denominator by y1/2. Of course, that will be multiplying by y inside the square root in the denominator.

    Notice that the answer is itself given in terms of an integral so you don't have to actually do any integration.
  8. Jul 23, 2008 #7
    yeah, so just to double check, multiplication of the 2 demoninators should be as follows..

    {[1+(1/y)]^1/2} * {y^1/2} = [y +y/y]^1/2

    basically, normal expansion, with the final answer all under the square root?
  9. Jul 23, 2008 #8
    Yes that is ok to do.. I was thinking something more simple than multiply top and bottom by [y^1/2], because it's not logical for me to do it that way (seems like a trick), instead I would combine the bottom and divide..

    1+1/y = y/y + 1/y = (y+1)/y (common denominators), so you get:

    [(y+1)/y]^1/2 = [(y+1)^1/2]/[y^1/2]

    now dy is divided by the above.. when you divide you multiply by the reciprocal to get:
    dy * [y^1/2] / [(y+1)^1/2]
  10. Jul 23, 2008 #9
    i now wish to show antidiff of q/[1+q^2]^1/2 = [1+q^2]^1/2 by using the change of variable q=sinh(theta).

    I find dq=cosh(theta)d(theta), sub everything in and find that the equation becomes
    antidiff sinh(theta)cosh(theta)/[1+sinh^2(theta)]^1/2 d(theta).

    I can change the top line to 1 and the bottom line to root (cosh^2(theta)), but I cannot see how I retreive the q.

  11. Jul 23, 2008 #10


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    You mean you want to find [tex]\int \frac{q}{\sqrt{1+q^2}} dq[/tex] or [tex]\int \sqrt{1+q^2} dq[/tex]?

    Why can't you cancel out the cosh theta above and below? Then you'll be left with a simpler term to integrate.
  12. Jul 24, 2008 #11
    I wish to prove:
    http://img167.imageshack.us/img167/1885/89082209id8.jpg [Broken]

    Yeah, I should have seen that before! So cancelling cosh(theta) top and bottom leaves me with the antiderivative of sinh(theta). My question now, how do I change back the variable to q?

    Normally, I would integrate sinh(theta) to give cosh(theta), and then sub back, but in this case I cannot sub back because after integration the sinh term is gone and I am left with cosh(theta). What is the correct way to get back q?
    Last edited by a moderator: May 3, 2017
  13. Jul 24, 2008 #12


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    Well if you wanted to prove that all you need to do is differentiate the right hand side, unless the question tells you to do it by integration by substitution.

    Make use of some hyperbolic trigo identities. How do you express cosh theta in terms of sinh theta ? And more importantly you can recover q by noting that [tex]\theta = \sinh^{-1} q[/tex]. This page should help:
  14. Jul 24, 2008 #13
    I wish to perform integration of lhs rather than diff rhs. I've been told the best way is to perform the substitution q=sinh(theta).

    anyhow, after integrating sinh(theta) I get cosh(theta)
    I figured I could do root (cosh^2(theta)) then swap that cosh^2(theta) with 1+sinh^2(theta), hence giving root[1+sinh^2(theta)], finally then I can swap the sinh(theta) back to q giving RHS.

    Sounds correct?
  15. Jul 24, 2008 #14


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    Yeah, it's ok as well.
  16. Jul 24, 2008 #15
    cool, as a lead on I now wish to evaluate the antidiff of (1+q²)^1/2 using the same change of variable q=sinh(θ).

    I substitute q and dq in and simplify to give antidiff cosh²(θ)dθ, then I proceed to use the identity cosh²(θ) = 1/2[1+cosh(2θ)], and antidiffing gives 1/2[θ+sinh(2θ)/2]

    obviously θ=sinhֿ¹(θ), but how will I get q back given the sinh(2θ) term?
  17. Jul 24, 2008 #16


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    Break up that sinh(2θ) term into sinhθ and coshθ. You can express coshθ in terms of sinhθ. Do that and make use of the Wikipedia page linked earlier to simplify your expression.
  18. Jul 24, 2008 #17
    hmmm, the inverse sinh term is bothering me...

    after swapping everything to sinh and then back to q, I get...

    = {sinhֿ¹(q) + q[1+q²]^1/2}/2
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