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Complicated (for me) differential equation problem

  1. Apr 20, 2015 #1
    1. I am supposed to find dx and dy. I think I am missing a step or a general idea. I spent quite some time figuring out what rules should I use and the only sequence I can think of is quotient rule and chain on the (x2 +y2)1/2 term. The answer that I find is ((xy(3x2+2y2))/(x2+y2)3/2 . On the other hand I found that wolfram alpha solves it with the y'(x) term that I cannot understand, since it should find the only x terms. Please find the attached wolfram alpha solution. I am not asking for a solution (step by step) just some guidance.

    Here is a link to the wolfram alpha's solution : http://i1375.photobucket.com/albums...x2ysqrtx2y2_--2015-04-20_1952_zpskiujuryl.jpg




    The solution given is:
    ((xy(x2+2y2))/(x2+y2)3/2

    Thanks you!

     

    Attached Files:

  2. jcsd
  3. Apr 20, 2015 #2

    Simon Bridge

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    You've provided the solution - but you also have to provide the problem. I had to go to a different webpage to discover it ... am I correct that the problem was:
    $$\frac{d}{dx} \frac{x^2y}{\sqrt{x^2+y^2}}$$ ... if so then you need all the rules:
    product, quotient, power, and chain rules - as well as those for implicit differentiation... so go revise.

    For instance, you can put ##f(x,y(x))=xy^2(x)## and ##g(x,y(x))=\sqrt{x^2+y^2(x)}## and use the quotient rule right off.
    This will give you a relationship between other derivatives which you have to use other rules to evaluate.

    i.e. $$\frac{d}{dx}xy^2 = y^2\frac{d}{dx}x + x\frac{d}{dx}y^2\\ \qquad = y^2 + x\left(\frac{d}{dy}y^2\right)\left(\frac{dy}{dx}\right)\\ \qquad = y^2 + 2xy\frac{dy}{dx}$$ ... see how that worked?
    The first line was just the product rule, the second line used the chain rule.

    Note: Wolfram gave you step-by-step instructions ... did you try following them?
    I also want to make sure that the derivative you are after is the exact derivative and not a partial derivative... i.e. do we understand y to be a function of x?
     
    Last edited: Apr 20, 2015
  4. Apr 21, 2015 #3
    Thanks for the reply! I tried to follow the wolfram's solution, but I got confused why they keep all the ys, I just thought that you can disregard the ys. I actually am after the partial derivative. First have to find ∂ƒ/∂xand after that for ∂ƒ/∂y. Sorry for the fact I missed to provide the problem. For some reason couldn't edit the 1st post, here is the problem:
    Find (∂ƒ/∂x , ∂ƒ/∂y) for the following function:
    ƒ(x,y) = x2y / √(x2+y2)
     
  5. Apr 21, 2015 #4

    Ray Vickson

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    You probably will not like my advice, but I suggest strongly that you avoid using programs such as Wolfram Alpha, and do everything by hand, at least at the beginning while you are still learning and practicing the methods. That really is the only sure way to learn the material.

    In this case, to evaluate ##\partial f(x,y)/\partial x##, just treat ##y## like a constant---that is what partial derivatives do. So now you need to differentiate ##g(x) = y x^2/\sqrt{x^2 + y^2}## with ##y = \text{const.}##. This is really not much different from differentiating something like ##g(x) = 2 x^2/\sqrt{x^2+4}##.
     
  6. Apr 21, 2015 #5
    Yes, I know its not a good practice, just wanted to see where I made a mistake, but it showed even more complicated solution and I got lost. Anyway, I found where I was wrong. It is at the differentiation of the y/√x2+y2 . I found confusing the notation with the substitution of u = x2 + y2 and all the notations d/du and du/dx . Thank you for the replies, hope I am walking in the right direction!
     
  7. Apr 23, 2015 #6

    Simon Bridge

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    Yep - with partial derivatives, it's much easier than when y has to be treated as a function of x and differentiated implicitly.
    The u-substitution, in this case, was just application of the chain rule... that would be the usual way to do it.
    Leibnitz notation can be tricky to get used to... one of the realizations you need is how it behaves like a fraction sometimes as in:
    $$\frac{dy}{dx} = \frac{du}{du}\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}$$ ... it's actually a more useful notation.

    Another handy trick is to make square roots into power of a half so: $$\frac{1}{\sqrt{x+y}}=(x+y)^{-1/2}$$ ... then it's chain-rule and power rule.
     
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