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Math Help - Help with a D.E., no idea what to do with it

  1. #1
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    Help with a D.E., no idea what to do with it

    (x^2)(y+1)dx + (y^2)(x-1)dy = 0

    1) it's not exact, I know that
    2) I tried turning it into an exact equation using Euler's multiplicating factor (not sure this is the right word in English), but couldn't find the factor
    3) tried the y/x=v substitution and ended up with:
    xv + 1 + xv^3 - v^3 + x^2v^2v' - xv^2v' = 0
    and I've no idea how to separate the x and v here
    4) tried separating the variables and ended up with this:
    \frac{(x-1)^2}{2} + 2(x-1) +ln|x-1| + \frac{(y+1)^2}{2} - 2(y+1) + ln|y+1| = C


    what do I do?

    thanks.
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  2. #2
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    Re: Help with a D.E., no idea what to do with it

    Quote Originally Posted by Greystoke View Post
    what do I do?
    I think your solution is correct as far as it goes.

    1. If you have an initial condition, you might plug that in.

    2. You should also differentiate your answer and see if you get the original DE back again.

    3. Define the interval of validity.
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  3. #3
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    Re: Help with a D.E., no idea what to do with it

    1. no initial conditions
    3. no intervals of validity in the text of the problem
    2. didn't differentiate the solution, will do

    I thought I was supposed to get a solution like "y = ....", not "ln|y+1| + etc = -ln|x-1| + etc +C"
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  4. #4
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    Re: Help with a D.E., no idea what to do with it

    Quote Originally Posted by Greystoke View Post

    I thought I was supposed to get a solution like "y = ....", not "ln|y+1| + etc = -ln|x-1| + etc +C"

    Most of the time we seek this form, but can we isolate y in your case?
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  5. #5
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    Re: Help with a D.E., no idea what to do with it

    I don't think so, at least I can't think of any way.

    Thanks.
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  6. #6
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    Re: Help with a D.E., no idea what to do with it

    It can't be done using basic algebra.

    I would use the function as it appears.
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  7. #7
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    Re: Help with a D.E., no idea what to do with it

    Quote Originally Posted by Greystoke View Post
    1. no initial conditions
    Then there's no need to find out what C is: you can just leave your solution as is.

    3. no intervals of validity in the text of the problem
    It wouldn't be in the text of the problem. What you want to do is write the original DE in the form dy/dx = f(x,y), and examine the continuity of f(x,y) and compare to the standard existence theorem. In addition, you need to look at the domains of the functions in your solution, and see if that further restricts your intervals of validity. Sometimes the interval of validity is further restricted after you find a solution.

    2. didn't differentiate the solution, will do
    Hint: implicit differentiation is handy when faced with a solution like you have here. Another hint: differentiate in such a manner that the constant disappears. That's a trick I learned from Danny here on the forum.

    I thought I was supposed to get a solution like "y = ....", not "ln|y+1| + etc = -ln|x-1| + etc +C"
    Looks like pickslides has answered this one perfectly adequately, as usual.
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