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Math Help - Solving dy/dx = (1+3x^2)/(3y^2-6y)

  1. #1
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    Solving dy/dx = (1+3x^2)/(3y^2-6y)

    dy/dx=(1+3x^2)/(3y^2-6y) with y(0)=1

    It was easy enough getting it to the solution:

    y^3-3y^2-x-x^3+2=0.

    Now it wants me to find the solution where it's valid. The answer in the back of the book is that |x|<1 and the hint was find where the integral curve has a vertical tangent. I'm not really sure how they arrived to this solution though.

    Thanks for any help
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  2. #2
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    You got that, I presume, by multiplying both sides by 3y^2- 6y and integrating. Of course, that is valid only as long as the denominator of the derivative is not 0- that is as long as 3y^2- 6y= 3y(y- 2)\ne 0. At points where it is 0, that is where y= 0 or y= 2, the denominator is 0 so the derivative does not exist (in other words, the graph has a vertical tangent). Since y lies between 0 and 2, this solution must be valid only for y between 0 and 2.

    Now, if you put y= 0 in your solution, you get -x- x^3+ 2= 0 and it is not too difficult to see that x= 1 is a solution to that. -x- x^2+ 2= -(x- 1)(x^2+ x+ 2) and x^2+ x+ 2= 0 has no real roots so x= 1 is the only real solution.

    If you put y= 2 in your solution, you get 8- 12- x- x^2+ 2= -2- x- x^2 and it is not difficult to see that x= -1 is the only real solution to that equation. Since x=0 is between -1 and 1, we see that this solution is valid only for x between -1 and 1- that is, |x|< 1.
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  3. #3
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    thanks for the explanation

    My brain is so fried today from all the math. Here's to you:
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