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huclogin

  • one year ago

The problem is that find the curve, its slope of tangent lines are twice as steep as the ray from the origin. The point is (x,y) I solve it like this:(dy/dx)=(2y/x) ∫dy/y=∫dx/x then ln|y|=2lnx+c (c is constant) then y=Ax^2(A=±e^c) I think that A is ambiguous up to the constant c, and the graph of the curve I'm aiming for, and the graph of the curves is a bunch of parabolas which is undefined at point (0,0). Did I make any mistakes? If I have any one of the curves to the left of y-axis and to the right connected together and take the derivative of it, can I always get dy/dx=2y/x?

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  1. huclogin
    • one year ago
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    T^T anyone here?

  2. phi
    • one year ago
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    it sounds right. though the parabola is defined at (0,0) however, we may want to exclude the origin?

  3. phi
    • one year ago
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    because the "ray from the origin" is undefined at (0,0)

  4. huclogin
    • one year ago
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    |dw:1438353168811:dw| It's the graph. I can not draw very well.

  5. huclogin
    • one year ago
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    Oh.. Maybe I can't call it parabolas, it's only similar to parabolas. Because when I solve the differential equations I divided it by x on the right-hand side.

  6. ganeshie8
    • one year ago
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    The domain must be restricted to either x<0 or x>0 because the slope is undefined at x=0. From the slope field you can see that as you travel along a particular curve toward 0, the moment you reach 0, you wont know which direction to go because there are too many directions : |dw:1438353913296:dw|

  7. phi
    • one year ago
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    ganeshie8 are you suggesting the solution is Ax^2 x<0 Bx^2 x>0 where A and B are non-zero constants.

  8. ganeshie8
    • one year ago
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    Nope, domain depends on the initial value If the initial value starts negative, we must restrict the domain to x<0 because we wont know which direction that particular solution goes after that singularity at x=0

  9. ganeshie8
    • one year ago
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    solution curves are half parabolas

  10. phi
    • one year ago
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    that sounds like philosophy. you could glue two different parabolas together, make sure one of them goes through the initial value, exclude (0,0) and you have a "solution" that meets the conditions.

  11. huclogin
    • one year ago
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    ganeshie8 Do you meas after I take the indefinite integral of the differential equation dy/y=2dx/x I can get a family of functions?

  12. ganeshie8
    • one year ago
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    Yes, the solutions are family of half parabolas, not full parabolas with a hole at (0, 0)

  13. huclogin
    • one year ago
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    ganeshie8 |dw:1438354495749:dw| If I connect the two curves(any curves, one from the left and one from the right) to gether and get a function then take the derivative of it, can I always get dy/dx=2y/x? Another question, I think that A is ambiguous up to the constant c, is it right?

  14. ganeshie8
    • one year ago
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    Yes for both, \(A=e^C\) is as much an arbitrary constant as \(C\) is

  15. ganeshie8
    • one year ago
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    but wait, do we need to worry that A is now positive always ?

  16. huclogin
    • one year ago
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    ganeshie8 NO, because after taking the integral of dy/y we get ln|y|

  17. ganeshie8
    • one year ago
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    Ahh right, that does the trick : \[\ln |y| = \ln|x^2| + C \implies |y| = e^Cx^2 \implies y = \pm e^Cx^2 = Ax^2 \] \(A\) can be negative too !

  18. huclogin
    • one year ago
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    ganeshie8 phi My problem is solved now.Thank you! ^ ^

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