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anonymous

  • one year ago

The slope field for a differential equation is shown in the figure. Determine the general solution of this equation

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    answer choices y=Cx^2 x=Cy^2 x^2 – y^2 = C^2 x2 + y2 = C^2

  3. anonymous
    • one year ago
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    okay so i know i have to reverse engineer this but i'm not doing too well

  4. ganeshie8
    • one year ago
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    |dw:1439493039926:dw|

  5. anonymous
    • one year ago
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    so i should just isolate the variables of each of the answer choices and derive each side right and graph all the diff equations i get and match it to the one im given , is that right?

  6. Astrophysics
    • one year ago
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    Could you not let x = 1 and y = 1 and then set it up as dy/dx that way you can see the slope easier?

  7. ganeshie8
    • one year ago
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    To my knowledge that slope field corresponds to a differential equation : \[y'=2x\] the family of solutions must look something like \[y = x^2+C\]

  8. anonymous
    • one year ago
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    http://www.mathscoop.com/calculus/differential-equations/slope-field-generator.php this might help

  9. anonymous
    • one year ago
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    and yes you are correct @ganeshie8 i just cant get the differential equation when onlly given the solution to a differential equation

  10. anonymous
    • one year ago
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    okay can you walk me through your answer ?

  11. Astrophysics
    • one year ago
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    Wait, I don't know if it's right, but what I was thinking is as I said above, let x = 1, y = 1 or what ever and just let dy/dx = x/y^2 and then plug in the values and just observe the points...not sure if that's a good method haha, ganeshie definitely knows more.

  12. ganeshie8
    • one year ago
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    the solution curves are the smooth curves obtained by joining the slope segments in the slope field : |dw:1439494028215:dw|

  13. ganeshie8
    • one year ago
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    Easy to see that they are family of parabolas that open up, under vertical translation

  14. anonymous
    • one year ago
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    okay let me rephrase the question, if i give you the solution to a differential equation (say x=Cy^2) can you find the differential equation it came from ?

  15. anonymous
    • one year ago
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    lol i know i dont know how to make the constant multiply :P

  16. ganeshie8
    • one year ago
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    x = Cy^2 1 = C2yy' x/1 = y/(2y') y' = y/(2x)

  17. anonymous
    • one year ago
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    okay and if we graph that we get

  18. anonymous
    • one year ago
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    so b is not our answer, so know tell me how you did that so i can do it to the rest of the answer choices XD

  19. ganeshie8
    • one year ago
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    since you have "one" arbitrary constant, you want to differentiate the solution just once

  20. anonymous
    • one year ago
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    okay so from x = Cy^2 to 1 = C2yy' the only thing you did was derive each side right ?

  21. ganeshie8
    • one year ago
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    right, just implicitly differentiate both sides with respect to x

  22. anonymous
    • one year ago
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    so what did you do from 1 = C2yy' to x/1 = y/(2y') ?

  23. ganeshie8
    • one year ago
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    x = Cy^2 ---------------(1) 1 = C2yy' --------------(2) next divide : (1)/(2)

  24. anonymous
    • one year ago
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    okay umm why ?

  25. ganeshie8
    • one year ago
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    your goal is to get rid of that arbitrary constant C dividing both equations does just that

  26. Astrophysics
    • one year ago
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    Nice explanation @ganeshie8

  27. anonymous
    • one year ago
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    okay i see so what do you do after you vet rid of the c ?

  28. anonymous
    • one year ago
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    get*

  29. ganeshie8
    • one year ago
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    you're done! the equation w/o arbitrary constant is the mother "differential equation" that represents the "family of solution curves" : x = Cy^2

  30. anonymous
    • one year ago
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    okay so let me try to do that for a different answer choice and tell me if i'm way off.

  31. ganeshie8
    • one year ago
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    the differential equation y' = y/(2x) represents the whole family of curves x = Cy^2

  32. ganeshie8
    • one year ago
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    Btw, none of the options are correct, but if I were you, I would tick option A

  33. anonymous
    • one year ago
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    um okay what makes you chose option 1 ?

  34. ganeshie8
    • one year ago
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    The correct answer is \(y = x^2 + C\) From the slope field it is clear that we must get solution curves that look same upto a vertical translation. My guess is that your professor cooked up this problem carelessly... Option 1 is close to parabolas, so..

  35. anonymous
    • one year ago
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    okay sounds about right :P , will let you know if its right in a little while when i turn it in

  36. ganeshie8
    • one year ago
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    good luck! please do share what the grader thinks is right

  37. anonymous
    • one year ago
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    will do ! thanks for all your help guys ! :D

  38. IrishBoy123
    • one year ago
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    you can re-write these things as level surfaces and i think you can nail it pretty solidly so if you take option B that has been discounted, you can say: \(x=Cy^2 \implies f(x,y) = x - Cy^2 = constant\) \(grad f = <1, -2Cy>\) so the slope \(dy/dx = \frac{-1}{-2Cy} = \frac{1}{2Cy}\) and it can't be that because dy/dx has no dependency on y and doesn't go ballistic at y = 0

  39. anonymous
    • one year ago
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    okay are you suggesting its option 1 then ?

  40. IrishBoy123
    • one year ago
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    i'd go A i can justify it, just lost a load of latex, so i might try post it again later

  41. IrishBoy123
    • one year ago
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    here's the observation: if you write \(y = mx + c\) as a closed surface , you get \(f(x,y) = y - mx = const\) so \(grad(f) = <-m, 1>\) so slope in normal xy-speak is \(-\frac{f_x}{f_y}\) that's not that surprising as \(df = f_x dx + f_y dy = 0 \implies \frac{dy}{dx} = -\frac{f_x}{f_y}\) do that for \(y=Cx^2\) and i think you get something that behaves like your graph.

  42. anonymous
    • one year ago
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    i was right !! :D thanks again !!

  43. IrishBoy123
    • one year ago
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    i reckon the "actual" right answer is: \(y=\frac{x^2}{2}+C \)

  44. anonymous
    • one year ago
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    i mean it not i

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