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anonymous

  • one year ago

If dy/dx = sin^2( piy/4) and y = 1 when x = 0, then find the value of x when y = 3.

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  1. freckles
    • one year ago
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    do you know how to solve the given differential equation?

  2. anonymous
    • one year ago
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    im not too sure how to manipulate dy/dx

  3. anonymous
    • one year ago
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    i know i have to isolate the variables and integrate and solve for y right?

  4. freckles
    • one year ago
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    \[\frac{dy}{dx}=\sin^2(\frac{\pi}{4} y) \\ \frac{dy}{\sin^2(\frac{\pi}{4} y)}=dx \\ \] like do you know how to ingrate both sides of this equation?

  5. anonymous
    • one year ago
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    well the right side is just x but im not to sure on integrating the left side

  6. freckles
    • one year ago
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    do you know the derivative of cot(x)?

  7. anonymous
    • one year ago
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    -csc^2(x)?

  8. freckles
    • one year ago
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    \[\frac{d}{dy} \cot(u)=-\csc^2(u) \cdot \frac{du}{dy} \\ \text{ where } u=u(y)\]

  9. freckles
    • one year ago
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    \[\int\limits_{}^{} \csc^2(\frac{\pi}{4} y) dy \\ u=\frac{\pi}{4} y \\ du=\frac{\pi}{4} dy \\ \int\limits \csc^2(u) \frac{4}{\pi} du\]

  10. freckles
    • one year ago
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    do you see how to continue

  11. anonymous
    • one year ago
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    why did you inverse the pi/4

  12. freckles
    • one year ago
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    because if du=pi/4 dy then dy=4/pi du

  13. freckles
    • one year ago
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    i just multiplied 4/pi on both sides to isolate dy

  14. anonymous
    • one year ago
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    okay okay hold up slow down , why are we integrating csc^2(piy/4)

  15. freckles
    • one year ago
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    because that was the left hand side of your equation once you separated the variables

  16. anonymous
    • one year ago
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    okay okay i see, give me a second to get my thoughts together

  17. freckles
    • one year ago
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    \[\frac{1}{\sin(x)}=\csc(x)\]

  18. anonymous
    • one year ago
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    yea but the sine was squared wouldn't that affect that

  19. freckles
    • one year ago
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    so you know that equation I just mentioned holds don't you think the equation will still hold if you square both sides

  20. freckles
    • one year ago
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    \[(\frac{1}{\sin(x)})^2=\csc^2(x) \\ \frac{1}{\sin^2(x)}=\csc^2(x)\]

  21. anonymous
    • one year ago
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    okay sorry i was thinking the square would be negative because it was in the denominator

  22. anonymous
    • one year ago
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    i see what you mean though

  23. anonymous
    • one year ago
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    okay so we are integrating \[\csc^2(\frac{ \pi y }{ 4 })\]

  24. anonymous
    • one year ago
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    oh you were doing substitution , i see

  25. anonymous
    • one year ago
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    okay so back to \[\int\limits\limits_{}^{} \csc^2(\frac{\pi}{4} y) dy \\ u=\frac{\pi}{4} y \\ du=\frac{\pi}{4} dy \\ \int\limits\limits \csc^2(u) \frac{4}{\pi} du \] should be do another substitution?

  26. freckles
    • one year ago
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    on what? that is just -4/pi*cot(u)+C where u=pi/4y

  27. anonymous
    • one year ago
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    \[\frac{ 4 }{ \pi } \int\limits (-\ln(\csc(x) + \cot(x)) + C)^2\]

  28. anonymous
    • one year ago
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    is that right?

  29. anonymous
    • one year ago
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    minus the integral sign

  30. freckles
    • one year ago
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    where did you get that answer

  31. freckles
    • one year ago
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    remember you yourself said the d/dx cot(x) =-csc^2(x)

  32. freckles
    • one year ago
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    or you could say -d/dx cot(x)=csc^2(x)

  33. anonymous
    • one year ago
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  34. anonymous
    • one year ago
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    are these wrong then?

  35. freckles
    • one year ago
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    \[\int\limits\limits\limits_{}^{} \csc^2(\frac{\pi}{4} y) dy \\ u=\frac{\pi}{4} y \\ du=\frac{\pi}{4} dy \\ \int\limits\limits\limits \csc^2(u) \frac{4}{\pi} du \\ \frac{-4}{\pi} \cot(u)+C \\ \frac{-4}{\pi} \cot(\frac{\pi}{4}y)+C\]

  36. freckles
    • one year ago
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    nope those all look great

  37. anonymous
    • one year ago
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    arghh i feel stupid

  38. freckles
    • one year ago
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    anyways we also have to integrate the other side

  39. freckles
    • one year ago
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    \[\frac{-4}{\pi} \cot(\frac{\pi}{4} y)+C=x\]

  40. anonymous
    • one year ago
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    -_- i was just typing that, okay so now just isolate y right

  41. freckles
    • one year ago
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    your second job is to find C for the point (x=0,y=1)

  42. freckles
    • one year ago
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    i would not solve for y i think that would cause more work then needed

  43. freckles
    • one year ago
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    besides you want to find x

  44. freckles
    • one year ago
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    for when y=3

  45. freckles
    • one year ago
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    which is the third part of the problem third and final

  46. anonymous
    • one year ago
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    \[C = \pi\]

  47. anonymous
    • one year ago
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    - pi

  48. freckles
    • one year ago
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    \[\frac{-4}{\pi} \cot(\frac{\pi}{4} y)+C=x \\ (0,1)=(x,y) \\ \frac{-4}{\pi}\cot(\frac{\pi}{4}(1))+C=0 \\ \frac{-4}{\pi} \cot(\frac{\pi}{4})+C=0 \\ \frac{-4}{\pi}(1)+C=0 \\ \frac{-4}{\pi}+C=0\]

  49. anonymous
    • one year ago
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    :(

  50. freckles
    • one year ago
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    it's okay... no frowns here!

  51. anonymous
    • one year ago
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    alright well c = 4

  52. freckles
    • one year ago
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    \[\frac{-4}{\pi} \cot(\frac{\pi}{4} y)+C=x \\ \frac{-4 }{\pi} \cot(\frac{\pi}{4} y)+\frac{4}{\pi}=x\] last step find x for when y is 3

  53. freckles
    • one year ago
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    well no

  54. freckles
    • one year ago
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    c=4/pi

  55. anonymous
    • one year ago
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    okay let me just take a breath , my frustration is getting the better of me here

  56. anonymous
    • one year ago
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    okay so now we just plug c into the solution and plug y = 3 and find x right

  57. freckles
    • one year ago
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    right

  58. freckles
    • one year ago
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    i will let you do that and I will come back in check in like 5 minutes or less

  59. anonymous
    • one year ago
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    okay im getting 8/pi

  60. freckles
    • one year ago
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    that sounds right

  61. freckles
    • one year ago
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    4/pi+4/pi 2*4/pi 8/pi

  62. anonymous
    • one year ago
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    okay great , thanks for the help sorry for being such a dummy :P

  63. freckles
    • one year ago
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    i'm going to go for tonight after this problem sleep time but before i go do you have any last questions on this problem?

  64. freckles
    • one year ago
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    you aren't a dummy probably just frustrated and trying to learn a new subject

  65. anonymous
    • one year ago
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    no im goo thanks again !

  66. freckles
    • one year ago
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    the one part is hard enough being frustrated makes it tad harder

  67. freckles
    • one year ago
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    anyways peace and i gave you a medal to award you for your effort

  68. anonymous
    • one year ago
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    :) night

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