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anonymous

  • 5 years ago

Lokisan! Please help me with more differential equations! :)

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  1. anonymous
    • 5 years ago
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    22xdy+11ydx+3ydy=0

  2. anonymous
    • 5 years ago
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    Loki! haha I'm back with more questions. :)

  3. anonymous
    • 5 years ago
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    Hehe, I see...

  4. anonymous
    • 5 years ago
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    Are you studying exact differential equations?

  5. anonymous
    • 5 years ago
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    I don't think so. I know what the answer is and its not a number. Its the formula of the constant

  6. anonymous
    • 5 years ago
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    Can you wait a little bit - I'm just finishing another post.

  7. anonymous
    • 5 years ago
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    Sure

  8. anonymous
    • 5 years ago
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    22xdy+11ydx+ydy=0 is the actual problem, Sorry

  9. anonymous
    • 5 years ago
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    ok

  10. anonymous
    • 5 years ago
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    i need paper

  11. anonymous
    • 5 years ago
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    haha alright

  12. anonymous
    • 5 years ago
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    oh, i don't have her question.

  13. anonymous
    • 5 years ago
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    Scotty, this looks worse than it is (sorry, I have someone else I'm helping here too).

  14. anonymous
    • 5 years ago
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    I'm hoping so. I think it fits into the trick of d(xy)=dy+ydx am I correct?

  15. anonymous
    • 5 years ago
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    Hang on...I misread it...I have to work something out. Sorry, I'll just finish with this other person and I won't be distracted.

  16. anonymous
    • 5 years ago
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    Ok so the problem is 22xdy+11ydx+3ydy=0

  17. anonymous
    • 5 years ago
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    I wish that 22 wasn't there. You sure it's not 11?

  18. anonymous
    • 5 years ago
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    Positive :/

  19. anonymous
    • 5 years ago
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    I know what the answer if, if that helps.

  20. anonymous
    • 5 years ago
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    yeah punch it in

  21. anonymous
    • 5 years ago
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    Its \[11xy^2+y^3=C\]

  22. anonymous
    • 5 years ago
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    Sorry, I keep being called away.

  23. anonymous
    • 5 years ago
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    It's alright..

  24. anonymous
    • 5 years ago
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    I'm thinking a substitution method. Just one sec...again called away...sorry. Won't be forever.

  25. anonymous
    • 5 years ago
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    I think I may have your method...just let me check properly.

  26. anonymous
    • 5 years ago
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    Ok cool

  27. anonymous
    • 5 years ago
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    I keep being called away! I'll just let you know what I'm thinking and you can play with it as well. Your equation, when rearranged, looks like this\[y'=-\frac{11y}{22x+y}=-\frac{11(\frac{y}{x})}{22+(\frac{y}{x})}\]after dividing the numerator and denominator by x.

  28. anonymous
    • 5 years ago
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    This is in the form for a type of differential equation that can be solved when we know that y' is some function of y/x; that is,\[y'=F(\frac{y}{x})\]

  29. anonymous
    • 5 years ago
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    We make the identification,\[v(x)=\frac{y}{x}\]so that\[y=xv\]

  30. anonymous
    • 5 years ago
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    and we take it from there. Is this familiar?

  31. anonymous
    • 5 years ago
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    Ok let me try to pmess with it

  32. anonymous
    • 5 years ago
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    Ok instead of that one... can you help me solve y'-9y=8e^x

  33. anonymous
    • 5 years ago
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    I'm doing this one for a timed quiz

  34. anonymous
    • 5 years ago
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    I figured out your other problem - I made an arithmetic error, but the method is correct.

  35. anonymous
    • 5 years ago
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    Yeah I'll do the second.

  36. anonymous
    • 5 years ago
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    Ok I think I saw where you were going with it and I gutted my way through it. Thanks

  37. anonymous
    • 5 years ago
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    Okay, integrating factor one - do you know?

  38. anonymous
    • 5 years ago
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    Not sure what that means..

  39. anonymous
    • 5 years ago
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    \[y'-9y=8e^x \rightarrow \frac{d (\mu y)}{dx}=\mu 8e^x\]

  40. anonymous
    • 5 years ago
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    \[\mu = e^{\int\limits_{}{}-9dx}=e^{-9x}\]

  41. anonymous
    • 5 years ago
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    so

  42. anonymous
    • 5 years ago
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    im sorry that should be y"

  43. anonymous
    • 5 years ago
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    \[e^{-9x}y=e^{-9x}.8e^x+c \rightarrow y=ce^{9x}-e^x\]

  44. anonymous
    • 5 years ago
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    y''-9y=8e^x ?

  45. anonymous
    • 5 years ago
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    Yes, sorry for the confusion. I miss typed.

  46. anonymous
    • 5 years ago
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    right do you just want the answer for the quiz and then the method

  47. anonymous
    • 5 years ago
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    Well. I want to learn but first and foremost I really need to pass this quiz

  48. anonymous
    • 5 years ago
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    I am writing it all down to try to understand though.

  49. anonymous
    • 5 years ago
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    \[y=c_1e^{3x}+c_2e^{-3x}-e^{x}\]

  50. anonymous
    • 5 years ago
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    I'll explain...

  51. anonymous
    • 5 years ago
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    Are you sure the + isn't a - between the first 2 terms?

  52. anonymous
    • 5 years ago
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    You have a second order differential equation which is linear, inhomogeneous (because the RHS is a function of x only) with constant coefficients. You need to solve inhom. equations by 1) solving the equivalent homogeneous equation first 2) finding a particular solution

  53. anonymous
    • 5 years ago
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    +

  54. anonymous
    • 5 years ago
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    In the end, it doesn't matter because it's absorbed into the constant.

  55. anonymous
    • 5 years ago
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    The only option I have with a + sign in it is... -2e^3x+e^-3x-e^x

  56. anonymous
    • 5 years ago
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    Ok so e^3x=e^-3x-e^x is the same thing?

  57. anonymous
    • 5 years ago
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    Oh, do you have initial conditions?

  58. anonymous
    • 5 years ago
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    Sorry when x=0

  59. anonymous
    • 5 years ago
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    Your equation has it's constants solve for

  60. anonymous
    • 5 years ago
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    Oh...you need to give me the whole question ;)

  61. anonymous
    • 5 years ago
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    Wait.... y=0 and y'=8 and x=0... I'm sorry I've been up all day doing this

  62. anonymous
    • 5 years ago
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    It's second order, so there should be 2 conds.

  63. anonymous
    • 5 years ago
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    ok

  64. anonymous
    • 5 years ago
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    c_1 = 2 and c_2 = -1

  65. anonymous
    • 5 years ago
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    \[y=2e^3x-e^{-3x}-e^x\]

  66. anonymous
    • 5 years ago
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    Yep that works!

  67. anonymous
    • 5 years ago
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    Good

  68. anonymous
    • 5 years ago
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    Thanks so much... I really feel low just asking for answers but I plan on taking the class again. If I fail I have to pay the whole tuition and I've been studying so hard and just not getting it lately :(

  69. anonymous
    • 5 years ago
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    It's because you're having to do it on your own - you're not being shown the ropes. Don't punish yourself.

  70. anonymous
    • 5 years ago
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    trying not to. Could you help me with 1 or 2 more? Then I'll leave you alone.

  71. anonymous
    • 5 years ago
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    yes

  72. anonymous
    • 5 years ago
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    How long do you have to do the quiz?

  73. anonymous
    • 5 years ago
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    i panic in tests - you shouldn't have told me, lol

  74. anonymous
    • 5 years ago
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    4xydx=(x^2+4)dy and that is the whole problem and not leaving anything out :)

  75. anonymous
    • 5 years ago
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    It's separable.

  76. anonymous
    • 5 years ago
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    \[4xydx=(x^2+4)dy \rightarrow \frac{4xy}{x^2+4}=\frac{dy}{dx} \rightarrow \frac{4x}{x^2+4}dx=\frac{dy}{y}\]

  77. anonymous
    • 5 years ago
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    You can integrate from there.

  78. anonymous
    • 5 years ago
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    Ok let me try

  79. anonymous
    • 5 years ago
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    \[y=c(x^2+4)^2\]

  80. anonymous
    • 5 years ago
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    yep thats what I got

  81. anonymous
    • 5 years ago
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    Aesome

  82. anonymous
    • 5 years ago
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    or, awesome

  83. anonymous
    • 5 years ago
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    Ok I think this should be the last one... I think I got the others actually.

  84. anonymous
    • 5 years ago
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    ok

  85. anonymous
    • 5 years ago
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    dy+2ydx=e^(-4x)dx

  86. anonymous
    • 5 years ago
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    It's an integrating factor one again.

  87. anonymous
    • 5 years ago
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    Yeah those are really difficult for me. Seperatable ones are the ones I'm more comfortable with...

  88. anonymous
    • 5 years ago
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    \[y=-\frac{1}{2}e^{-4x}+ce^{-2x}\]

  89. anonymous
    • 5 years ago
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    I'm just skipping to the solution because of the test...

  90. anonymous
    • 5 years ago
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    The closest answer I have that looks like that is e^(4x)+Ce^(2x)

  91. anonymous
    • 5 years ago
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    hmmm...that's the solution to the equation you gave me...

  92. anonymous
    • 5 years ago
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    let me write it in the formula so it looks better maybe it looks funny on text

  93. anonymous
    • 5 years ago
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    \[dy+2ydx=e^{-4x}dx \rightarrow y'+2y=e^{-4x}\]?

  94. anonymous
    • 5 years ago
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    \[dy-2ydx=6e^(5x) dx\]

  95. anonymous
    • 5 years ago
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    that dan 5x is super script

  96. anonymous
    • 5 years ago
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    so you gave me the wrong question before?

  97. anonymous
    • 5 years ago
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    Im a failure sorry I'm rushing

  98. anonymous
    • 5 years ago
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    \[y(x) = c_1 e^{2 x}+2 e^{5 x}\]

  99. anonymous
    • 5 years ago
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    is that one available?

  100. anonymous
    • 5 years ago
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    yes just not written that way but yeah

  101. anonymous
    • 5 years ago
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    I had actually guesed that one after trying to work it myself. Hurray for small miracles.

  102. anonymous
    • 5 years ago
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    so you've passed?

  103. anonymous
    • 5 years ago
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    Well... not 100% sure. I think I got the rest of them right.

  104. anonymous
    • 5 years ago
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    Do you still want to look at the first one?

  105. anonymous
    • 5 years ago
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    For dy+2ydx=e^(-4x)dx I got.... y=-1/2e^(-4x)+ce^(-2x)

  106. anonymous
    • 5 years ago
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    that's what I got..?

  107. anonymous
    • 5 years ago
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    and thats the only other one I kind of guessed on. I do want to look at those though yes.

  108. anonymous
    • 5 years ago
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    No I did that one myself

  109. anonymous
    • 5 years ago
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    good

  110. anonymous
    • 5 years ago
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    I think its right, not 100% sure though

  111. anonymous
    • 5 years ago
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    It's right

  112. anonymous
    • 5 years ago
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    if

  113. anonymous
    • 5 years ago
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    it's the one I did before...and it is...it's right.

  114. anonymous
    • 5 years ago
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    You're fine at this!

  115. anonymous
    • 5 years ago
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    Oh ok, sorry I'm scattered. ok I'm submitting it!

  116. anonymous
    • 5 years ago
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    What about your first question?

  117. anonymous
    • 5 years ago
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    How much time do you have?

  118. anonymous
    • 5 years ago
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    5 mins left

  119. anonymous
    • 5 years ago
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    the first one was a practice but I do want to know how to do it

  120. anonymous
    • 5 years ago
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    ok

  121. anonymous
    • 5 years ago
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    How about.... is y'=2y(x-3) seperable

  122. anonymous
    • 5 years ago
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    I said yes.

  123. anonymous
    • 5 years ago
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    yes, separable

  124. anonymous
    • 5 years ago
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    ok and 2xdx+ysinxdy=sinxdx

  125. anonymous
    • 5 years ago
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    I got y^2=x-2sinx+C

  126. anonymous
    • 5 years ago
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    that's separable too - is that how you solved it?

  127. anonymous
    • 5 years ago
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    Yes

  128. anonymous
    • 5 years ago
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    that one would have taken a bit of effort

  129. anonymous
    • 5 years ago
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    Ok is y'=y/x y=cx

  130. anonymous
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    yes

  131. anonymous
    • 5 years ago
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    ok well it looks all good then :)

  132. anonymous
    • 5 years ago
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    Thank you so much for that man.....

  133. anonymous
    • 5 years ago
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    that's okay - i hope you pass

  134. anonymous
    • 5 years ago
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    have you heard of a site called wolframalpha? www.wolframalpha.com

  135. anonymous
    • 5 years ago
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    Ok so that original problem..

  136. anonymous
    • 5 years ago
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    You can type in things and it will calculate, like your diff. eqs

  137. anonymous
    • 5 years ago
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    but it won't show you what to do, that's the only thing

  138. anonymous
    • 5 years ago
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    I have, I don't know how to type these kinds of problems in though.... could you show me how?

  139. anonymous
    • 5 years ago
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    I'm just thinking, if you have more of these tests, you can have a look at that as a resource

  140. anonymous
    • 5 years ago
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    I did use it before, I have 2 more tests actually.

  141. anonymous
    • 5 years ago
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    y'-2y=e^(-4x) is what you can type in for one of your d.e.'s above. It will give you the answer.

  142. anonymous
    • 5 years ago
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    Use primes on the y's...

  143. anonymous
    • 5 years ago
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    You don't have to tell it to solve for y; it guesses. If it doesn't, then it's a form it doesn't recognize.

  144. anonymous
    • 5 years ago
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    Hmm.... but even for stuff like... 22xdy+11ydx+3ydy=0?

  145. anonymous
    • 5 years ago
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    Yeah, divide everything on your paper by dx so you can have (dy/dx) which you can equate as y'

  146. anonymous
    • 5 years ago
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    lol... thats a genius way to do it.

  147. anonymous
    • 5 years ago
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    \[22x \frac{dy}{dx}+y+3y \frac{dy}{dx}=0 \rightarrow 22xy'+y+3yy'=0\]

  148. anonymous
    • 5 years ago
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    I don't know why they keep giving you the equation in that form...it's weird.

  149. anonymous
    • 5 years ago
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    In the form I gave you? Thats the way my teacher gives me them....

  150. anonymous
    • 5 years ago
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    Yeah, I know...weird.

  151. anonymous
    • 5 years ago
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    Almost all my questions are like that is that not how they are normally done?

  152. anonymous
    • 5 years ago
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    lol, no

  153. anonymous
    • 5 years ago
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    Hmm why would she try to teach like that then?

  154. anonymous
    • 5 years ago
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    Maybe she's getting you ready to think in terms of differentials or something. Really, it doesn't matter. Just stick to what she does.

  155. anonymous
    • 5 years ago
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    did you pass this quiz?

  156. anonymous
    • 5 years ago
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    Ok well I really appreciate it. You might be the only reason I passed that quiz... I'm sure of it.

  157. anonymous
    • 5 years ago
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    I won't know for a day or two, but I'm sure I did.

  158. anonymous
    • 5 years ago
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    Cool, hope so...don't want to be spending more money on tuition...

  159. anonymous
    • 5 years ago
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    god I know. I have...2 more quizes and a final. Then I'll be done with this class.

  160. anonymous
    • 5 years ago
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    Good luck then. If you need help, look up an old post and send something. I'll get an e-mail about it.

  161. anonymous
    • 5 years ago
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    Ok I will man. thanks. Hopefully if you're around I can ask for your help like this again I would star you 1000 times if I could

  162. anonymous
    • 5 years ago
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    No worries. have a go at that first one. The method of substitution v=y/x should work. You end up with a differential equation in v which is separable...and you're comfortable with that ;)

  163. anonymous
    • 5 years ago
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    I'm going to go offline for a while. I have some stuff to do. Good luck!

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spraguer (Moderator)
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