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anonymous

  • 5 years ago

Show how to find the particular solution of the differential equation, dy/dx+5y=9, x=0 when y=1

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  1. anonymous
    • 5 years ago
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    I could use a LaPlace transform right?

  2. anonymous
    • 5 years ago
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    Yeah.. and it's pretty easy.

  3. anonymous
    • 5 years ago
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    Ok the only problem I'm having is, all I have to use is the formula L(y')=sL(y)-y(0) correct?

  4. anonymous
    • 5 years ago
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    Yeah.

  5. anonymous
    • 5 years ago
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    ok so I'm having trouble figuring out how to incorperate x=0 and y=1

  6. anonymous
    • 5 years ago
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    x=0 when y=1 means that y(0)=1

  7. anonymous
    • 5 years ago
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    Oh.. duh.

  8. anonymous
    • 5 years ago
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    so I would just rewrite... \[s L(y)-y(0)+5L(y)=L(9)\]

  9. anonymous
    • 5 years ago
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    OK.. what are the L(y) and L(9)?

  10. anonymous
    • 5 years ago
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    Not 100% sure on L(y) but I know L(9)=9/s

  11. anonymous
    • 5 years ago
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    Would it just be y/s ?

  12. anonymous
    • 5 years ago
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    or no, I could solve for L(y)

  13. anonymous
    • 5 years ago
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    I think I have it

  14. anonymous
    • 5 years ago
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    Yeah.. you could do so, or you could write it as Y(s)

  15. anonymous
    • 5 years ago
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    You know that you have to transform it back to the time domain.

  16. anonymous
    • 5 years ago
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    L(y)=(9/s+1)/(s+5)

  17. anonymous
    • 5 years ago
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    are you sure about the simplification you made?

  18. anonymous
    • 5 years ago
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    oh I see it.. it's right.

  19. anonymous
    • 5 years ago
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    I don't know how to do fractions notation using the equation thing or I would. :)

  20. anonymous
    • 5 years ago
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    I think I need to grab a pen and a paper!

  21. anonymous
    • 5 years ago
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    Well you so far have: \[Y(s)={9/s+1 \over s+5}\] you can multiply by s (since it's not 0).. you will get: \[{9+s \over s(s+5)}\]

  22. anonymous
    • 5 years ago
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    Oh right to clean it up. Ok so now how do I evaluate for the particular solution?

  23. anonymous
    • 5 years ago
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    \[{9+s \over s(s+5)}= {A \over s}+ {B \over s+5}\] how are you with partial fractions, can you find A and B?

  24. anonymous
    • 5 years ago
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    not good at ALL with partial fractions...

  25. anonymous
    • 5 years ago
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    it's easy.. multiply both sides of the last equation by s(s+5)

  26. anonymous
    • 5 years ago
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    You mean A(s(s+5)/s+B(s(s+5)/(s+5) ?

  27. anonymous
    • 5 years ago
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    Almost there!! something should be cancelled, right? And din't forget the left side of the equality sign.

  28. anonymous
    • 5 years ago
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    don't*

  29. anonymous
    • 5 years ago
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    A(s+5)+B(s)

  30. anonymous
    • 5 years ago
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    err

  31. anonymous
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    that should equal to?

  32. anonymous
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    9+s/s(s+5)=A(s+5)+B(s)

  33. anonymous
    • 5 years ago
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    nearly

  34. anonymous
    • 5 years ago
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    you multiply both sides by s(s+5)

  35. anonymous
    • 5 years ago
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    Oh right ok so its

  36. anonymous
    • 5 years ago
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    9+s=A(s+5)+B(s)

  37. anonymous
    • 5 years ago
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    Hey lokisan :).. sorry I took over. we seem to be useless wjen you're around :P

  38. anonymous
    • 5 years ago
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    you're doing fine :D

  39. anonymous
    • 5 years ago
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    I'm 'half' watching...

  40. anonymous
    • 5 years ago
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    there is more than one way find A and B now.. take s to be zero and find the value of A.

  41. anonymous
    • 5 years ago
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    If I did something wrong, Please let me know :)

  42. anonymous
    • 5 years ago
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    you're on the right track

  43. anonymous
    • 5 years ago
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    A=9/5

  44. anonymous
    • 5 years ago
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    Then I put that back into the original for B?

  45. anonymous
    • 5 years ago
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    or A and B for s...

  46. anonymous
    • 5 years ago
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    Yeah.. or just plug s=-5

  47. anonymous
    • 5 years ago
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    b=-4/5

  48. anonymous
    • 5 years ago
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    now what does Y(s) equal to?

  49. anonymous
    • 5 years ago
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    I'm kind of confused.. so I now put int he values of A and B back into 9+s=A(s+5)+B(s) to solve for the s?

  50. anonymous
    • 5 years ago
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    Do you know why we used the partial fractions in the first place?

  51. anonymous
    • 5 years ago
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    So we could evaluate what s was when x=0 and y=1

  52. anonymous
    • 5 years ago
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    No.. because we want to write an expression for Y(s) that can be easily transformed back to our original function.

  53. anonymous
    • 5 years ago
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    don't forget we are looking for the particular solution of the DE.

  54. anonymous
    • 5 years ago
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    My instructor never used the notation as Y(s) so I think thats where I'm losing you.

  55. anonymous
    • 5 years ago
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    Oh oh I think I understand... we're looking to change 9+s/s(s+5)

  56. anonymous
    • 5 years ago
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    \[L(y)={9+5 \over s(s+5)}={9 \over 5}.{1 \over s}-{4 \over 5}.{1 \over s+5}\]

  57. anonymous
    • 5 years ago
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    Isn't it 9+s

  58. anonymous
    • 5 years ago
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    oh yeah 9+s sorry

  59. anonymous
    • 5 years ago
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    now focus on the right part, and try to transform it back to y(x).

  60. anonymous
    • 5 years ago
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    Ugh I understand what you mean, I just am unaware of the method. I don't do inverse laplace right?

  61. anonymous
    • 5 years ago
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    You do nothing but the inverse Laplace.

  62. anonymous
    • 5 years ago
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    So like... 1/s=1 etc?

  63. anonymous
    • 5 years ago
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    Yeah.

  64. anonymous
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    not 1/s=1.. but the inverse laplace of 1/s=1 or more accurately u(x), (which is the unit step function)

  65. anonymous
    • 5 years ago
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    How do I solve the 9/5 and -4/5

  66. anonymous
    • 5 years ago
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    laplace transform and its inverse are both linear operations, that's L[af(t)]=aL[f(t)]

  67. anonymous
    • 5 years ago
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    Ugh I was understanding it and now I'm lost. I understand to use the table but this is just beyond me now.

  68. anonymous
    • 5 years ago
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    You should not worry about these constants.. just keep them as they are.

  69. anonymous
    • 5 years ago
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    Am I complicating things here? :(

  70. anonymous
    • 5 years ago
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    So just worry about the 1/s and 1/s+5?

  71. anonymous
    • 5 years ago
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    No you're doing good. I'm sure its just me.

  72. anonymous
    • 5 years ago
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    yeah.

  73. anonymous
    • 5 years ago
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    you already did 1/s.. what about 1/(s+5)?

  74. anonymous
    • 5 years ago
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    I just think you're not familiar with notation.

  75. anonymous
    • 5 years ago
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    e^-at is the transform...

  76. anonymous
    • 5 years ago
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    e^-5t then right?

  77. anonymous
    • 5 years ago
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    You know what you need to do, though.

  78. anonymous
    • 5 years ago
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    Yes

  79. anonymous
    • 5 years ago
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    Me?

  80. anonymous
    • 5 years ago
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    No, Scotty.

  81. anonymous
    • 5 years ago
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    Exactly Scotty.

  82. anonymous
    • 5 years ago
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    Sorry...

  83. anonymous
    • 5 years ago
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    It's OK :)

  84. anonymous
    • 5 years ago
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    Ok so to clean it up I use use what I have left right?

  85. anonymous
    • 5 years ago
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    Which basically will end up e^5t

  86. anonymous
    • 5 years ago
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    How?

  87. anonymous
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    Your answer should have e^(-5t) in it.

  88. anonymous
    • 5 years ago
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    9/5*1-4/5*e^(-5t)

  89. anonymous
    • 5 years ago
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    It's in x not in t.

  90. anonymous
    • 5 years ago
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    Yes, Scotty.

  91. anonymous
    • 5 years ago
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    But in x.

  92. anonymous
    • 5 years ago
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    I'm used to doing them in t...sorry.

  93. anonymous
    • 5 years ago
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    so e^-(5x)

  94. anonymous
    • 5 years ago
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    so final answer is: \[y(x)={9 \over 5}- {4 \over 5} e ^{-5x}\] right?

  95. anonymous
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    We all are used to t.

  96. anonymous
    • 5 years ago
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    perfect

  97. anonymous
    • 5 years ago
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    Ok I was rushing through on the simplification.

  98. anonymous
    • 5 years ago
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    I'm 12 years old and what is this? \[\frac{\mathbb{d}y}{\mathbb{d}x} = 9 - 5y \implies \int \frac{1}{9-5y} \mathbb{d}y = \int \mathbb{d}x \implies c -\frac{1}{5} \ln 9 - 5y = x\] x = 0 when y = 1 \[\implies c = \frac{1}{5} \ln 4 \implies x = \ln \frac{4}{9-5y} \implies y = \frac{9-4e^{-5x}}{5}\]

  99. anonymous
    • 5 years ago
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    What?

  100. anonymous
    • 5 years ago
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    He needed to use Laplace.

  101. anonymous
    • 5 years ago
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    So that just gave me the general solution right? Now I evaluate the particular?

  102. anonymous
    • 5 years ago
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    He solved the problem without using the Laplace transform.

  103. anonymous
    • 5 years ago
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    Just kidding guys, no need to bully me :(

  104. anonymous
    • 5 years ago
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    LOL

  105. anonymous
    • 5 years ago
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    FWIW, he said ' I could us..', implying it wasn't necessary. And it isn't.

  106. anonymous
    • 5 years ago
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    Hey Micheal!!

  107. anonymous
    • 5 years ago
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    oh man...

  108. anonymous
    • 5 years ago
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    You spelt my name wrong. It's 'Isaac Newton'

  109. anonymous
    • 5 years ago
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    sure...

  110. anonymous
    • 5 years ago
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    Oh yeah?

  111. anonymous
    • 5 years ago
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    Im confused on what just happened.

  112. anonymous
    • 5 years ago
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    No. But you did spell my real name wrong.

  113. anonymous
    • 5 years ago
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    :(

  114. anonymous
    • 5 years ago
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    Sorry Michael!

  115. anonymous
    • 5 years ago
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    :D

  116. anonymous
    • 5 years ago
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    ...

  117. anonymous
    • 5 years ago
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    So you just solved my problem doing tradidional integration then? Is that what I gather?

  118. anonymous
    • 5 years ago
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    Yeah.. He used separating variables method to solve the DE.

  119. anonymous
    • 5 years ago
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    You could have done it either way. You were on about Laplace, so Anwar answered as such.

  120. anonymous
    • 5 years ago
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    ^^ Thank you!! You saved me :)

  121. anonymous
    • 5 years ago
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    I see, its not solved all the way though right? I have C and X, I just need to use those to solve y?

  122. anonymous
    • 5 years ago
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    Is this a differential equation course or what exactly?

  123. anonymous
    • 5 years ago
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    Its a Calc II with LaPlace and Diff Eq in it.

  124. anonymous
    • 5 years ago
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    A pain in the retriceonline course to be exact, where my teacher refuses to help me which is why Im so lost.

  125. anonymous
    • 5 years ago
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    :( Oh dear.

  126. anonymous
    • 5 years ago
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    Yes.... so what do I need to do to finish using the method you did newton?

  127. anonymous
    • 5 years ago
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    It's OK, I finished it (I only posted the answer because you already had it).

  128. anonymous
    • 5 years ago
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    So, Scotty, are you okay with using Laplace now?

  129. anonymous
    • 5 years ago
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    Yes better. I need to practice a bit more I think :/

  130. anonymous
    • 5 years ago
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    Sure.

  131. anonymous
    • 5 years ago
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    It's a really useful method. You transform a differential equation into an algebraic one which can be easier to solve. Plus, you're initial conditions are taken into account in the process.

  132. anonymous
    • 5 years ago
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    *your

  133. anonymous
    • 5 years ago
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    Damn! Was about to correct that. I'll get you next time...

  134. anonymous
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    Yeah I can see how it would be useful, I just need a little more practice recognizing the transform tables and how to apply them. Also I guess my notation is different than the traditional method

  135. anonymous
    • 5 years ago
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    haha

  136. anonymous
    • 5 years ago
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    You should also practice working them out yourself (i.e. that table entries).

  137. anonymous
    • 5 years ago
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    I wouldn't veen know how to start that, I'll just read some textbooks. Anyways off to the next problem on my review. Thanks guys

  138. anonymous
    • 5 years ago
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    np. I'm off to bed. Good work, Anwar :)

  139. anonymous
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    ...and Scotty :)

  140. anonymous
    • 5 years ago
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    Oh thanks :)

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