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Simple diff eq that i cant figure out

  1. Aug 30, 2007 #1
    1. The problem statement, all variables and given/known data
    y ' [x] = xy
    y[0] = 1


    2. Relevant equations
    none


    3. The attempt at a solution

    I posted this somewhere else and 2 people got the answer is y = e^x

    I am getting something else though.

    I get up to here:

    dy/dx = xy

    1/y dy = x dx

    ln y = (x^2)/2 + C

    that is what i get to, but here is what the other person got from this point:

    y = e^[ (x^2/2) + C ] (Im not confused on e^ln^y = y fyi)

    then he simplified that to:

    y = Ce^x (I DONT understand that property of natural logs)

    1 = C * e^0

    C = 0

    y = e^x

    anyone care to explain? dont worry if you know a better way and start from scratch, ignore mine and this other guys answer if you want.
     
  2. jcsd
  3. Aug 30, 2007 #2

    VietDao29

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    Homework Helper

    Yup, so far, so good. :wink:

    Yup, this is correct. Can you get this step? We have:

    [tex]\ln y = x \Rightarrow e ^ {\ln y} = e ^ x \Rightarrow y = e ^ x[/tex]

    Nope, this is wrong. You can simply check it by differentiate y with respect to x. If y = ex, then, y' = ex = y, not xy.

    Here's some properties of exponential that you should remember:
    (1). [tex]\alpha ^ {\beta + \gamma} = \alpha ^ \beta \times \alpha ^ \gamma[/tex]

    (2).[tex]\alpha ^ {\beta - \gamma} = \frac {\alpha ^ \beta}{\alpha ^ \gamma}[/tex]

    (3).[tex]\alpha ^ {-\gamma} = \alpha ^ {0 - \gamma} = \frac{\alpha ^ 0}{\alpha ^ \gamma} = \frac{1}{\alpha ^ \gamma}[/tex]

    (4).[tex]\alpha ^ {\beta \times \gamma} = \left( \alpha ^ \beta \right) ^ \gamma[/tex]

    (5).[tex]\alpha ^ {\frac{\beta}{\gamma}} = \sqrt[\gamma]{\alpha ^ \beta}[/tex]

    Back to your problem:

    [tex]y = e ^ {\frac{x ^ 2}{2} + C}[/tex]

    Using (1), we have:

    [tex]\Rightarrow y = e ^ C \times e ^ {\frac{x ^ 2}{2}}[/tex] since C is a constant, we have eC is also a constant, we denote it D, we have:

    [tex]\Rightarrow y = D \times e ^ {\frac{x ^ 2}{2}}[/tex]

    Can you go from here? Can you find D? :)
     
    Last edited by a moderator: Aug 30, 2007
  4. Aug 30, 2007 #3
    haha no, i cant actually. please continue.
     
  5. Aug 30, 2007 #4
    Use the condition y(0) = 1 to find D.
     
  6. Aug 30, 2007 #5
    ok, D = 0 then
     
  7. Aug 30, 2007 #6
    No, because that forces y=0 for all x, which certainly does not satisfy y(0)=1.
     
  8. Aug 31, 2007 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    In [tex]y= De^{\frac{x^2}{2}}[/tex], replace y with 1, replace x with 0, and solve for D.
     
  9. Sep 3, 2007 #8
    Just so you don't get thrown off

    y = e^[ (x^2/2) + C ]

    does not simplify down to

    y = Ce^x (What happened to the x^2 then?)
     
  10. Sep 3, 2007 #9

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, that was pointed out before.
    [tex]y= e^{\frac{x^2}{2}+ C[/tex]
    does simplify to
    [tex]y= Ce^{\frac{x^2}{2}[/tex]
    But this new "C" is not the same as the previous "C"!

    Now if y= 1 when x= 0, that becomes
    [tex]y= 1= Ce^{0}[/itex]
    Solve that for C.
     
  11. Sep 6, 2007 #10
    ok, C = 1 then

    what is next?
     
  12. Sep 6, 2007 #11
    plug in C=1 into y=Ce^[(x^2)/2]

    and you'll get your final answer
     
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