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Seperable Differnetial Equation Question?

  1. Jan 26, 2012 #1
    Find General Solution of the Differential Equation If Possible Find Explicit Solution

    Problem:
    xy'=2y


    My attempt at solving:
    xy'=2y

    (1/2y)*x*y'=0

    1/2 integral 1/y dy = integral of 1/x dx

    1/2 * ln |y| = ln |x|

    I am not sure If I did this right? I am just wondering if this is correct or if I messed up somewhere.
     
  2. jcsd
  3. Jan 26, 2012 #2

    Curious3141

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    Homework Helper

    Where did that zero come from? Don't you mean 1?

    OK, but you're missing a constant term. I'd suggest putting the constant term as [itex]\frac{1}{2}\ln|k|[/itex] on the RHS - you'll see why when you simplify the expression.

    Of course, you need to simplify the expression by taking antilogs, etc.
     
  4. Jan 26, 2012 #3
    I was able to get the equation to

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

    I am rather lost on how to get rid of the logarithm.

    How do you do an anti-log?
     
  5. Jan 26, 2012 #4

    Curious3141

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    Homework Helper

    Remember to keep the absolute value terms (|y| and |x|).

    Start by simplifying that 2ln x term with a rule of logs, namely [itex]n\log{x} = \log{x^n}[/itex].

    Then take e^ (e to the power of) both sides. Remember the rules of exponents, i.e. a^(b+c) = (a^b).(a^c).

    You should now have eliminated the logs, and have an equation with only |y| on the LHS and a single term in terms of |x| on the RHS. Now you should be able to see why I advised you put the constant in that particular form. But no matter, just replace that constant expression with another of your choosing, since it's an arbitrary constant after all.

    Now remove the modulus on both sides. Keep in mind that |-x| = |x|, so be careful with this step.
     
  6. Jan 26, 2012 #5
    I managed to figure out that one but now I am lost on this one...

    I figured out how this one is done but I don't understand why I couldn't do the problem the way I originally tried to solve it.

    The problem is:
    y'=(1+y^2)*e^x

    I went and distributed the e^x's
    y' = e^x + y^2*e^x

    moving y^2 over by division:
    y^-2*dy/dx = e^x+e^x
    y^-2*dy/dx = 2e^x

    and got after rearranging:
    y^-2dy = 2e^x dx

    Did the integrals of both sides

    Ended up with:
    -y^-1 = 2e^x+C

    but after you re-arrange its nothing like the solution that it should be of:
    tan(e^x+C)


    Why didn't this work?
     
  7. Jan 26, 2012 #6
    You have to be careful
    y' = e^x + y^2*e^x
    ≠ y^-2*dy/dx = e^x+e^x
     
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