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Separable Equation Trouble

  1. Feb 16, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the solution to the differential equation. Which passes through the point (0,e).

    [itex]\frac{dy}{dx}[/itex] = [itex]\frac{11xy}{(ln(y))^{10}}[/itex]

    2. Relevant equations

    I can get through the integration part but I am lost when it comes to using the ln rules to find the solution.

    3. The attempt at a solution

    First step I move the variables to each side.

    [itex]\frac{(ln(y))^{10}}{y}[/itex] dy = 11x dx

    I then integrate each side:

    ∫[itex]\frac{(ln(y))^{10}}{y}[/itex] dy = ∫11x dx

    ∫[itex]\frac{(ln(y))^{10}}{y}[/itex] dy, I used sub method. u = ln(y), du = 1/y dy -> dy = ydu

    ∫[itex]\frac{u^{10}}{y}[/itex] * y du, so the y's cancel out. making it:

    ∫ u[itex]^{10}[/itex] du = [itex]\frac{u^{11}}{11}[/itex] = [itex]\frac{ln(y)^{11}}{11}[/itex]

    The other side is simple integration:

    ∫11x dx = [itex]\frac{11x^{2}}{2}[/itex], this gives:

    [itex]\frac{ln(y)^{11}}{11}[/itex] = [itex]\frac{11x^{2}}{2}[/itex] + c, since it passes through
    the point (0,e), I find my constant value c right away.

    y(0)=e =[itex]\frac{ln(e)^{11}}{11}[/itex] = [itex]\frac{11(0)^{2}}{2}[/itex] + c

    y(0)=e = 1 = 0 + c, so c = 1.

    I then multiplied the 11 by both sides and got:

    (ln(y))[itex]^{11}[/itex] = [itex]\frac{121x^{2}}{2}[/itex] + 11

    This is where I am getting confused. I do not know how to use the ln rules on this one. The power of 11 is confusing me and I can not find an example like this one. Any help would be nice. thanks.
     
  2. jcsd
  3. Feb 16, 2014 #2

    rock.freak667

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    You can leave your answer as it is, you aren't really required to make y the subject. Otherwise you'd need to take the 11th root of both sides of the equation.
     
  4. Feb 16, 2014 #3
    Wow. I am stupid. Haha. I can't believe I did not think of that. For some reason the exponent of 11 was messing me up. I'll take the 11th root of each side then put it in the form ln(y) = ........ because that is what it asks for.

    Thanks a lot man. I have being doing differential equations all day and my brain is running on low.
     
  5. Feb 16, 2014 #4

    Dick

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    You also slipped up when you found the constant c=1. If you substitute x=0 and y=e into your final expression it doesn't work. What happened?
     
  6. Feb 16, 2014 #5
    I put it in my calculator wrong. I included the power of 11 in the e, so I had ln(e[itex]^{11}[/itex])= 11. It should be (ln(e))[itex]^{11}[/itex]=1. So c would be 1/11. So then my final answer would be

    ln(y) = ([itex]\frac{121x^{2}}{2}[/itex]+1)[itex]^{\frac{1}{11}}[/itex]. I could go further but thats all the question asked for.

    Thanks for pointing that out. I am starting to make simple mistakes so its about time I stop and take a break.
     
  7. Feb 16, 2014 #6

    Dick

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    Yeah, take a break. Needing a calculator to find ##(ln(e))^{11}## is already a bad sign.
     
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