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Solving Differential Equations

  1. Oct 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Solve the following: xdx = y[tex]^{2}[/tex]dy

    2. Relevant equations
    Fundamental theorem of calculus - thanks gabbagabbahey


    3. The attempt at a solution
    [tex]\frac{dy}{dx}[/tex]=[tex]\frac{x}{y^{2}}[/tex]

    [tex]\int[/tex][tex]\frac{dy}{dx}[/tex] = [tex]\int[/tex][tex]\frac{x}{y^{2}}[/tex]

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

    So with the question, ive integrated both sides to find y^2 on the bottom, so should i rearrange to find y on the LHS and the x on the RHS?
     
    Last edited: Oct 20, 2010
  2. jcsd
  3. Oct 20, 2010 #2

    gabbagabbahey

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    Unless NA is your abbreviation for the fundamental theorem of calculus, you are mistaken :wink:


    What happens if you integrate both sides of the equation [itex]x=y^2\frac{dy}{dx}[/itex] with respect to [itex]x[/itex]?
     
  4. Oct 20, 2010 #3
    Hi gabbagabbahey,

    Sorry, i am just getting to grips with the funky math features on the site, so my translation from paper to web isn't vry good. First of all i put in a monster mistake on the web, working on correcting this now.
     
  5. Oct 20, 2010 #4
    Okay, more or less fixed now..... thanks again Gabb
     
  6. Oct 20, 2010 #5
    Right, so carrying on from my first post,

    if i rearrange for y, i get:

    y = [tex]\sqrt{\frac{x^{2}}{2}}[/tex]
     
  7. Oct 20, 2010 #6

    Char. Limit

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    Your starting point is

    [tex]x dx = y^2 dy[/tex]

    Right?

    You can just integrate both sides right there. Start with that.
     
  8. Oct 20, 2010 #7
    Ok, so if i integrate both sides i get:

    [tex]\frac{x^{2}}{2}[/tex] = [tex]\frac{y^{3}}{3}[/tex]

    and y = [tex]\sqrt[3]{\frac{3x^{2}}{2}}[/tex]

    Surely this isn't right is it? I don't recall coming across cubic roots in 'this particular' module. (I am simply looking for familiarity with the rest of the course)
     
  9. Oct 20, 2010 #8

    Char. Limit

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    You forgot the plus C.
     
  10. Oct 20, 2010 #9
    Ahh okay fair enough Char.Limit, i am confused as to why re-arranging the initial equation (as i did earlier) yields such a different result?
     
  11. Oct 20, 2010 #10

    Char. Limit

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    It's mainly because you can't integrate x/y^2 dx, because y also depends on x.

    So you can see why you couldn't integrate...

    [tex]\int \frac{x}{y^2(x)} dx[/tex]
     
  12. Oct 20, 2010 #11
    Oh i see now. So with questions like these, i should generally keep all the y's on one side and the x's on the other? Ill attempt another question and post it to see if i have my head in the right place. Thank you Char.Limit, and as ever, you make a good point.
     
  13. Oct 20, 2010 #12

    Char. Limit

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    No problem. Glad I could help.
     
  14. Oct 20, 2010 #13
    1. The problem statement, all variables and given/known data

    Solve the following: [tex]\frac{dy}{dx}[/tex] = [tex]\frac{1 + y}{1 + x}[/tex]

    2. Relevant equations
    Fundamental theorem of calculus - thanks gabbagabbahey


    3. The attempt at a solution

    Re-arranging to get y terms and x terms on opposite sides:

    [tex]\int[/tex][tex]\frac{dy}{1 + y}[/tex] = [tex]\int[/tex][tex]\frac{dx}{1 + x}[/tex]

    Which gives:

    ln(x + 1) = ln(y + 1)

    and so:

    y = x

    Does this work out?
     
  15. Oct 20, 2010 #14

    Char. Limit

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    Close, but you still need to remember the +C.
     
  16. Oct 20, 2010 #15
    Damn, every time. A habit i need to get used to.
     
  17. Oct 20, 2010 #16

    Char. Limit

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    No real problem. You just have to make sure you do it.
     
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