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Simple differential equation

  1. Jul 22, 2014 #1
    1. The problem statement, all variables and given/known data

    a) Write (x21)y'+2xy as the derivative of a product
    b) Solve (x21)y'+2xy=e-x

    3. The attempt at a solution

    a) I use the product rule backwards and get

    ((x2+1)y)'

    b) I exploit what i just found out...

    (x21)y'+2xy=((x2+1)y)'

    and get...

    e-x=((x2+1)y)'

    integrate on both sides...

    ∫e-xdx=∫((x2+1)y)'dx

    -e-x+C=(x2+1)y

    and get that...

    y=(C-e-x)(x2+1)-1

    this is the correct answer according to the book.

    What i am curious about is in the step marked with bold text. I wrote dx at the end just by guessing. Why couldnt i just have written dy instead?
     
  2. jcsd
  3. Jul 22, 2014 #2
    You could have integrated both sides with respect to y if you really wanted to, but while the left side would then become ye-x, there is no theorem that helps us easily integrate the right side with respect to y. We are only able to integrate it with respect to x because we found in step one that it is the derivative with respect to x of (x2 + 1)y, and the fundamental theorem of calculus then assures us that its integral with respect to x is (x2 + 1)y plus an arbitrary constant.
     
  4. Jul 22, 2014 #3
    Thank you!
     
  5. Jul 22, 2014 #4

    ehild

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    y is function of x and the comma ' means derivative with respect to x. y ' means dy/dx. If f=(x2+1)y then
    f '= df/dx . You get f if you integrate f '. ∫f ' dx= f

    Formally you can handle the problem as if df/dx was a simple fraction. df/dx = e-x, multiply both sides by dx df= e-x dx and put the integral symbol at the front:

    ∫ (df/dx) dx = ∫e-x dx

    ehild
     
  6. Jul 23, 2014 #5

    HallsofIvy

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    You mean (x2+ 1)y'. That puzzled me for a while!

     
  7. Jul 23, 2014 #6
    Sorry!
     
  8. Jul 23, 2014 #7
    This is only true in this case right? Its more just an assumtion we make based on the look of the task?
     
  9. Jul 23, 2014 #8
    Yes. In most texts on ordinary differential equations, y is assumed to be the dependent variable and x is assumed to be the independent variable as a matter of notation only, usually established by the author in the first chapter. Therefore y' is assumed to always mean y'(x), or dy/dx in these texts only.
     
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