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Separable Differential Equations

  1. Nov 3, 2008 #1
    1. The problem statement, all variables and given/known data

    I'm having trouble understanding my class notes from a lecture on separable differential equations.

    I would like to solve the equation g(y)y' = f(x)

    3. The attempt at a solution

    g(y)y' = f(x), G(x), F(x) exists and are continuous

    The left side is the derivative of G(y(x)) and the right is F(x) + C

    [tex]\frac{d}{dx} G(y(x)) = \frac{d}{dx} F(x) + C[/tex]

    G'(y(x))y'(x) = g(y(x))y'(x)

    [tex]\frac{d}{dx} (F(x) + C) = F'(x) + 0 = f(x)[/tex]

    G(y(x)) = F(x) + C

    So do you simply do

    G-1(G(y(x))) = y(x) = G-1(F(x)) + G-1(C) ?
     
  2. jcsd
  3. Nov 3, 2008 #2

    gabbagabbahey

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    Assuming [itex]G^{-1}[/itex] exists and is a 1:1 map, then yes....Some functions do not have inverses that uniquely determine y(x), for instance [itex]y^2(x)=F(x)+C \Rightarrow y(x)= \pm \sqrt{F(x)+C}[/itex] which is not a single valued function.
     
  4. Nov 3, 2008 #3
    hello, I'm also taking a class on ODE but i have a problem -i use An Intro course in Diff. eq.'s by Zill - that i get a nonsense result here is the eq:

    sin3x + 2y(cos3x)^3 = 0 (here ^ is to raise a power.how are u raising powers & all the mathematical writings?)

    the eq in standard form look: (y^2)'=2ydy= -sin3x dx/2(cos 3x)^3.

    the last result i get which is nonsense ofcourse is: y^2 = -1/6(cos3x)^2. another result includes tan3x but is still negative.

    so y^2 is negative which is impossible. is the result right? I think there's a problem with the D.E. given.

    hope u can help. thx
     
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