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Math Help - solving a first order differential equation using separation of variables

  1. #1
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    solving a first order differential equation using separation of variables

    this is a rather elementary problem for diff. eq, and i solved it correctly using separation of variables, but there is a reason that the answer y=4/(x^2) is only valid for x<0.

    dy/dx=y^(3/2)
    dy/(y^3/2) = dx
    integrate both sides to get....
    -2*y^(-1/2)= x + C
    -2=(x+C)*y^(1/2)
    y^(1/2) = -2/(x+C)
    y = 4/[(x+C)^2]

    so...for some reason, when i solved for y explicitly in terms of x, I shold have been able to predict that the solution y = 4/x^2 is only valid for x<0

    but I cant figure out what step in my algebra should have led me to conclude that the solution y = 4/x^2 is only valid for x<0.

    any help appreciated.
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  2. #2
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    If y^{3/2} > 0 , then \frac{dy}{dx} > 0 so the solution curve will have a positive slope.

    Now look at the graph of your 'solution', y = 4/x^{2}.

    It has a vertical asymptote at x = 0 and its slope is negative for all x>0.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by BobP View Post
    ... if y^{3/2} > 0 , then \frac{dy}{dx} > 0 so the solution curve will have a positive slope...
    Let's write the DE as...

    y^{'} = \sqrt {y^{3}} (1)

    If  y<0 clearly the (1) has no solution if x \in \mathbb {R}. If y>0 the derivative is expressed as 'square root' so that it can be as well positive or negative and the choice depends from the 'iniatial condition' y(x_{0}) = y_{0} with y_{0} \ge 0. The function y=\frac{4}{x^{2}} is then solution of (1) \forall {x \ne 0}. The (1) has another interesting properties: the value of y^{'} doesn't depend fron x. That means that if f(x) is solution of (1) f(x+\xi) with \xi arbitrary is also solution of (1), so that for each initial condition y(x_{0}) = y_{0} with y_{0} > 0 we have two solution of (1) of the form y= \frac{4}{(x - \xi)^{2}}, one with positive slope and the other with negative slop in x=x_{0}...

    Kind regards

    \chi \sigma
    Last edited by chisigma; August 31st 2009 at 05:41 AM. Reason: 4/x^2 and not 1/x^2 is solution of (1)...
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  4. #4
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    I would have thought that with normal usage, the RHS of the equation would be taken to be positive. However, if that is not the case aren't we looking at two different differential equations here y' = +y^{3/2} and y' = -y^{3/2} ?

    With regards to the solution y = 4/x^{2}, (BTW y = 1/x^{2} is not a solution), dy/dx = -8/x^{3}, so if this is to be a solution of the first of these equations it is necessary that x<0 while if it is to be a solution of the second equation we need x>0.
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  5. #5
    MHF Contributor chisigma's Avatar
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    Thanks to your indication i was able to correct the error... \frac{4}{x^{2}} and not \frac{1}{x^{2}} is solution od the DE...


    y^{'}= \sqrt{y^{3}} (1)

    How silly of me! ... The problem is therefore that the omission of the initial conditions in the usual form y(x_{0})=y_{0} with y_{0} >0. What I have demonstrated is that with the usual approach the (1) has in general two solution satisfiyng the condition y(x_{0})=y_{0} with y_{0} >0. As example we take y(x)=\frac{4}{x^{2}} which is solution of the (1) with initial conditions y(2)= 1. Unfortunatley it is easy to verifiy that y(x)= \frac{4}{(x-4)^{2}} is also solution of (1) with the same initial condition...

    Kind regards

    \chi \sigma
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