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Looking for an explanation of a simple Lipshitz condition

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  1. Jul 25, 2016 #1
    1. The problem statement, all variables and given/known data
    gif.gif
    Has solution
    gif.gif
    It then goes on to state the solution blows up at gif.gif , which I understand.

    My issue is when I do the solution I get
    gif.gif

    (Working)
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  2. jcsd
  3. Jul 25, 2016 #2

    Stephen Tashi

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    You seem to be treating [itex] y [/itex] as if it were a constant instead of a function of [itex] t [/itex].
    We can't assume [itex] \int y dt = (y)(t) = yt [/itex] because, in this problem the antiderivative needed is [itex] \int y(t) dt [/itex] where [itex] y(t) [/itex] is an unknown function.

    For example, if [itex] y(t) = e^t [/itex] then [itex] \int y(t) dt = e^t + c \ne (y(t))(t) [/itex].
     
  4. Jul 25, 2016 #3
    If you could humour me as an engineer having to take a maths course and is struggling to remember concepts taught along time ago, How could I prove their answer, or what should I read up on to be able to show this proof?
     
  5. Jul 25, 2016 #4

    Ray Vickson

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    Wrong: ##y## is a function of ##t## because, after all, you have ##\dot{y} = dy/dt## in your equation to start with! So, from ##dy/dt = y^2## you have
    [tex] \frac{dy}{y^2} = dt \: \Longrightarrow -y^{-1} = t + c, [/tex]
    or
    [tex] \frac{1}{y} = -c-t \: \Longrightarrow y = \frac{-1}{t+c} [/tex]
    Set ##c = -1/y_0## and see what you get.

    Anyway, if all you want to do is to see why ##y = y(t) = y_0/(1-y_0 t)## is a solution, just plug it into your differential equation ##dy/dt = y^2## and see if it works.
     
  6. Jul 25, 2016 #5

    Stephen Tashi

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    Does the problem simply want to you show that ## y(t) = \frac{y0}{1 - y0 t} ## is a solution ? You do that by substituting that function in for ## y ## in the differential equation. (Of course you also have to substitute the derivative of ## \frac{y0}{1 - y0 t}## in for ##\dot{y}## , which means remembering enough calculus to compute that derivative ).


    When you substitute a solution into a differential equation, the equation should become an identity - i.e. the two sides should be the same function when you simplify things.


    Solving differential equations when you aren't given a solution, often involves guessing it. If you have to find the solution by yourself, start by reviewing your knowledge of derivatives with emphasis on noting how derivatives of a function y can be expressed in terms of the original function y.

    For example:

    ## \dot{y} = 5y## corresponds to the derivative of ## e^{5t} ## is ## 5 e^{5t} ##.

    Using "## D ##" to denote the operation of taking a derivative, ## De^{kt} = k e^{kt} ##. So when you are asked to solve a differential equation ## \dot{y} = k y ## (where ## k ## is a constant) you should be able to guess the answer is an exponential function.


    Ray Vickson illustrated the technique of solving a "separable" differential equation. You can find videos on the web that explain that technique in detail.
     
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