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Math Help - Straight forward ODE

  1. #1
    No one in Particular VonNemo19's Avatar
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    Straight forward ODE

    Hi again. Here's one.

    (a) Without solving, explain why the intitial value problem

    \frac{dy}{dx}=\sqrt{y}, y(x_0)=y_0

    has no solution for y_0<0.

    (b) Solve the initial value problem in part (a) for y_0>0 and find the largest interval I on which the solution is defined.


    My attempt:

    (a) Because the only way to arrive at a radical through differentiation is to begin with one, and any expression involving a radical (even root) with nothing outside of the radical must always give a positive result over the reals.

    (b) I get

    y=\frac{1}{4}(x+c)^2.

    If I set y_0=\frac{1}{4}(x_0+c)^2, I get c=\pm2\sqrt{y}-x_0.

    But, I won't evfen attempt to find the interval of definition because the book doesn't include the plus or minus. They have c=2\sqrt{y}-x_0.

    What's the deal here?
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  2. #2
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    <br />
\frac{dy}{dx}=\sqrt{y}<br />

    <br />
\frac{dy}{\sqrt{y}}=dx<br />

    <br />
2 \sqrt{y}=x+C<br />

    <br />
2 \sqrt{y_0}=x_0+C<br />

    Making square you add one more root.
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  3. #3
    No one in Particular VonNemo19's Avatar
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    OK, I unserstand that, but why is it wrong to find an explicit formula for y in this case? It seems clear to me that if y=\frac{1}{4}(x+c)^2, then \frac{dy}{dx}=\frac{1}{2}(x+c)=\sqrt{y} which is the DE.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    OK, I unserstand that, but why is it wrong to find an explicit formula for y in this case? It seems clear to me that if y=\frac{1}{4}(x+c)^2, then \frac{dy}{dx}=\frac{1}{2}(x+c)=\sqrt{y} which is the DE.
    ...and the interval I on which the solution is defined would then be I=(-\infty,\infty). I really don't understand what is wrong with this.
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  5. #5
    Behold, the power of SARDINES!
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    Quote Originally Posted by VonNemo19 View Post
    ...and the interval I on which the solution is defined would then be I=(-\infty,\infty). I really don't understand what is wrong with this.
    The key idea is that the domain of your solution does not have to be the same as the interval of existence and uniqueness of the solution to an ODE.

    We know that if  \displaystyle \frac{dy}{dx}=f(x,y) and

    \displaystyle f(x,y) \text{ and } \frac{\partial x}{\partial y} are continuous [a,b] \times [c,d] and (x_0,y_0) is an interior point of the above rectangle.
    Then there exists an interval (x_0-h,x_0+h) \subset [a,b] such that a solution exists and is unique.

    Now for your problem \displaystyle \frac{\partial f}{\partial y}=\frac{1}{2\sqrt{y}} is discontinous at (x,0)

    Now given an initial point (x_0,y_0); y_0 > 0 then the solution "is"

    \displaystyle y_0=\frac{1}{4}\left( x_0+c\right)^2 \implies c=-x_0 \pm 2y_0

    \displaystyle y=\frac{1}{4}\left(x-x_0 +2y_0 \right)^2

    Things are already starting to look off because now we have a non unique solution and here is the kicker notice what happens when x=x_0-2y_0

    This gives y(x_0-2y_0)=0 Now I can make another solution as follows

    \displaystyle y(x)=\begin{cases} \frac{1}{4}\left(x-x_0 +2y_0 \right)^2, \text{ if } x > x_0-2y_0 \\ 0, \text{ if } x \le x_0-2y_0\end{cases}

    Notice this new function is continuous and differentiable on all of \mathbb{R} but is not unique. So we run into problems where \displaystyle \frac{\partial f}{\partial y} is undefined.

    So the largest interval with a unique solution is either (-\infty,x_0-2y_0)\quad (x_0-2y_0,x_0+2y_0) \quad (-x_0+2y_0,\infty)\quad
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