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I'm unable to fully understand how to solve differential equations systematically. I don't know when to impose a condition of discontinuity and what exactly the conditions of the solution to an IVP problem are.

Alright, the problem consists of four parts:

(a) Verify that $ y^2-2y = x^2-x+c $ is an implicit solution for the differential equation $$ (2y-2)y'=2x-1 $$

(b) Find a member of the one-parameter family in part(a) that satisfies $y(0)=1$

(c) Use your result in part(b) to find an explicit function $y=\phi(x)$ that satisfies $y(0)=1$ and give the domain of $\phi(x)$.

(d) Is $y=\phi(x)$ a solution of the initial-value problem? If so, give its interval I of definition; if not, explain.

What I have tried:

(a) This part is straightforward, just differentiate $ y^2-2y = x^2-x+c $.

(b) Setting $x=0$ and $y=1$ gives $c=1$, thus the solution is $y^2-2y = x^2-x+1$. But I was just wondering why can $y$ be 1? Because when I move $y^2-2y$ to the right I have $$ y'=\frac{2x-1}{2y-2} $$ which is not defined when $y=1$. I think there must be a misconception.(?)

(c) Okay, I suppose there are solutions that satisfy $y(0)=1$ (or pass through a point $(0,1)$). (Do they really exist?) Now solving $y^2-2y = x^2-x+1$ we get $$y=1+ \sqrt{x(x-1)}, y=1- \sqrt{x(x-1)}$$ and the domain for both explicit functions is $(-\infty,0)$ and $(1,\infty)$ .

(d) A solution to the initial-value problem has to be differentiable and the interval of definition has to contain the initial point. So, do I understand correctly that each $\phi(x)$ in part(c) is not a solution of the initial-value problem because it isn't differentiable and doesn't even pass the point $(0,1)$? But, still, are they solutions of the differential equation?

Any help would be greatly appreciated.

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    $\begingroup$ Answer to (b): y(0)=1 is not incompatible with $y'(0)=\infty$. You can have infinite slope at x=0, which immediately bends over to finite slope at x>0, like in $y=\sqrt{x}$. $\endgroup$
    – Joachim W
    Commented Mar 11, 2016 at 9:26
  • $\begingroup$ In (c) there must be an error under one of the roots, otherwise both functions have the same domain. $\endgroup$
    – Joachim W
    Commented Mar 11, 2016 at 9:28
  • $\begingroup$ What kind of an error do you mean? I think both functions pass through the point (0,1), I suppose? $\endgroup$
    – IgNite
    Commented Mar 11, 2016 at 9:35
  • $\begingroup$ There is a typo in the wording of the question. If you differentiate $y^2-2y=x^2-x+c$ you get $(2y-2)y'=2x-1$ which is not what you wrote : $(2y-y)y'=2x-1$. $\endgroup$
    – JJacquelin
    Commented Mar 11, 2016 at 9:43
  • $\begingroup$ Oh, sorry. I will edit that in a minute. Thanks for pointing out. $\endgroup$
    – IgNite
    Commented Mar 11, 2016 at 9:48

1 Answer 1

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(b) - First things first, the fact that y'(1) is not defined, tells you that y is not differentiable at x=1, and little more. The second comment is that when you devide by (2y - 2) you are tacitly assuming that $y \neq 1$.

(c) This might just me but isn't the domain of $$ y = 1 + \sqrt{x(x - 1)}$$ equal to $( -\infty, 0] \cup [1, \infty) $ which has 0 as a boundary point.

(d) The solutions pass through the point, but are differentiable only in the interior of the function domains. I am a little rusty here, but I seem to recall that only Lipshitz continuity was necessary for existence and uniqueness of a solution...

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  • $\begingroup$ I was just wondering that..shouldn't it be (-\infty,0) and (1,\infty)? Because at the boundary the function is not differentiable and therefore y' is not defined? $\endgroup$
    – IgNite
    Commented Mar 11, 2016 at 9:53
  • $\begingroup$ I'm not sure what you mean, but the function $y = 1 + \sqrt{x(x - 1)}$ has the given domain where it is real, so y is defined there... $\endgroup$
    – TSU
    Commented Mar 11, 2016 at 9:56
  • $\begingroup$ Aha, I see. The domain you meant is the domain of the function. And, are you saying that, if the function is to be a solution to the DE, it has to be differentiable--which it is only in the interior of the function domains? $\endgroup$
    – IgNite
    Commented Mar 11, 2016 at 9:57
  • $\begingroup$ I think this is it: en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem $\endgroup$
    – TSU
    Commented Mar 11, 2016 at 9:58
  • $\begingroup$ Alright alright.. I think I will have to read about this more $\endgroup$
    – IgNite
    Commented Mar 11, 2016 at 10:02

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