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.
(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...