When solving the differential equation:

$dy/dx = y^2$, with $y(0) = 1$

I've found

$y = 1/(1-x)$ as the solution.

The problem asks then for an explanation to why $x=3/2$ is an invalid point to consider when graphing the solution.

The solutions manual says that although the formula for $y$ makes sense at $x=3/2$:

$y(3/2) = -2$;

it is not consistent with with the rate of change interpretation of the differential equation. It continues saying that the function is defined, continuous and differentiable for $-\infty < x < 1$.

It is clear to me that at $x=1$ both $y$ and $dy/dx$ are undefinded, but I cannot understant why the solution y does not apply for values of $x$ higher than $1$.

  • $\begingroup$ This problem is presented in problem set for MIT Open Courses, Single Variable Calculus, Unit 3 - Integration, Section 3F - Differential equations: separation of variables, problem 3F-3. (included for future google searches). ocw.mit.edu/courses/mathematics/… , $\endgroup$ – Ivan Koshelev Aug 20 '18 at 20:45

Most likely, the book is trying to say that, though the equation $$y=\frac{1}{1-x}$$ is a solution to $\frac{dy}{dx}=y^2$ and $y(0)=1$, and is, in fact, the unique solution to that over all of $\mathbb{R}$ minus a point, the value of $y(\frac{3}2)$ lacks an interpretation in terms of what the differential equation is trying to represent; in particular, the function has two "branches" - it is discontinuous at $x=1$, which splits it. The left branch, in $(-\infty,1)$ could be found just by drawing a curve satisfying the differential equation. But the right branch doesn't mean the same thing, since our initial conditions do not suggest any point on that branch - it satisfies $\frac{dy}{dx}=y^2$, but there's a lot of functions which do, and the only justification we could have for choosing the one equaling $\frac{1}{1-x}$ (as opposed to other solutions $\frac{1}{c-x}$ for other $c$) is that this is the only one with an asymptote lining up to the left branch - but this reasoning is far removed from "$y$ increases at a rate in proportion to its square", since now we've invented a whole new branch of the function based on the reasoning that we must define $y$ everywhere - but if we only considered $y$ to be a curve increasing at a rate proportional to its square, there's no way we could get to this, since we get to $\infty$ at $x=1$, and then it's not obvious how we could proceed.


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