1
$\begingroup$

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 $1$.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.