4
$\begingroup$

I am reading this text http://www.math-cs.gordon.edu/courses/ma225/handouts/sepvar.pdf to justify the method to solve first order seperable differentiable equations, where we are told first told that: enter image description here

and then: enter image description here

Now, while I can understand 1), I am struggling to understand how exactly the integral on the left hand side of 3) surmounts to $$\int n(y) dy$$ since $\frac{dy}{dx}$ can't be treated as a fraction.

$\endgroup$
  • $\begingroup$ @mvw Can you please clarify? $\endgroup$ – Reinhild Van Rosenú Jul 7 '15 at 13:34
  • 2
    $\begingroup$ I believe it's just a matter of substitution. You have $\int_0^x n\big(y(\tilde x)\big) y'(\tilde x) \mathrm d \tilde x $ and substitute $ t := y(\tilde x), \mathrm d t = y(\tilde x)'\mathrm d \tilde x$ to obtain $\int_{y(0)}^{y(x)} n(t)\mathrm d t $ $\endgroup$ – krvolok Jul 7 '15 at 13:37
  • 2
    $\begingroup$ Note that it is incredibly misleading to say "integrate both sides with respect to $x$" as the text does. You are integrating with respect to some parameter, such that the limits of integration are $x_0$ (usually set to zero) and $x$. This way, integration results in an expression like $F(y(x),y(x_0))=G(x,x_0)$, which can hopefully be inverted to give $y(x)=H(x,x_0,y(x_0))$. For some reason, many texts present it in such a way, especially those aimed at applications or physics. But the objective is to have $x$ as the limit of integration, so that you can invert for the solution! $\endgroup$ – krvolok Jul 7 '15 at 13:47
  • $\begingroup$ @krvolok Thanks a lot, I was just actually wondering if it intuitively makes sense to integrate both sides with respect to both sides. $\endgroup$ – Reinhild Van Rosenú Jul 7 '15 at 13:52
  • $\begingroup$ @krvolok I was wondering how we can justify in this case that dt equals dy, or can be treated as dy? Do we treat y as t? $\endgroup$ – Reinhild Van Rosenú Jul 7 '15 at 14:00
5
$\begingroup$

you consider $y$ as a function of $x$ then $\frac{dy}{dx}= y'(x)$

to compute your integral just consider the change of variable formula:

$$\int n(y(x))y'(x)\, dx = \int n(y) dy $$

$\endgroup$

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.