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.

  • $\begingroup$ @mvw Can you please clarify? $\endgroup$ – Reinhild Van Rosenú Jul 7 '15 at 13:34
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    $\begingroup$ 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
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    $\begingroup$ 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

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:

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


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