Change of variable during integrating So we have the following differential equation: $$\dfrac{dy}{dx} = y^{2/3}$$
So the logical next step would be $$dy\space y^{-2/3} = dx $$
So what I'd now do is simply integrate both sides, but in my text book it says $\int_0^x ds   $ instead of simply $\int dx$. What is the reason for this?
 A: If you have the condition $y\left(0\right)=1
 $, then you can use definite integrals. We have $$dyy^{-2/3}=dx\Rightarrow\int_{1}^{y}u^{-2/3}du=\int_{0}^{x}ds
 $$ and so $$3\left(y^{1/3}-1\right)=x.
 $$ When you use indefinite integral, you have infinite solution; in this case we know the initial condition, so we can use the definite integral.
A: You should know that going from $$\frac{dy}{dx} = h(y)g(x)$$
to $$dy\,h(y) = dx\,g(x)$$
and then taking indefinite integrals "with respect to $y$ on LHS and wrt $x$ on RHS", is something of a mnemonic device. Here's how the technical procedure goes, in your case: $$\begin{align}y'(x) &= y(x)^{2/3} \implies \\[1ex] \implies y'(x) y(x)^{-2/3} &= 1 &(1)\\[1ex] \implies \int_{x_0}^xy'(s)y(s)^{-2/3} \,ds &= \int_{x_0}^x 1\,ds \end{align}$$
We could have integrated over any interval, and the particular letter we choose for the variable of integration doesn't matter. For example, we could have kept the letter $x$ as the argument of $y$ and integrated over $[s_0,s]$, or $[a,b]$ or whatever. The point is that $(1)$ is an equality between functions, so the integral of said functions over any interval must also be equal. The rest is just convenient naming so that we end up with familiar looking expressions.
Let us continue though. On the left use the change of variables $t = y(s)$, with which $dt = y'(s)ds$ (another mnemonic, at least until we speak of differential geometry!). $$\int_{y(x_0)}^{y(x)}t^{-2/3}\,dt = \int_{x_0}^x1\,ds \implies \\ \implies 3\left(y(x)^{1/3}-y(x_0)^{1/3}\right) = x-x_0 \\[1ex] \implies 3y(x)^{1/3} =x-x_0+3y(x_0)^{1/3} $$
If we name the constant value $-x_0+3y(x_0)^{1/3}$ as $A$, for example, then we have that $$y(x) = \frac 19(x+A)^3$$
which is what you would have gotten, had you used the usual procedure of indefinite integration.
It seems your book takes this, the technical approach, with $x_0 = 0$. The advantage of this is that we don't have to determine constants after getting to an expression like the above, rather they are incorporated automatically when we substitute $y(x_0)$.
