ODE, and Questioning Method I have a question about the method used to answer this question:

IVP, and ODE problem
$$\frac{dy}{dx}=ye^{-x^2} \ \ \ \ \ \ y(4)=1 $$
\begin{align}
\frac{dy}y&=\int e^{-x^2}dx \\
\ln(y)&=\int(e^{-x^2})dx \\
\end{align}

Then this is where I get confused the work suddenly jumps to the following:

\begin{align}
\frac{dy}y&=e^{-x^2}dx \\
\frac{1}{y}\frac{dy}{dt}dt&=e^{-x^2}dx \\
\int_4^x\frac{1}y\frac{dy}{dt}dt&=\int_4^xe^{-t^2}dt \\
\ln(\lvert(y(t)\rvert)\Bigg\vert_4^x&=\int_4^xe^{-t^2}dt
\end{align}

The final answer is the following:
$$y=e^{\int_4^xe^{-t^2}dt}$$
My Question
My question is could someone explain the use of the dummy variables, and how the method actually works? Because I am confused about the final answer to the ODE.
 A: You can get for any function $g(y)$ and differentiable function $y(x)$ the formula
$$
g(y(x))-g(y(4))=\int_4^xg'(y(s))y'(s)\,ds
$$
by the fundamental theorem and the chain rule of differentiation. Now insert the ODE $y'=ye^{-x^2}$ of which $y(x)$ is a solution,
$$
g(y(x))-g(y(4))=\int_4^xg'(y(s))y(s)e^{-s^2}\,ds.
$$
This is in general useless, as the right side still contains the unknown solution function $y$. However in the special case $g'(y(s))y(s)=1$ the right side reduces to a simple quadrature, the integral of a known function. This condition implies that the function $y$ does not change the sign of its value inside the considered interval. As $y(4)=1$, the sign is positive.
To achieve this, demand more generally $g'(y)=\frac1y$. As this is a well-known integration task, one easily finds $g(y)=\ln|y|$. (plus arbitrary integration constants that cancel in the next step.)
Now insert that into the first formula to finally get
$$
\ln y(x)-\underbrace{\ln y(4)}_{=0} =\int_4^xe^{-s^2}\,ds.
$$
All the other formulas and calculation paths are just short-hands, known as "method of separation of variables", to replace the reference to substitution and chain rules with a more intuitive computation method.
A: Continue with $\frac {dy}y=e^{-x^2}dx $ and rewrite it as
$$d(\ln y(t))=e^{-t^2}dt $$
Then, integrate over the limits $[4,x]$,
$$\int_4^x d(\ln y(t)) = \int_4^x e^{-t^2}dt $$
$$\ln y(x) - \ln y(4) = \int_4^x e^{-t^2}dt $$
Plug in the initial condition $y(4) = 1$ to get
$$\ln y(x)= \int_4^x e^{-t^2}dt $$
or,
$$y(x)=e^{\int_4^xe^{-t^2}dt}$$
