derivativesintegration
Often while solving an equation of the form
$$ f(y) \frac{dy}{dx} = g(x) $$
we integrate on both sides to obtain
$$
\int f(y) \frac{dy}{dx} \, dx = \int g(x) \, dx.
$$
We normally write it as
$$
\int f(y) \, dy = \int g(x) \, dx.
$$
But I don't understand how we replace $ \int f(y) \frac{dy}{dx} \, dx $ with $ \int f(y) \, dy $ on the LHS.
Is there a way to show that
$$
\int f(y) \frac{dy}{dx} \, dx = \int f(y) \, dy
$$
The derivative $\frac{dy}{dx}$ looks like it's a fraction, and in many cases behaves like it was a fraction, but it's important to remember that...
The trick you have been thought for solving first order ODEs simply does not extend to higher order ODEs. This is a common misconception and it's important to learn the limitations of this trick.
To give you an answer beyond "this simply does not work" let's go through the reason why it works to "multiply by ${\rm d}t$" for first order ODEs and then show what happens for a second order ODE.
Given $$\frac{dy}{dt} = y$$ we divide by $y$, multiply by $dt$ and integrate to get $$\int \frac{{\rm d}y}{y} = \int {\rm d}t \implies \log(y) = t + c\implies y = C e^t$$ The rigorous way of doing this is to divide the ODE by $y$ (assuming $y\not=0$) and integrate the ODE using the fundamental theorem of calculus and the chain rule on the left hand side which allows us to write $$\int \frac{1}{y}\frac{dy}{dt}{\rm d}t = \int \frac{d\log y}{dt}{\rm d}t = \log(y) + C$$ which is the same we would get if treated $\frac{dy}{dt}$ as it was a fraction. Now if we try to do the same procedure (the rigorous one) on the second order ODE $\frac{d^2y}{dt^2} = y$ then we get $$\int \frac{1}{y}\frac{d^2y}{dt^2}{\rm d}t = \int {\rm d}t \implies \int \frac{d(y')}{y} = t + c$$ Now we see the problem here. In order to solve the integral $\int \frac{dy'}{y}$ we would need to know $y$ as a function of $y' \equiv \frac{dy}{dt}$. But this is not known to us unless we already know the solution we are trying to find. There is simply not enough information about $y$ available to us at this stage to be able to solve this integral which shows that this way of solving the ODE does not work like it did in the first order case.
There are many other examples of where treating the derivative as a fraction fails. One that comes to mind comes from the implicit function theorem where we have an equation $F(x,y(x)) = 0$ and can deduce $\frac{dy}{dx} = -\frac{\partial F/\partial x}{\partial F/\partial y}$ (notice the minus sign which is not there is we treat the right hand side derivatives as fractions). For more information see for example When not to treat $dy/dx$ as a fraction in single-variable calculus? and Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio?
Best Answer
This is just integration by substitution. Note for $$ \int f(y)y' dx $$ we can set ${u = y(x)}$ to obtain $$ \rightarrow \int f(u)du $$ but ${u=y}$, so $$ \rightarrow \int f(y)dy $$