In separable differential equations of the form:
$ M(x,y) dx + N(x,y) dy =0$
We can separate variables and then integrate both sides to get the solution $y = f(x)$ to the differential equation.
In such an equation, when integrating, we cannot consider $y$ constant because what we're looking for is a solution curve in which $y$ is dependent on $x$ which has slope $ \frac {dy}{dx}$.
On the other hand, if we have an exact ( lets just assume all the conditions are met for it to be exact) equation of the same form,
$ M(x,y) dx + N(x,y) dy =0$
What we do is try to find a function $ \psi(x,y) = c$ that satisfies the exact differential equation, but the difference here ( and that's where I'm confused ) is that when we integrate for example,
$ M(x,y) = \left(\frac{\partial \psi}{\partial x}\right)$
When we integrate that last expression, we hold $y$ constant! Why are we doing that? Is it because we're solving for an implicit function in $x$ and $y$ [ $ \psi = c]$? But if we're solving for an implicit function then what's the meaning of $ \frac{dy}{dx}$ then? I hope my question was clear, thanks.
Best Answer
We hold $y$ constant because we are solving for differential equation $M(x,y)=\psi_x(x,y) = \frac{\partial\psi}{\partial x}$ where $\psi_x$ is the partial derivative of $\psi$ in respect to $x$. If integrating in respect one variable (in this case, $x$), we treat all other variables as constants, a lesson learned in multivariable calculus.
I also want to stress that the form $M(x,y)dx + N(x,y)dy = 0$ does not give the best picture of what is going on. Another way to think about exact equations is in the form $$\psi_x(x,y) + \psi_y(x,y)\frac{dy}{dx} = 0$$ We can work backwards from the form $\psi(x,y) = 0$ to get to here. We begin by treating $y$ as some function of $x$, which we will call $y(x)$. Then using the chain rule from multivariable calculus we can derive the following: \begin{align*} \frac{d}{dx}\psi(x,y(x)) &= \frac{d}{dx}C \\ \psi_x(x,y)+\frac{d\psi}{dy}\frac{dy}{dx} &= 0 \\ \psi_x(x,y) + \psi_y(x,y)\frac{dy}{dx} &= 0 \end{align*}