O que seria uma equação diferencial parcial de primeira ordem?

The differential equations of the type $y'(x)=f(x,y(x))$ are called first-order equations because they only involve first derivatives. This single equation encompasses many different varieties of differential equations, so we will address only specific cases in this article (namely, when $f(x,y(x))$ has low powers of $y(x)$).

To abbreviate notation, we will denote $y(x)$ by $y$, with the understanding that $y$ depends on $x$. For simplicity, all functions are assumed to be continuous and integrable.

First consider the case where $f(x,y)$ doesn't depend on $y$. In other words, $f(x,y)=g(x)$. Then, the solution to the differential equation $y'=g(x)$ is given by the integral $$\begin{equation} y= \int g(x) dx. \end{equation}$$

Our first case had no powers of $y$ in $f(x,y)$. As you might guess, our next case involves linear terms of $y$ in $f(x,y)$.

We now consider the case where $f(x,y)=g(x)-h(x)y$, which are linear first-order ordinary differential equations. It turns out that, for sufficiently nice $g(x)$ and $h(x)$, these can always be solved!

Note that simply integrating both sides will not suffice , because we would get $$\begin{equation} y=\int g(x) dx - \int h(x) y\text{ } dx. \end{equation}$$

As in algebra, we cannot immediately find $y$ if it appears on both sides.

We want to solve the differential equation \begin{equation} y '=g(x)-h(x)y, \end{equation} which has both a $y'$ and a $y$.

We've seen this before when using the product rule! For example, we know that $$\begin{equation} \frac{d}{dx} \left(x^2y\right)=y\frac{d}{dx}\left(x^2\right) + x^2 \frac{dy}{dx}, \end{equation}$$ which involves both $y$ and $y'$. So the motivation for solving our differential equation is to somehow incorporate a reverse product rule. If we can bundle the $y$ and $y'$ terms as the derivative of a product, we can perhaps simplify our problem.

Well, the easiest way to combine the $y$ and $y'$ terms is to put them on the same side:

\begin{equation} y'+h(x)y=g(x). \end{equation}

Let's remind ourselves of the product rule: $\frac{d}{dx} \left(p(x) y \right)= p'(x) y + p(x) y'$. Notice that both the $y$ and the $y'$ terms in the product rule have functions of $x$ as coefficients. In our differential equation, the $y'$ term seems pretty lonely. Let's try to fix that by multiplying by a function of $x$. Since we don't know what such function will be best, let's keep everything as general as possible and multiply by a function $\mu(x)$. We obtain:

\begin{equation} \mu(x) y'+\mu(x)h(x)y=\mu(x)g(x), \end{equation}

and our goal is to collapse the left-hand side using the product rule. Supposing that the left-hand side can be written as $\frac{d}{dx} \left(\mu(x) y \right)$, equate the coefficients of the $y$ and $y'$ terms.

By the product rule, we have that $$\begin{equation} \mu'(x)=\mu(x)h(x) \rightarrow \int \frac{\mu'(x)}{\mu(x)}=\int h(x) \rightarrow \mu(x)=e^{\int h(x) dx}, \end{equation}$$

by separation.

We have found a good definition for $\mu(x)$. Letting it be the above, we have simplified our differential equation to \begin{equation} \frac{d}{dx} \left(\mu(x)y\right)=\mu(x)g(x). \end{equation}

Now the right hand side is only a function of $x$ and the left-hand side is the derivative of a function. We can thus apply the same idea as in the first case: just integrate both sides!

We obtain that $$\begin{equation} y=\frac{1}{\mu(x)} \int \mu(x)g(x) dx , \end{equation}$$

where $\mu(x)$ is defined as above. As long as $g(x)$ and $h(x)$ are integrable, we're done! The function $\mu(x)$ is called an integration factor.