# Definition

An *ordinary differential equation* (ODE) contains one or more derivatives of
an unknown function of one variable.

\begin{align}
&y' = \cos x \\
&y'' + 9y = 0 \\
&x^2 y''' y' + 2e^x y^4 = (x^2 + 2)y^2
\end{align}

The *order* of an ODE is its highest derivative. First-order ODEs in *implicit*
form look like:

\begin{equation}
F(x, y, y') = 0
\end{equation}

and in *explicit* form look like:

\begin{equation}
y' = f(x, y)
\end{equation}

When do we see first-order ODEs in chemical engineering?

- Unsteady mass/energy balances
- Chemical kinetics

## Direction fields and initial value problems

ODEs have families of solutions due to integration constants. The direction
field (or slope field) is obtained by plotting the slope at (*x*, *y*) from
$y' = f(x, y)$, and it shows how a point will evolve.

````{example} Direction field
Given the first-order ODE

\begin{equation}
y' = xy \quad y(0) = 1
\end{equation}

Draw the slope field, then verify that

\begin{equation}
y = c e^{x^2 / 2}
\end{equation}

is a general solution.

---

The direction field is

```{image} ./_images/directionfield.png
:alt: Directional field diagram
:width: 300px
:align: center
```

Now, let's verify the proposed solution. The first derivative of *y* is

\begin{equation}
y' = c x e^{x^2/2}
\end{equation}

Substituting *y* and $y'$ in the differential equation

\begin{align}
y' &= x y \\
c x e^{x^2 / 2} &= x (c e^{x^2 / 2})
\end{align}

shows this is indeed a solution.
````

To obtain a specific solution, an *initial condition* specifying a point
$y(x_0) = y_0$ is needed. A first-order ODE with an initial condition is called
an *initial value problem*.

```{example} Applying initial condition
For the first-order ODE above, find the solution when $y(0) = 1$.

---

Substitute $x = 0$ and $y = 1$ into the general solution $y = c e^{-x^2/2}$
to determine $c$:

\begin{equation}
1 = y(0) = c e^0 = c
\end{equation}

This gives the solution

\begin{equation}
y = e^{x^2/2}
\end{equation}
```