First Order ODEs: Notation and Theorems

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Section 8.2 First Order ODEs: Notation and Theorems

First order ODEs (Section 8.1) are among the simplest. For fairly simple smoothness conditions, a solution is not only guaranteed to exist, but is also guaranteed to be unique. The technical theorem is stated below, but you may want to regard this as a Don't Worry about It theorem. It tells you that you can just go ahead and try to solve it by one of the methods listed in the next couple of sections.

Notation for First Order ODEs.

There are a number of different methods for solving these equations. Which one you may want to use depends on the form of the equation. Listed here are some common forms of first order ODEs:

Standard Form:

\begin{equation} \frac{dy}{dx}=f(x,y)\tag{8.2.1} \end{equation}

You can ALWAYS write a first order ODE in standard form. This form is used primarily in the statement of theorems about existence and uniqueness of solutions.

Separable Form: If you can write

\begin{equation} f(x,y)=h(x)g(y),\tag{8.2.2} \end{equation}

then Eqn(8.2.1) is called separable. It is NOT always possible to do this. See Section 8.3 for a discussion of this method.

Differential Form: If you write

\begin{equation} f(x,y)=-\frac{M(x,y)}{N(x,y)},\tag{8.2.3} \end{equation}

then Eqn(8.2.1) becomes

\begin{equation} M(x,y)\, dx + N(x,y)\, dy =0\tag{8.2.4} \end{equation}

There are many ways to divide \(f(x,y)\) into \(M(x,y)\) and \(N(x,y)\text{.}\) Choose a way that is helpful for the problem at hand. In particular, you can ALWAYS choose \(M(x,y)\) and \(N(x,y)\) so that (8.2.4) is an exact differential and the equation can be solved by simple integration. See Section 8.4 for a discussion of this method.

Theorem 8.2.1. First-Order ODEs: Uniqueness Theorem.

If \(f\) and \(\frac{\partial f}{\partial y}\) are continuous in a rectangle \(\vert x-x_0\vert \le a\text{,}\) \(\vert y-y_0\vert\le b\text{,}\) then there exists an interval about \(x_0\) in which the initial value problem \(y^{\prime}=f(x,y)\text{,}\) \(y(x_0)=y_0\) has a unique solution.