# The Existence/Uniqueness of Solutions to General First Order Differential Equations

Recall from The Existence/Uniqueness of Solutions to First Order Linear Differential Equations page that a linear differential equation $\frac{dy}{dt} + p(t) y = g(t)$ has a unique solution to the initial value problem $y(t_0) = y_0$ provided that both $p(t)$ and $g(t)$ are continuous functions on an open interval $I = (\alpha, \beta)$ that contains $t_0$.

Of course, not all first order differential equations are linear. The following theorem is an extension of that found on the page mentioned above that describes the existence and uniqueness of solutions to initial value problems of nonlinear differential equations.

Theorem 1: Let $f = f(t, y)$ and $\frac{\partial}{\partial y} f(t, y)$ be continuous functions on an open rectangle $I = (\alpha, \beta) \times (\gamma, \delta) = \{ (t, y) : \alpha and let $(t_0, y_0) \in I$. Then for some interval $(t_0 – h, t_0 + h) \subseteq (\alpha, \beta)$ there exists a unique solution $y = \phi (t)$ to the differential equation $\frac{dy}{dt} = f(t, y)$ and satisfies the initial value condition $y(t_0) = y_0$. |

*Note that if the differential equation $\frac{dy}{dt} = f(t, y)$ is linear, then $\frac{dy}{dt} = -p(t)y + g(t)$, and so if $f(t, y) = -p(t)y + g(t)$, then $\frac{\partial}{\partial y} f(t, y) = -p(t)$. Therefore if $\frac{\partial}{\partial y} f(t, y)$ is continuous on $I$ then $p(t)$ is continuous on $(\alpha, \beta)$, and if $f(t, y)$ is continuous on $I$ then $g(t)$ is continuous on $(\alpha, \beta)$, and so a unique solution to $\frac{dy}{dt} = f(t, y)$ is guaranteed.*

*We should also note that if $f = f(t, y)$ is continuous on an open rectangle $I$ and $\frac{\partial}{\partial y} f(t, y)$ is NOT, then we are still guaranteed the existence of a solution, however, the lack of continuity of $\frac{\partial f}{\partial y}$ does not guarantee us the UNIQUENESS of our solution.*

Let’s now look at some examples regarding Theorem 1.

## Example 1

**Using Theorem 1, determine the largest interval containing a unique solution to the initial value problem $t(t – 4) \frac{dy}{dt} + y = 0$ with the initial condition $y(2) = 1$.**

We first rewrite our differential equation by dividing by $t(t-4)$ to get:

(1)

Then we have that $\frac{dy}{dt} = f(t, y) = -\frac{1}{t(t – 4)}$. This function is continuous for $t \neq 0$ and $t \neq 4$. Furthermore, $\frac{\partial f}{\partial y} = 0$ which is continuous everywhere. Therefore, the possible intervals for the unique solution $\phi (t)$ to the given initial value problem are $(- \infty, 0)$, $(0, 4)$ or $(4, \infty)$.

The initial condition $y(2) = 1$ species that $2$ must be contained in the interval, and so $(0, 4)$ is the largest interval containing the unique solution to this initial value problem.

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