# Existence-Uniqueness Theorem for Homogeneous First-Order Differential Equation

## Theorem

Let $P \left({x}\right)$ be a continuous function on an open interval $I \subseteq \R$.

Let $a \in I$.

Let $b \in \R$.

Let $f \left({x}\right) = y$ be a function satisfying the differential equation:

$y' + P \left({x}\right) y = 0$

and the initial condition:

$f \left({a}\right) = b$

Then there exists a unique function satisfying these initial conditions on the interval $I$.

That function takes the form:

$f \left({x}\right) = b e^{-A \left({x}\right)}$

where:

$\displaystyle A \left({x}\right) = -\int_a^x P \left({t}\right) \mathrm d t$

## Proof

### Existence

Differentiating $f \left({x}\right) = b e^{-A \left({x}\right)}$ with respect to $x$:

 $\displaystyle f' \left({x}\right)$ $=$ $\displaystyle b e^{-A \left({x}\right)} \cdot \left({-A' \left({x}\right)}\right)$ $\displaystyle$ $=$ $\displaystyle -b P \left({x}\right) e^{-A \left({x}\right)}$ $\displaystyle$ $=$ $\displaystyle -P \left({x}\right) f \left({x}\right)$

So thedifferential equation becomes:

$f' \left({x}\right) + P \left({x}\right) f \left({x}\right) = -P \left({x}\right) f \left({x}\right) + P \left({x}\right) f \left({x}\right) = 0$

For the initial condition:

 $\displaystyle f \left({a}\right)$ $=$ $\displaystyle b e^{-A \left({a}\right)}$ $\displaystyle$ $=$ $\displaystyle b e^{-\int_a^a P \left({x}\right) \ \mathrm d x}$ $\displaystyle$ $=$ $\displaystyle b e^0$ $\displaystyle$ $=$ $\displaystyle b$

Thus such a function exists satisfying the conditions.

$\Box$

### Uniqueness

Suppose that $f$ is a function satisfying the initial conditions.

Let $g \left({x}\right) = f \left({x}\right) e^{A \left({x}\right)}$.

By the Product Rule:

 $\displaystyle g' \left({x}\right)$ $=$ $\displaystyle f' \left({x}\right) e^{A \left({x}\right)} + f \left({x}\right) e^{A \left({x}\right)}\cdot A' \left({x}\right)$ $\displaystyle$ $=$ $\displaystyle e^{A \left({x}\right)} (f' \left({x}\right) + P \left({x}\right) f \left({x}\right))$ $\displaystyle$ $=$ $\displaystyle 0$

So $g \left({x}\right)$ must be constant.

Therefore:

$g \left({x}\right) = g \left({a}\right) = f \left({a}\right) e^{A \left({a}\right)} = f \left({a}\right) = b$

From this, we conclude that:

$f \left({x}\right) = g \left({x}\right) e^{-A \left({x}\right)} = b e^{-A \left({x}\right)}$

$\blacksquare$