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

## Theorem

Let $\map P x$ be a continuous function on an open interval $I \subseteq \R$.

Let $a \in I$.

Let $b \in \R$.

Let $\map f x = y$ be a function satisfying the differential equation:

$y' + \map P x y = 0$

and the initial condition:

$\map f a = b$

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

That function takes the form:

$\map f x = b e^{-\map A x}$

where:

$\ds \map A x = -\int_a^x \map P t \rd t$

## Proof

### Existence

Differentiating $\map f x = b e^{-\map A x}$ with respect to $x$:

 $\ds \map {f'} x$ $=$ $\ds b e^{-\map A x} \cdot \paren {-\map {A'} x}$ $\ds$ $=$ $\ds -b \map P x e^{-\map A x}$ $\ds$ $=$ $\ds -\map P x \map f x$

So thedifferential equation becomes:

$\map {f'} x + \map P x \map f x = -\map P x \map f x + \map P x \map f x = 0$

For the initial condition:

 $\ds \map f a$ $=$ $\ds b e^{-\map A a}$ $\ds$ $=$ $\ds b e^{-\int_a^a \map P x \rd x}$ $\ds$ $=$ $\ds b e^0$ $\ds$ $=$ $\ds b$

Thus such a function exists satisfying the conditions.

$\Box$

### Uniqueness

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

Let $\map g x = \map f x e^{\map A x}$.

By Product Rule:

 $\ds \map {g'} x$ $=$ $\ds \map {f'} x e^{\map A x} + \map f x e^{\map A x} \cdot \map {A'} x$ $\ds$ $=$ $\ds e^{\map A x} \paren {\map {f'} x + \map P x \map f x}$ $\ds$ $=$ $\ds 0$

So $\map g x$ must be constant.

Therefore:

$\map g x = \map g a = \map f a e^{\map A a} = \map f a = b$

From this, we conclude that:

$\map f x = \map g x e^{-\map A x} = b e^{-\map A x}$

$\blacksquare$