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

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

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:

Thus such a function exists satisfying the conditions.

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:

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