Existence-Uniqueness Theorem for Homogeneous First-Order Differential Equation

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

Let $a \in I$.

Let $b \in \R$.

Let $f(x)=y$ be a function satisfying the differential equation: $$y' + P(x)y = 0$$ and the initial condition $$f(a)=b$$

Then, there is a unique function satisfying these initial conditions on the interval $I$. That function takes the form $$f(x) = be^{-A(x)}$$ where $$A(x) = -\int_a^x P(t)dt$$

Existence
It is easily verifiable that the function $$f(x) = be^{-A(x)}$$ exists and satisfies the differential equation and initial condition provided $P(x)$ is continuous (allowing the use of the Fundamental Theorem of Calculus). For the differential equation,

$ \begin{align*} f'(x) &= be^{-A(x)}\cdot(-A'(x)) \\ &= -bP(x)e^{-A(x)} \\ &= -P(x)f(x) \end{align*} $

So the differential equation becomes $f'(x) + P(x)f(x) = -P(x)f(x) + P(x)f(x) = 0$.

For the initial condition,

$ \begin{align*} f(a) &= be^{-A(a)} \\ &= be^{-\int_a^a P(x)dx} \\ &= be^0 \\ &= b \end{align*} $

Thus, such a function exists satisfying the conditions.

Uniqueness
Suppose that $f$ is a function satisfying the initial conditions. Let $g(x) = f(x)e^{A(x)}$. Then, by the Product Rule

$ \begin{align*} g'(x) &= f'(x)e^{A(x)} + f(x)e^{A(x)}\cdot A'(x) \\ &= e^{A(x)}(f'(x) + P(x)f(x)) \\ &= 0 \end{align*} $

So $g(x)$ must be constant. Therefore, $g(x) = g(a) = f(a)e^{A(a)} = f(a) = b$. From this, we conclude that $f(x) = g(x)e^{-A(x)} = be^{-A(x)}$.