# Existence-Uniqueness Theorem for First-Order Differential Equation It has been suggested that this page or section be merged into Solution to Linear First Order Ordinary Differential Equation. (Discuss)

## Theorem

Let $P$ and $Q$ be continuous functions on some open interval $I \subseteq \R$.

Let $a \in I$.

Let $b \in \R$.

There is a unique function $f(x)=y$ on $I$ that satisfies the differential equation

$y' + P(x)y = Q(x)$ along with the initial condition
$f(a)=b$

This function is

$\displaystyle f(x) = be^{-A(x)} + e^{-A(x)}\int_a^x Q(t) e^{A(t)} dt$ where $\displaystyle A(x) = \int_a^x P(t)dt$

## Proof

### Existence

If $P$ and $Q$ are continuous, then they are integrable, and we may use the Fundamental Theorem of Calculus on definite integrals involving these functions.

 $\displaystyle f'(x)$ $=$ $\displaystyle -bP(x)e^{-A(x)} - P(x)e^{-A(x)}\int_a^x Q(t) e^{A(t)}dt + e^{-A(x)} Q(x) e^{A(x)}$ by the Product Rule $\displaystyle$ $=$ $\displaystyle -P(x) (be^{-A(x)} + e^{-A(x)}\int_a^x Q(t) e^{A(t)}dt) + Q(x)$ by factoring $\displaystyle$ $=$ $\displaystyle -P(x) f(x) + Q(x)$ by Substitution

Therefore, $f'(x) + P(x) f(x) = Q(x)$. For the initial condition,

 $\displaystyle f(a)$ $=$ $\displaystyle be^{-A(a)} + e^{-A(a)} \int_a^a Q(t) e^{A(t)}dt$ by Substitution of $a$ into $f(x)$ $\displaystyle$ $=$ $\displaystyle be^0 + e^0 \cdot 0$ by Integral on Zero Interval $\displaystyle$ $=$ $\displaystyle b$ Simplification using arithmetic

### Uniqueness

Suppose $f$ is a solution to the differential equation and satisfies the initial condition. Let $g(x) = f(x) e^{A(x)}$.

 $\displaystyle g'(x)$ $=$ $\displaystyle f'(x) e^{A(x)} + P(x) f(x) e^{A(x)}$ by the Product Rule $\displaystyle$ $=$ $\displaystyle e^{A(x)} Q(x)$ by Substitution

Moreover, $g(a) = b$. By the Second Fundamental Theorem of Calculus,

$\displaystyle g(x) = \int_a^x Q(t) e^{A(t)} dt + b$

Furthermore, $\displaystyle f(x) = g(x) e^{-A(x)}$. Therefore, we can conclude that

$\displaystyle f(x) = be^{-A(x)} + e^{-A(x)}\int_a^x Q(t) e^{A(t)} dt$

$\blacksquare$