# 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 real functions on some open interval $I \subseteq \R$.

Let $a \in I$.

Let $b \in \R$.

There exists a unique function $\map f x = y$ on $I$ that satisfies the linear first order ordinary differential equation:

$(1): \quad y' + \map P x y = \map Q x$

along with the initial condition:

$\map f a = b$

This function is:

$\ds \map f x = be^{-\map A x} + e^{-\map A x} \int_a^x \map Q t e^{\map A t} \rd t$

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

## Proof

### Existence

Because $P$ and $Q$ are continuous, they are integrable.

Hence we may use the Fundamental Theorem of Calculus.

 $\ds \map {f'} x$ $=$ $\ds -b \map P x e^{-\map A x} - \map P x e^{-\map A x} \int_a^x \map Q t e^{\map A t} \rd t + e^{-\map A x} \map Q x e^{\map A x}$ Product Rule $\ds$ $=$ $\ds -\map P x \paren {b e^{-\map A x} + e^{-\map A x} \int_a^x \map Q t e^{\map A t} \rd t} + \map Q x$ by factoring $\ds$ $=$ $\ds -\map P x \map f x + \map Q x$ by substitution

Therefore:

$\map {f'} x + \map P x \map f x = \map Q x$

For the initial condition:

 $\ds \map f a$ $=$ $\ds b e^{-\map A a} + e^{-\map A a} \int_a^a \map Q t e^{\map A t}dt$ substituting of $a$ into $\map f x$ $\ds$ $=$ $\ds b e^0 + e^0 \cdot 0$ Integral on Zero Interval $\ds$ $=$ $\ds b$ simplification

$\Box$

### Uniqueness

Let $f$ be a solution to $(1)$ which satisfies the initial condition.

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

 $\ds \map {g'} x$ $=$ $\ds \map {f'} x e^{\map A x} + \map P x \map f x e^{\map A x}$ Product Rule $\ds$ $=$ $\ds e^{\map A x} \map Q x$ by Substitution

Moreover:

$\map g a = b$
$\ds \map g x = \int_a^x \map Q t e^{\map A t} \rd t + b$

Furthermore:

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

Therefore, we can conclude that:

$\ds \map f x = b e^{-\map A x} + e^{-\map A x} \int_a^x \map Q t e^{\map A t} \rd t$

$\blacksquare$