# Existence-Uniqueness Theorem for First-Order Differential Equation

From ProofWiki

## Contents

## 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.

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

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle -P(x) (be^{-A(x)} + e^{-A(x)}\int_a^x Q(t) e^{A(t)}dt) + Q(x)\) | \(\displaystyle \) | \(\displaystyle \) | by factoring | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle -P(x) f(x) + Q(x)\) | \(\displaystyle \) | \(\displaystyle \) | by Substitution |

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

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle f(a)\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle be^{-A(a)} + e^{-A(a)} \int_a^a Q(t) e^{A(t)}dt\) | \(\displaystyle \) | \(\displaystyle \) | by Substitution of $a$ into $f(x)$ | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle be^0 + e^0 \cdot 0\) | \(\displaystyle \) | \(\displaystyle \) | by Integral on Zero Interval | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle b\) | \(\displaystyle \) | \(\displaystyle \) | Simplification using arithmetic |

### Uniqueness

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

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

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle e^{A(x)} Q(x)\) | \(\displaystyle \) | \(\displaystyle \) | 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

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

$\blacksquare$

## Sources

- Tom M. Apostol:
*Calculus Volume 1*(1967): $\S 8.3$