# First Order ODE/dy = k y dx

## Theorem

Let $k \in \R$ be a real number.

The first order ODE:

$\dfrac {\d y} {\d x} = k y$

has the general solution:

$y = C e^{k x}$

## Proof 1

 $\ds \dfrac {\d y} {\d x}$ $=$ $\ds k y$ $\ds \leadsto \ \$ $\ds \int \dfrac {\d y} y$ $=$ $\ds \int k \rd x$ Separation of Variables $\ds \leadsto \ \$ $\ds \ln y$ $=$ $\ds k x + C'$ Primitive of Reciprocal, Primitive of Constant $\ds \leadsto \ \$ $\ds y$ $=$ $\ds e^{k x + C'}$ $\ds$ $=$ $\ds e^{k x} e^{C'}$ $\ds$ $=$ $\ds C e^{k x}$ putting $C = e^{C'}$

$\blacksquare$

## Proof 2

Write the differential equation as:

$y' - k y = 0$

Taking Laplace transforms:

$\laptrans {y' - k y} = \laptrans 0$

From Laplace Transform of Constant Mapping, we have:

$\laptrans 0 = 0$

We also have:

 $\ds \laptrans {y' - k y}$ $=$ $\ds \laptrans {y'} - k \laptrans y$ Linear Combination of Laplace Transforms $\ds$ $=$ $\ds s \laptrans y - \map y 0 - k \laptrans y$ Laplace Transform of Derivative

So:

$\paren {s - k} \laptrans y = \map y 0$

Giving:

$\laptrans y = \dfrac {\map y 0} {s - k}$

So:

 $\ds y$ $=$ $\ds \map { {\mathcal L}^{-1} } {\frac {\map y 0} {s - k} }$ Definition of Inverse Laplace Transform $\ds$ $=$ $\ds \map y 0 \map { {\mathcal L}^{-1} } {\laptrans {e^{k x} } }$ Linear Combination of Laplace Transforms, Laplace Transform of Exponential $\ds$ $=$ $\ds \map y 0 e^{k x}$ Definition of Inverse Laplace Transform

Setting $C = \map y 0$ gives the result.

$\blacksquare$

## Proof 3

 $\ds \dfrac {\d y} {\d x}$ $=$ $\ds k y$ $\ds \leadsto \ \$ $\ds \dfrac {\d y} {\d x} - k y$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds e^{-k x} \paren {\dfrac {\d y} {\d x} - k y}$ $=$ $\ds 0$ Solution to Linear First Order Ordinary Differential Equation: $e^{\int -k \rd x} = e^{-k x}$ $\ds \leadsto \ \$ $\ds \map {\dfrac \d {\d x} } {y e^{-k x} }$ $=$ $\ds 0$ Solution to Linear First Order Ordinary Differential Equation continued $\ds \leadsto \ \$ $\ds y e^{-k x}$ $=$ $\ds C$ $\ds \leadsto \ \$ $\ds y$ $=$ $\ds C e^{k x}$

$\blacksquare$

## Also represented as

This first order ODE can also be represented as:

$\dfrac {\d y} {\d x} + k y = 0$

from which the solution is:

$y = C e^{-k x}$