# First Order ODE/dy = k y dx

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## 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:

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

## Also see

## Sources

- 1958: G.E.H. Reuter:
*Elementary Differential Equations & Operators*... (previous) ... (next): Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 1$. The first order equation: $\S 1.1$ Introduction: $(5)$ - 1972: George F. Simmons:
*Differential Equations*... (previous) ... (next): $1$: The Nature of Differential Equations: $\S 1$: Introduction: $(4)$