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
\(\ds \dfrac {\d y} {\d x}\) | \(=\) | \(\ds k y\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \dfrac {\d y} y\) | \(=\) | \(\ds \int k \rd x\) | Solution to Separable Differential Equation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \ln y\) | \(=\) | \(\ds k x + C'\) | 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:
\(\ds y\) | \(=\) | \(\ds \invlaptrans {\frac {\map y 0} {s - k} }\) | Definition of Inverse Laplace Transform | |||||||||||
\(\ds \) | \(=\) | \(\ds \map y 0 \invlaptrans {\laptrans {e^{k x} } }\) | 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 general 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: $(4)$