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