First Order ODE/dy = k y dx/Proof 2
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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
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.
$\blacksquare$