First Order ODE/dy = k y dx/Proof 3

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

\(\ds \dfrac {\d y} {\d x}\)

\(=\)

\(\ds k y\)

\(\ds \leadsto \ \ \)

\(\ds \dfrac {\d y} {\d x} - k y\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds e^{-k x} \paren {\dfrac {\d y} {\d x} - k y}\)

\(=\)

\(\ds 0\)

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

\(\ds \leadsto \ \ \)

\(\ds y e^{-k x}\)

\(=\)

\(\ds C\)

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds C e^{k x}\)

$\blacksquare$

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: $\S 1.2$ The integrating factor