First Order ODE/x dy = k y dx

Theorem

The first order ODE:

$(1): \quad x \rd y = k y \rd x$

has the general solution:

$y = C x^k$

Proof

$(1)$ can be expressed as:

\(\ds x \rd y\)

\(=\)

\(\ds k y \rd x\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds k \int \dfrac {\d x} x\)

Solutions to Separable Differential Equation

\(\ds \leadsto \ \ \)

\(\ds \ln y\)

\(=\)

\(\ds k \ln x + \ln C\)

\(\ds \)

\(=\)

\(\ds \map \ln {C x^k}\)

\(\ds y\)

\(=\)

\(\ds C x^k\)

$\blacksquare$