First Order ODE/y dx + x dy = 0

Theorem

The first order ODE:

$(1): \quad y \rd x + x \rd y = 0$

has the general solution:

$x y = C$

Proof

$(1)$ can be expressed as:

\(\ds x \rd y\)

\(=\)

\(\ds -y \rd x\)

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds C x^{-1}\)

First Order ODE: $x \rd y = k y \rd x$

\(\ds \leadsto \ \ \)

\(\ds x y\)

\(=\)

\(\ds C\)

multiplying through by $x$

$\blacksquare$