Quasilinear Differential Equation/Examples/x + y y' = 0/Proof 1

Theorem

The first order quasilinear ordinary differential equation over the real numbers $\R$:

$x + y y' = 0$

has the general solution:

$x^2 + y^2 = C$

where:

$C > 0$

$y \ne 0$

$x < \size {\sqrt C}$

with the singular point:

$x = y = 0$

Proof

Let us rearrange the equation in question:

\(\ds x + y y'\)

\(=\)

\(\ds 0\)

\(\ds \leadstoandfrom \ \ \)

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

\(=\)

\(\ds -x\)

\(\ds \leadstoandfrom \ \ \)

\(\ds y \rd y\)

\(=\)

\(\ds \paren {-1} x \rd x\)

This is in the form:

$y \rd y = k x \rd x$

where $k = 1$.

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

$y^2 = \paren {-1} x^2 + C$

from which the result follows.

$\blacksquare$