Quasilinear Differential Equation/Examples/x + y y' = 0/Proof 1
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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.
$\blacksquare$