Quasilinear Differential Equation/Examples/x + y y' = 0
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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 1
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.
$\blacksquare$
Proof 2
Observe that from Derivative of Power and Chain Rule for Derivatives:
- $\dfrac {\map \d {x^2 + y^2} } {\d y} = 2 x + 2 y \dfrac {\d y} {\d x}$
Hence:
- $\dfrac {\map \d {x^2 + y^2} } {\d y} = 0$
So from Derivative of Constant:
- $x^2 + y^2 = C$
where $C$ is abitrary.
$\blacksquare$
Explicit Solution
The general solution of $x + y y' = 0$ is implicitly over the real numbers.
We have:
\(\ds x^2 + y^2\) | \(=\) | \(\ds y\) | Quasilinear Differential Equation: $x + y y' = 0$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y^2\) | \(=\) | \(\ds C - x^2\) | where $C - x^2 \ge 0$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \pm \sqrt {C - x^2}\) | where $\size {\sqrt C} \ge x$ |
But when $y = 0$, the derivative $y'$ of the ordinary differential equation $x + y y' = 0$ needs to be infinite for non-zero $x$.
Hence:
- $-\sqrt C < x < \sqrt C$
$\blacksquare$
Solution Curves
Its solution curves can be presented as:
Sources
- 1978: Garrett Birkhoff and Gian-Carlo Rota: Ordinary Differential Equations (3rd ed.) ... (previous) ... (next): Chapter $1$ First-Order Differential Equations: $1$ Introduction: Example $1$