Quasilinear Differential Equation/Examples/x + y y' = 0/Explicit Solution
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Theorem
The first order quasilinear ordinary differential equation:
- $x + y y' = 0$
has a general solution which can be expressed explicitly as:
- $y = \pm \sqrt {C - x^2}$
over the domain:
- $-\sqrt C < x < \sqrt C$
Proof
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:
- $-\sqrt C < x < \sqrt C$
$\blacksquare$
Sources
- 1978: Garrett Birkhoff and Gian-Carlo Rota: Ordinary Differential Equations (3rd ed.) ... (previous) ... (next): Chapter $1$ First-Order Differential Equations: $1$ Introduction: Example $1$