Quasilinear Differential Equation/Examples/x + y y' = 0/Explicit Solution

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 $1$