Quasilinear Differential Equation/Examples/x + y y' = 0/Proof 2

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

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.

$\blacksquare$

Sources

• 1978: Garrett Birkhoff and Gian-Carlo Rota: Ordinary Differential Equations (3rd ed.) ... (previous) ... (next): Chapter $1$ First-Order Differential Equations: $1$ Introduction: Example $1$