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:


Solution-Curves-x+yy'=0.png


Sources