What is the solution of this Ordinary Differential Equation?

Question

How can I find $y(x)$ that solve this:
$$ x(y^2 - 6) - (x^2 - 8)yy' = 0$$ $$ y(\sqrt{8}) = -\sqrt{6} $$

Answer 1

We separate the variables, in $$x(y^2 - 6) - (x^2 - 8)yy' = 0$$

to get $$ \frac {ydy}{y^2-6} = \frac {xdx}{x^2-8}$$

Upon integration and solving for $y$ we get $$ y=-\sqrt {6+c(x^2-8)}$$

Where c is an arbitrary constant as long as the radicand stays non-negative.

Answer 2

$$x(y^2 - 6) - (x^2 - 8)yy' = 0$$ $$x(y^2 - 6) - (\frac 12 x^2 - 4)2yy' = 0$$ $$x(y^2 - 6) - (\frac 12 x^2 - 4)(y^2)' = 0$$

you have the form $f'g-fg'$ Divide by $g^2$ and get the form $(\frac f g)'=0$

for $x \pm \ne \sqrt 8$ $$\left (\frac {y^2-6}{\frac 12 x^2-4} \right)'= 0$$

Integrate ...

Answer 3

From the initial equation we draw

$$y(x)=\pm\sqrt6\iff x=\sqrt8\lor y'(x)=0.$$

Hence,

$$y(x)=-\sqrt6$$ fulfills the conditions.

We also draw

$$x=0\iff y(x)=0\lor y'(x)=0$$

and

$$x=\pm\sqrt8\iff y(x)=\pm\sqrt 6.$$

Now, assuming that $x^2\ne8\land y^2(x)\ne 6$, the integration of the separable equation yields

$$y^2(x)-6=C(x^2-8).$$

Finally we have the solutions

• $y(x)=-\sqrt6$,

• $y(x)=\begin{cases}x=\sqrt8\to -\sqrt6,\\x\ne\sqrt8\to \pm\sqrt{6+C(x^2-8)}\end{cases}$, where the positive sign must be rejected to ensure differentiability.

This is summarized by

$$y(x)=-\sqrt{6+C(x^2-8)},$$ a family of half-hyperbolas and -ellipses with the special cases

$$y(x)=-\sqrt6, \\y(x)=-\frac{\sqrt3}2|x|.$$