What is the ifferential equation of the family of hyperbolas: $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $x^2/a^2 + y^2/b^2 = 1$ ?

Question

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $x^2/a^2 + y^2/b^2 = 1$

Diff. w.r.t. x,
$2 \frac{x}{a^{2}} - 2 \frac{y}{b^{2}} \frac{d y}{d x} = 0$ $2x/a^2-2y/b^2dy/dx=0$

$\frac{x}{a^{2}} - \frac{y}{b^{2}} \frac{d y}{d x} = 0$ $x/a^2-y/b^2dy/dx=0$

$\frac{y}{b^{2}} \frac{d y}{d x} = \frac{x}{a^{2}}$ $y/b^2dy/dx=x/a^2$

$\frac{y}{x} \frac{d y}{d x} = \frac{b^{2}}{a^{2}}$ $y/xdy/dx=b^2/a^2$

or $(yy')/x=b^2/a^2$

Diff. w.r.t. x again.
$(yy''+(y')^2-yy')/x^2=0$

$yy''+(y')^2-yy'=0$

But the answer given in the book is
$xyy''+x(y')^2-yy'=0$

Where is my mistake?

1252 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-16T08:29:00

When u diff. w.r.t. x for the second time,

$(yy')/x=b^2/a^2$

Applying the quotient rule first.
$=>(x(yy')' - yy'(x)')/x^2=0$

now, when applying the product rule to $(yy')'$
$=>(x(yy" + y'y') - yy')/x^2=0$

$=>(x(yy" + y'y') - yy')=0$

$=>xyy" + x(y')^2 - yy'=0$

Hence, this matches the answer in your book :)

Answer 2 · 2018-02-16T08:45:41

My mistake was in differentiating the second time.
$(x(yy''+(y')^2)-yy')/x^2=0$

$(x(yy''+(y')^2)-yy')/x^2=0$

$xyy''+x(y')^2-yy'=0$