Question

What is the ifferential equation of the family of hyperbolas: `x^2/a^2 + y^2/b^2 = 1`?

`x^2/a^2 + y^2/b^2 = 1`

Diff. w.r.t. x,
`2x/a^2-2y/b^2dy/dx=0`

`x/a^2-y/b^2dy/dx=0`

`y/b^2dy/dx=x/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?

Answer 1

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

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

Explanation:

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

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