What is the ifferential equation of the family of hyperbolas: x^2/a^2 + y^2/b^2 = 1x2a2+y2b2=1?

x^2/a^2 + y^2/b^2 = 1x2a2+y2b2=1

Diff. w.r.t. x,
2x/a^2-2y/b^2dy/dx=02xa22yb2dydx=0

x/a^2-y/b^2dy/dx=0xa2yb2dydx=0

y/b^2dy/dx=x/a^2yb2dydx=xa2

y/xdy/dx=b^2/a^2yxdydx=b2a2

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?

2 Answers
Feb 16, 2018

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 :)

Feb 16, 2018

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