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

Question

#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?

1122 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#