Solve the differential equation #xyy' +xyy'= y^2 +1#?

1 Answer
Mar 18, 2017

# y = +-sqrt(Bx -1)#

Explanation:

We have:

# xyy' +xyy'= y^2 +1 #
# :. 2xyy' = y^2 +1 #
# :. y/(y^2 +1 )y' = 1/(2x)#

This is a First Order separable DE, so we can separate the variables to get;

# int \ y/(y^2 +1 ) \ dy = int \ 1/(2x) \ dx#

The RHS is trivial and for the LHS we can use a substitution:
Let #u=y^2+1 => (du)/dy = 2y #, or # int 1/2 ... \ du = int ... y\ dy #

Substituting we get

# int \ (1/2)/(u) \ du = int \ 1/(2x) \ dx#

We can now integrate to get:

# 1/2ln|u| = 1/2 ln |x| + A#
# :. ln|u| = ln |x| + 2A#
# :. ln|u| = ln |x| + lnB# (say)
# :. ln|u| = ln B|x| #
# :. u = Bx #
# :. y^2+1 = Bx #
# :. y^2 = Bx -1#
# :. y = +-sqrt(Bx -1)#