Question

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

Answer 1

`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)`