Well it depends. what the problem


so the ODE is indeed exact and there is a solution of the form
. We have




With
, we have

so

√196s² = √196 times √s²
s² is the square of 's'
196 is the square of 14
So both can easily come out of the radical.
√196s² = <u>14s</u>
In the default window of a graphing calculator, there is only one intersection that you see.
However, if you zoom out, you will see that they are 3 intersections to the pair of equations.