We can solve this using trig. Use the Law of Cosines (III below) to identify the wind's speed. Use the Law of Sines (IV) to solve for the wind's angle relative to the heading. (Beware that law of Sines will often produce two values and Excel, for instance, defaults to the acute one, which would not be right in this case.)
This assumes that you already have the distance and course between start lat/lon and end lat/lon. This could get really tricky if you are trying to solve this over a long distance where the spherical nature of the earth means that your ground track and heading a changing continuously throughout the flight. On a flight from LAX to JFK you will depart on a track of 66° and will be on a track of 93° when you arrive.
That makes wind calculation over a long flight meaningless... Wind correction is really a zonal phenomenon meaning it's only useful to solve in a specific small segment of a flight. And thus we skip the spherical coordinate complexity in this illustration.
I'll use my smartboard and avoid an hour spent in MathJax.
