8t^2 * y'' + (y')^3 = 8ty' , t > 0
The Attempt at a Solution
I tried using the substitution v = y' to get:
8t^2 * v' + v^3 = 8tv
I rewrote it in the form 8t^2 * dv/dt + v^3 = 8tv, and then moved the v^3 to the other side to get 8t^2 * dv/dt = 8tv - v^3.
I then multiplied both sides by dt to get:
8t^2 dv = (8tv - v^3) dt
Try to get it in exact form.
8t^2 dv - (8tv - v^3) dt = 0
8t^2 dv + (8tv + v^3) dt = 0
But when I test for exactness, it doesn't work, because 16t doesn't equal 3v^2 + 8t. Did I mess up somewhere, or is there some way to continue?