Has anyone done 18.03 course? I am having trouble understanding the relation between the isocline and the second derivative test. In Recitation 2, the solutions say that the isocline m=1/2 is a global min, but I am confused as to how they got to that conclusion. Thanks!

Question

anonymous · Answer

I watched that whole class

anonymous · Answer

They set slope to some number then solve for isocline

anonymous · Answer

Thanks!  My question was geared more towards how they came to the conclusion that a certain isocline is the global minimum for the function. $$y \prime = x-2y$$ leads to $$y \prime \prime = 1-2y \prime$$ and since $$y \prime = 0 $$ at critiical points, I can see that the y'' is positive and therefore a min. However, I don't see how they made it a global min instead of just a local min.

anonymous · Answer

for local min , it needs to be in boundary

anonymous · Answer

http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/recitations/MIT18_03S10_rec_02_sol.pdf I'm referring to number 5. I don't understand the explanation past "However" in the last paragraph. :(