Differential equations question 10 pts

Consider the differential equation : dy/dt = t + cosy

1) A solution curve passes through the point (-1, pi). What is its slope at this point?
I got the slope to be -2.

2) Show that the second derivative of every solution satisfies: d^(2)y/dt^2 = 1 - tsiny - 0.5sin2y.

3) A solution curve passes through (0, pi/2). Prove this curve has a relative minimum at (0,pi/2).

Show work for any answers! I will pick a best answer, thanks.

1) Let t = -1 and y = pi.

dy/dt = -1 + cos pi = -1 + -1 = -2

2)
d^2 y/dt^2 = 1 - sin y * dy/dt
d^2 y/dt^2 = 1 - sin y * (-t + cos y)
d^2 y/dt^2 = 1 + t sin y - sin y cos y

Note that sin 2y = 2 sin y cos y so that (1/2) sin 2y = sin y cos y.

d^2 y/dt^2 = 1 + t sin y - (1/2) sin y cos y

3) Remember in calculus that f has a minimum at a if f '' (a) > 0. To show that the curve has a minmum at (0, pi/2) we just need to show that d^2 y/dt^2 > 0 at (0, pi/2).

d^2 y/dt^2 = 1 + 0 sin pi/2 - (1/2) sin pi/2 cos pi/2 = 1 + 0*1 - (1/2)*1*0 = 1 > 0.

1) Correct.
2) d²y/dt² = 1 - sin(y)(dy/dt)... hmm do we know dy/dt?
3) Is that point critical, and what's the second derivative test tell us?