Show that all solutions to a 1-order ODE satisfies all solutions to a 2-order ODE???

[malenkylizards]

text of the problem:
Consider the differential equation
\(\displaystyle dy/dx = x + sin y\)
(a) A solution curve passes through the point (1,Pi/2). What is its slope at this point?
(b) Argue that every solution curve is increasing for x > 1.
(c) Show that the second derivative of every solution satisfies
\(\displaystyle d^2y/dx^2=1 + x cos y + 1/2 sin 2y\)
(d) A solution curve passes through (0,0). Prove that this curve has a relative minimum at (0,0).

my attempt at a solution:
I solved (a) and (b) already, to the best of my knowledge.
(a) I simply plugged in the values here, since the slope is equivalent to dy/dx. dy/dx = 1 + sin(pi/2) = 1 + 1 = 2
(b) I figured that if {d^2y}/{dx^2} > 0 when x > 1, then dy/dx would be increasing for all x>1. So I took a partial derivative of dy/dx, and it evaluated to 1. 1 > 0, QED.
(c) I'm not sure how to proceed here. In class, we haven't been taught how to actually find a solution yet, merely check a given solution. So I can assume that finding the set of solutions need not be part of solving this problem!
(d) I am guessing/hoping that once I understand (c), this problem will make sense.

Thanks for your help!

 

[burakaltr]

(a) y' = x + Siny

at x=1;y=pi/2

what am i asked, y'(1)

Plug in values y'(1)=(1+SinPi/2)

(b) y'=x+Siny when x>1 , how does Sine function behave;ie its limits-->Solve

(c) (y')'=right side diffentiated wrt " x " ; You take one more derivative of both sides; for y' you get Y'' on the left; you show right side

(d)f'(0)=0 ; Prove by (c) that f''(0)=0