Question 2 Not yet answered Marked out of 100 P Flag question In order to solve sin 2zdz+cos3ydy = 0 with y(/2) = x/3 we first write cos 3ydy=-sin 2zdr and sin 3y=cos 22 +0 Setting and x-1/2 we find that It follows that 1 1 sin 3y=cos 27+=cos (2). We can solve for Indeed, y=A- arcsin (3 cosz) where We can also solve this equation by using another method. We let and It follows from that the equation is Then we integrate After differentiating the last expression we obtain that Hence, we get where cis equal to again. integrate A=m/3 to get Y sin(2x) 3/4 113 cos(2x) 3+ M = sin 2x N = cos 3y. My = N with respect to x to get -- cos 2z+ C(y). 2 C(y) = sin 3y. 31 2 y=m/3 sin cos2x = c differentiate Acm/2 1/2 F-1/2 exact