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**Unformatted text preview: **Math 415.01 - Dr. Huseyin Coskun - Solutions to HW 2- 10:30 recitation (Vutha) January 19, 2011 The homework will be graded for 25 pts. (Completion: 5 pts , Graded questions: 20 pts) (2.2.19) The given equation sin(2 x ) dx + cos(3 y ) dy = 0 is seperable. Seperating variables, we can write it as- cos(3 y ) dy = sin(2 x ) dx Integrating both sides, we get- 1 3 sin(3 y ) =- 1 2 cos(2 x ) + C From the given initial condition, we know that y = π 3 when x = π 2 . ie.- 1 3 sin( π ) =- 1 2 cos( π ) + C = ⇒ C =- 1 2 Therefore we find- 1 3 sin(3 y ) =- 1 2 cos(2 x ) + 1 2 = ⇒ sin(3 y ) = 3 2 (cos(2 x ) + 1) = 3(cos 2 ( x )) (Using the double angle formula) Applying arcsin on both sides, we get two solutions : y = arcsin(3 cos 2 ( x )) 3 and y = π- arcsin(3 cos 2 ( x )) 3 Out of these only the second one satisfies the initial value. The domain on the solution depends on the domain of arcsin. In fact, we will need- 1 < 3 cos 2 ( x ) < 1 which happens only when- 1 √ 3 < cos( x ) < 1 √ 3 1 (2.3.22) (a) The differential equation for the upward motion is(2....

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