Y'=ysin^2(x) with initial condition? Help!
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Y'=ysin^2(x) with initial condition? Help!

[From: ] [author: ] [Date: 13-03-06] [Hit: ]
.. Any help would be so wonderful!x = 1/2,......
The initial condition is that x=.5 and y=1 for the above derivative. The problem asks for what y is? ( the original y, that is). I have absolutely no clue how to do this problem, and it isn't in our book... Any help would be so wonderful! :)

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dy/dx = y sin²x

Separate variables:
dy/y = sin²x dx

Integrate both sides
∫ dy/y = ∫ sin²x dx
ln|y| = ∫ (1/2 − 1/2 cos(2x)) dx
ln|y| = 1/2 x − 1/4 sin(2x) + C

Now we use initial condition to find C
x = 1/2, y = 1
1/2 (1/2) − 1/4 sin(1) + C = ln(1)
(1−sin(1))/4 + C = 0
C = −(1−sin(1))/4 = (sin(1)−1)/4

ln|y| = 1/2 x − 1/4 sin(2x) + (sin(1)−1)/4

y = e^[(2x−sin(2x)+sin(1)−1)/4]
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