Question #cd49c

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Answer 1 · 2015-01-24T00:08:50

First of all, let's write $y'$ as $dy/dx$ . The expression becomes

$dy/dx = \sin(x)/\sin(y)$

Multiply both sides for $\sin(y) dx$ and obtain

$\sin(y)\ dy = \sin(x)\ dx$

integrating, one has

$\cos(y) = \cos(x)+ c$

and thus

$y = \cos^{-1}(\cos(x)+c)$

This is the general solution of the problem, and we can fix the constant $c$ , given the condition $y(0)=\pi/4$ . In fact,

$y(0)=\cos^{-1}(\cos(0)+c) = \cos^{-1}(1+c)=\pi/4$

which means

$1+c=1/\sqrt{2}$ , and finally

$c=1/\sqrt{2} -1$ .

Your solution is thus

$y = \cos^{-1}(\cos(x)+1/\sqrt{2} -1)$