Question

Question #cd49c

Answer 1

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)`