How do I solve the initial-value problem: ##y’=sinx/siny;y(0)=π/4##

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)##

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