# How do I solve the initial-value problem: y’=sinxsiny;y(0)=π4�′=sin�sin�;�(0)=π4

First of all, let’s write y‘�′ as dydx����. The expression becomes

dydx=sin(x)sin(y)����=sin(�)sin(�)

Multiply both sides for sin(y)dxsin(�)�� and obtain

sin(y) dy=sin(x) dxsin(�) ��=sin(�) ��

integrating, one has

cos(y)=cos(x)+ccos(�)=cos(�)+�

and thus

y=cos−1(cos(x)+c)�=cos-1(cos(�)+�)

This is the general solution of the problem, and we can fix the constant c�, given the condition y(0)=π4�(0)=�4. In fact,

y(0)=cos−1(cos(0)+c)=cos−1(1+c)=π4�(0)=cos-1(cos(0)+�)=cos-1(1+�)=�4

which means

1+c=12–√1+�=12, and finally

c=12–√−1�=12-1.

Your solution is thus

y=cos−1(cos(x)+12–√−1)