# How do you solve the initial-value problem y'=sinx/siny where y(0)=π/4?

Mar 14, 2015

You can separate yuor variables $x$ and $y$ and integrate:

Mar 14, 2015

The answer is: $y = \arccos \left(\cos x - 1 + \frac{\sqrt{2}}{2}\right)$.

$y ' = \sin \frac{x}{\sin} y \Rightarrow \sin y \mathrm{dy} = \sin x \mathrm{dx} \Rightarrow \int \sin y \mathrm{dy} = \int \sin x \mathrm{dx} \Rightarrow$

$- \cos y = - \cos x + c$ and now let's find $c$;

$- \cos \left(\frac{\pi}{4}\right) = - \cos 0 + c \Rightarrow - \frac{\sqrt{2}}{2} = - 1 + c \Rightarrow$

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

The solution is:

$- \cos y = - \cos x + 1 - \frac{\sqrt{2}}{2} \Rightarrow \cos y = \cos x - 1 + \frac{\sqrt{2}}{2}$

$y = \arccos \left(\cos x - 1 + \frac{\sqrt{2}}{2}\right)$.