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

2 Answers
Mar 14, 2015

You can separate yuor variables x and y and integrate:
enter image source here

Mar 14, 2015

The answer is: y=arccos(cosx-1+sqrt2/2).

y'=sinx/sinyrArrsinydy=sinxdxrArrintsinydy=intsinxdxrArr

-cosy=-cosx+c and now let's find c;

-cos(pi/4)=-cos0+crArr-sqrt2/2=-1+crArr

c=1-sqrt2/2.

The solution is:

-cosy=-cosx+1-sqrt2/2rArrcosy=cosx-1+sqrt2/2

y=arccos(cosx-1+sqrt2/2).