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

Question

12470 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-03-14T20:52:34

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

Answer 2 · 2015-03-14T21:39:35

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)#.