Check my work please: First order ODE

cos(y)sin(t) dy/dt = cost(t)sin(y)

gives is cos(y)/sin(y) dy = cos(t)/sin(t) dt

integrating both sides of cot(y)dy = cot(t)dt we've got:
ln(siny) = ln(sint) + c , at this point we exponentiate both sides and this is where I need help.

Question

Check my work please: First order ODE

cos(y)sin(t) dy/dt = cost(t)sin(y)

gives is cos(y)/sin(y) dy = cos(t)/sin(t) dt

integrating both sides of cot(y)dy = cot(t)dt we've got: 
ln(siny) = ln(sint) + c   ,  at this point we exponentiate both sides and this is where I need help.

anonymous · Answer

I get sin(y) = sin(t) + c

and then arcsin both sides. This gives me y = arcsin(sin(t)+c)

but that doesn't match up with the answer. Does anyone see where I go wrong?

myininaya · Answer

Did you mean to put that extra t in the problem?

myininaya · Answer

Assuming that extra t isn't there, that isn't suppose to be +c.

myininaya · Answer

Hint-->e^(a+b)=e^a*e^b

anonymous · Answer

hmm, so, when I exponentiate I should sin(t)*e^c , and we can just call that new costant something like c', so the right side is then arcsin(sint*c')

anonymous · Answer

thwarted by algebra as usual haha, thanks for looking at this for me!

myininaya · Answer

Yep. Was that the answer you were trying to match in your book?

anonymous · Answer

indeed!

myininaya · Answer

Ok, cool stuff! :)