Originally Posted by

**zhupolongjoe** Consider:

e^(-y)(dy/dt)+2 cost t=0

a) Solve it

I did this and got -e^(-y)=-2sint +C, or $\displaystyle \text{y=-ln(2sin t}$$\displaystyle \color{red}\text{)+C} $

b) Does every value of C correspond to a solution curve? If not, which do and which do not and why?

I'm not sure about this....I don't think so, since we have a negative ln....this one doesn't make too much sense to me though since 2 sin t will always be between -2 and 2, hence the only values -ln(2 sin t) can take are between -ln(-2) (which is not real) and -ln(2)...or I guess more appropriately -ln(1) and -ln(2)....

c) Do all the solutions have the same domain? Explain.

Probably not, but not sure.

Thanks!