Question

Find all solutions of (sin x)(sin y)+(cos y)y'=0.

Answer 1

@SithsAndGiggles

Answer 2

Looks separable, doesn't it?

Answer 3

\[\begin{align*}
\sin x\sin y+\cos y~y'&=0\\
\cos y~dy&=-\sin x\sin y~dx\\
\cot y~dy&=-\sin x~dx
\end{align*}\]

Answer 4

Wait a minute. Let me work it out.

Answer 5

If you're not familiar with the integral result of \(\displaystyle\int\cot x~dx\), consider it as \(\displaystyle\int\frac{\cos x}{\sin x}~dx\), then use a substitution.

Answer 6

I'm familiar, it's ln abs(sin y)=cos x+C, right? But how do I solve for y?

Answer 7

Yes that's correct. Take the exponential of both sides:
\[\large e^{\ln|\sin y|}=e^{\cos x+C}~~\iff~~\sin y=e^Ce^{\cos x}~~\iff~~\sin y=Ce^{\cos x}\]
Then take the inverse sine.

Answer 8

so the answer is y=sin^-1 (ce^cos x), right?

Answer 9

Yes I think so. Usually, an implicit solution is acceptable. I would have stopped at \(\sin y=\cdots\), but stick to your instructions.

Answer 10

Okay, thank you!

Answer 11

You're welcome! Checking with Wolfram: http://www.wolframalpha.com/input/?i=Sin%5Bx%5D*Sin%5By%5D%2BCos%5By%5D*y%27%3D0

They skip the step where you have \(e^C=C\), but it's still right.