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# Solve the initial-value problem $y' = (\sin x) \sin y, y(0) = \pi/2,$ and graph the solution (if your CAS does implicit plots).

## $$\frac{d y}{d x}=\frac{\sin x}{\sin y}, \quad y(0)=\frac{\pi}{2}$$

#### Topics

Differential Equations

### Discussion

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

So when we have a separable differential equation, what we do is we want to get Y on the same side as its derivative and X. On the other side. And also it will be helpful to rewrite Y. Prime. It's dy dx. So we'll do all that right now. First to get wide on the left side by multiplying both sides by side. Y. To get signed Y. Times Y. Prime. Or dy dx is equal to sine X. Now we essentially multiply both sides by dx. And now we can integrate. And on the left side you'll get uh well integral sign is negative coast. So negative coast Y is equal to same deal on this side, negative coast X plus an arbitrary constant. We can call C. And we can multiply both sides by negative one to get rid of those net post two negatives to get coast Y. Coast coast X minus C. And now we need to find what years using the initial value. And we get um I'm just gonna write it here. Um So we're gonna put pile over two in place of Y. And uh zero in place of X. So we get coast of poverty Is equal to the coast of zero -C. Because the power to his wine ah coast of europe, Sorry to coast over two is 0. Coast of zero is 1 and therefore uh C. Is one. So um our answer is coast wise. You pull the coast X minus one. Uh Note that we can't we can't take the uh our coast or inverse coast whatever you choose to call it of of these. I say why reason being um that Coast X would be uh coast X -1 would have to be restricted to being between Um zero started between -1 and one. When in fact, if we take any possible value of X, uh it should be able to be between negative two and zero Uh because of this negative one. So we um are not going to do that. So this is our final answer. And then we're going to show a graph of of this um relation and that looks like this mm

The University of Western Ontario

#### Topics

Differential Equations

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp