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

Question

Isabella Ackerman · Accepted Answer

You can separate yuor variables #x# and #y# and integrate:

Emma Aldridge · Answer

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)#.

Evelyn Agard · Answer

undefined To solve the initial-value problem $ y' = \frac{\sin(x)}{\sin(y)} $ with $ y(0) = \frac{\pi}{4} $, you can separate variables and then integrate.

First, separate variables by multiplying both sides by $ \sin(y) $ and dividing both sides by $ y' $:
$$ \sin(y) \, dy = \sin(x) \, dx $$

Next, integrate both sides:
$$ \int \sin(y) \, dy = \int \sin(x) \, dx $$

After integrating, you get:
$$ -\cos(y) = -\cos(x) + C $$

Where $ C $ is the constant of integration.

Now, apply the initial condition $ y(0) = \frac{\pi}{4} $:
$$ -\cos\left(\frac{\pi}{4}\right) = -\cos(0) + C $$
$$ -\frac{\sqrt{2}}{2} = -1 + C $$
$$ C = -\frac{\sqrt{2}}{2} + 1 $$

So, the particular solution to the initial-value problem is:
$$ -\cos(y) = -\cos(x) -\frac{\sqrt{2}}{2} + 1 $$

Solving for $ y $, you get:
$$ y = \arccos\left(\cos(x) +\frac{\sqrt{2}}{2} - 1\right) $$

This is the solution to the initial-value problem $ y' = \frac{\sin(x)}{\sin(y)} $ with $ y(0) = \frac{\pi}{4} $.