- #1

- 372

- 0

[tex]\sin (2x) \: dx + \cos (3y) \: dy = 0 \qquad y \left( \frac{\pi}{2} \right) = \frac{\pi}{3}[/tex]

I can't see exactly where my mistake is, but I do know that my answer is not correct. My textbook gives:

[tex]y=\frac{1}{3} \left[ \pi - \arcsin \left( 3 \cos ^2 x \right) \right]\mbox{.}[/tex]

Here's what I've done:

[tex]\sin (2x) \: dx + \cos (3y) \: dy = 0[/tex]

[tex]\sin (2x) \: dx = - \cos (3y) \: dy [/tex]

[tex]\int \sin (2x) \: dx = - \int \cos (3y) \: dy [/tex]

[tex]-\frac{1}{2} \cos (2x) = -\cos (y) \sin (y) + \mathrm{C}[/tex]

[tex]\cos (2x) = 2\cos (y) \sin (y) + \mathrm{C}[/tex]

Applying the initial condition at this point gives:

[tex]\mathrm{C} = -\frac{\sqrt{3}}{2} - 1[/tex]

Thus, we have the following.

[tex]\cos (2x) = 2\cos (y) \sin (y) -\frac{\sqrt{3}}{2} - 1[/tex]

[tex]\cos (2x) = \sin (2y) -\frac{\sqrt{3}}{2} - 1[/tex]

[tex]\sin (2y) = \cos (2x) +\frac{\sqrt{3}}{2} + 1[/tex]

[tex]\sin (2y) = 2\cos ^2 (x) - 1 +\frac{\sqrt{3}}{2} + 1[/tex]

[tex]\sin (2y) = 2\cos ^2 (x) +\frac{\sqrt{3}}{2}[/tex]

[tex]y = \frac{1}{2} \arcsin \left[ 2\cos ^2 (x) +\frac{\sqrt{3}}{2} \right][/tex]

As you can see, I don't get the same answer.

Any help is highly appreciated.