# Differential equations/initial-value problem

1. Mar 30, 2005

Here's the initial-value problem I'm trying to solve:

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

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

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

Here's what I've done:

$$\sin (2x) \: dx + \cos (3y) \: dy = 0$$

$$\sin (2x) \: dx = - \cos (3y) \: dy$$

$$\int \sin (2x) \: dx = - \int \cos (3y) \: dy$$

$$-\frac{1}{2} \cos (2x) = -\cos (y) \sin (y) + \mathrm{C}$$

$$\cos (2x) = 2\cos (y) \sin (y) + \mathrm{C}$$

Applying the initial condition at this point gives:

$$\mathrm{C} = -\frac{\sqrt{3}}{2} - 1$$

Thus, we have the following.

$$\cos (2x) = 2\cos (y) \sin (y) -\frac{\sqrt{3}}{2} - 1$$

$$\cos (2x) = \sin (2y) -\frac{\sqrt{3}}{2} - 1$$

$$\sin (2y) = \cos (2x) +\frac{\sqrt{3}}{2} + 1$$

$$\sin (2y) = 2\cos ^2 (x) - 1 +\frac{\sqrt{3}}{2} + 1$$

$$\sin (2y) = 2\cos ^2 (x) +\frac{\sqrt{3}}{2}$$

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

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

Any help is highly appreciated.

2. Mar 30, 2005

### dextercioby

Are u claiming that

$$-\int \cos 3y \ dy=-\cos y \sin y +C$$

If so,on what grounds?

Daniel.

3. Mar 30, 2005

### HallsofIvy

Staff Emeritus
If $$\integral sin 2x dx= -\frac{1}{2}cos(2x)$$

why isn't $$\integral cos 3x dx= \frac{1}{3}sin(3x)$$?

4. Mar 30, 2005

### dextercioby

U mean with integral signs & integration constants,Halls,right?:uhh:

Daniel.

5. Mar 30, 2005

What I really did previously was $$-\int \cos (2y) \: dy = - \cos (y) \sin (y) + \mathrm{C}$$, which does not apply in this case. In fact, I missed one digit in the integrand. Thanks for pointing it out.

I've just fixed that mistake, but not quite the problem. Here's what I have now:

$$\sin (2x) \: dx + \cos (3y) \: dy = 0$$

$$\sin (2x) \: dx = - \cos (3y) \: dy$$

$$\int \sin (2x) \: dx = - \int \cos (3y) \: dy$$

$$-\frac{1}{2} \cos (2x) = -\frac{1}{3} \sin (3y) + \mathrm{C}$$

Applying the initial condition $$y \left( \frac{\pi}{2} \right) = \frac{\pi}{3}$$ gives $$\mathrm{C} = \frac{1}{2}$$.

$$-\frac{1}{2} \cos (2x) = -\frac{1}{3} \sin (3y) + \frac{1}{2}$$

$$\sin (3y) = \frac{3 + 3\cos (2x)}{2}$$

$$\sin (3y) = \frac{3 + 3\left( 2\cos ^2 x - 1 \right)}{2}$$

$$\sin (3y) = 3\cos ^2 x$$

$$y = \frac{1}{3} \arcsin \left( 3\cos ^2 x \right)$$

I'm still confused!!

6. Mar 30, 2005

### dextercioby

The last formula

$$y=\frac{1}{3}\arcsin\left(3\cos^{2}x\right)$$

does not verify the initial condition,becasue the range of "arcsin" is $\left[-\frac{\pi}{2},+\frac{\pi}{2}\right]$...

$$\sin 3y=3\cos^{2} x$$

I advise you to leave it in this implicit form or try to put in the one the problem is giving...

Daniel.

7. Mar 30, 2005

Okay, I just need to make some corrections in my last formula because of the problem with the range of arcsin. That would make it

$$y = \frac{\pi}{3} - \frac{1}{3} \arcsin \left( 3\cos ^2 x \right)$$

Thank you

8. Mar 30, 2005

### dextercioby

It's good that u realized how to solve this trancendental equation

$$\sin x = a$$

Daniel.