Differential equations/initial-value problem

[DivGradCurl]

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.

 

[dextercioby]

Are u claiming that

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

If so,on what grounds?

Daniel.

 

[HallsofIvy]

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

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

 

[dextercioby]

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

Daniel.

 

[DivGradCurl]

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!

 

[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]...

However,your penultimate formula is correct

\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.

 

[DivGradCurl]

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

 

[dextercioby]

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

\sin x = a

Daniel.