Differential equations/initial-value problem

  • #1
372
0
Here's the initial-value problem I'm trying to solve:

[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.
 

Answers and Replies

  • #2
dextercioby
Science Advisor
Homework Helper
Insights Author
13,162
725
Are u claiming that

[tex] -\int \cos 3y \ dy=-\cos y \sin y +C [/tex]

If so,on what grounds?

Daniel.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
966
If [tex]\integral sin 2x dx= -\frac{1}{2}cos(2x)[/tex]

why isn't [tex]\integral cos 3x dx= \frac{1}{3}sin(3x)[/tex]?
 
  • #4
dextercioby
Science Advisor
Homework Helper
Insights Author
13,162
725
U mean with integral signs & integration constants,Halls,right?:uhh:

Daniel.
 
  • #5
372
0
What I really did previously was [tex]-\int \cos (2y) \: dy = - \cos (y) \sin (y) + \mathrm{C}[/tex], 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:

[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) = -\frac{1}{3} \sin (3y) + \mathrm{C}[/tex]

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

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

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

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

[tex]\sin (3y) = 3\cos ^2 x[/tex]

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

I'm still confused!!
 
  • #6
dextercioby
Science Advisor
Homework Helper
Insights Author
13,162
725
The last formula

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

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

However,your penultimate formula is correct

[tex] \sin 3y=3\cos^{2} x [/tex]

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

Daniel.
 
  • #7
372
0
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

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

Thank you
 
  • #8
dextercioby
Science Advisor
Homework Helper
Insights Author
13,162
725
It's good that u realized how to solve this trancendental equation

[tex] \sin x = a [/tex]

Daniel.
 

Related Threads on Differential equations/initial-value problem

Replies
4
Views
15K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
11
Views
2K
Replies
9
Views
14K
  • Last Post
Replies
4
Views
2K
Replies
4
Views
1K
  • Last Post
Replies
2
Views
2K
Replies
3
Views
3K
Top