- #1

- 782

- 1

I get

[tex]1 = C\sqrt{\cos 2y} \cos x[/tex]

My init conditions were y(0) = pi/2 so I end up getting C as [itex]\sqrt{-1}[/itex] but the equation should be

[tex]-1 = \cos 2y \cos^2 x[/tex]

Where's my error?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter cscott
- Start date

- #1

- 782

- 1

I get

[tex]1 = C\sqrt{\cos 2y} \cos x[/tex]

My init conditions were y(0) = pi/2 so I end up getting C as [itex]\sqrt{-1}[/itex] but the equation should be

[tex]-1 = \cos 2y \cos^2 x[/tex]

Where's my error?

- #2

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 966

That is, did you forget the absolute value?

- #3

- 782

- 1

HallsofIvy said:

That is, did you forget the absolute value?

Wouldn't I then have to add absolute signs around both cosines? Then C would equal 1 instead of -1.

I'm not sure where 1/u comes up... can't I just integrate tan x to get -ln |cos x| and tan 2y to get -1/2 ln |cos 2y|?

- #4

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 966

[tex] \int tan x dx+ \int tan 2y dy= -ln|cos x|- \frac{1}{2}ln|cos 2y|+ C[/tex]

Of course, you didn't tell us what the actual problem is so I have no idea how you would decide what C is!

- #5

- 782

- 1

Yes, but the "1/u" is where ln|cos x| comes from: tanx= sin x/cos x.

Oops, I should have seen that.

Of course, you didn't tell us what the actual problem is so I have no idea how you would decide what C is!

Oh, but I did state the initial conditions in my OP! Here they are again: [itex]y(0) = \pi/2[/itex]

I worked from where you left off (where we both agree) and I got

[tex]C^2 = |\cos 2y||\cos^2 x|[/tex]

So with the initial conditions

[tex]C^2 = |-1||1| \Leftrightarrow C = \sqrt{1}[/tex]

I can get -1 out of that, but why choose it over 1? Also, the answer stated in my OP (as given in my book) has no absolute value signs, why's that?

Thanks for your help.

Last edited:

- #6

benorin

Homework Helper

- 1,328

- 122

[tex] \int \tan x dx+ \int \tan 2y dy = -\ln |\cos x|- \frac{1}{2}\ln |\cos 2y|+ C=0[/tex]

notice that since

[tex]-1\leq \cos x\leq 1 \Rightarrow 0\leq |\cos x| \leq 1\Rightarrow -\infty \leq \ln |\cos x| \leq 0 \Rightarrow 0 \leq -\ln |\cos x| \leq \infty ,[/tex]

and likewise we have [tex] 0 \leq - \frac{1}{2}\ln |\cos 2y| \leq \infty[/tex] so that

[tex] 0 \leq -\ln |\cos x| - \frac{1}{2}\ln |\cos 2y| \leq \infty[/tex]

but

[tex] -\ln |\cos x| - \frac{1}{2}\ln |\cos 2y| +C=0[/tex]

so it must be that [tex]C\leq 0,[/tex] so choose the value [tex]C=-1[/tex].

- #7

- 782

- 1

Thanks for your help.

Share: