Question

@zepdrix
This one feels tricky - I don't know how to get the y's on one side...

Answer 1

\[y' \cosh^2{x}=\sin^2{y}\]
y(0) = pi/2

Answer 2

division ....

Answer 3

lol :)

Answer 4

What I meant with getting them on one side, was getting the y's there and how to use them then haha

I get
\[\frac{1}{\sin^2y} \text{dy}=\frac{1}{\cosh^2y}\text{dx}\]
but don't know how to integrate that

Answer 5

the cosh^2 has an x, sorry, not a y

Answer 6

Remember your trig identities?

\(\large \dfrac{1}{\sin\theta}=?\)

Answer 7

The right side looks a little tricky...
I'd have to brush up on my hyperbolic identities...

I'm assuming that gives us \(\large sech^2x\) though..

Answer 8

arent the hyper trigs "e"-able?

Answer 9

They're e-able.. but doesn't that put a bunch of e's in the denominator?
I think that will be a little tricky.. hmm :O

Maybe it's worth trying.

Answer 10

Ahh!

\[\csc^2y \text{ dy} = sech^2x\text{ dx}\]?

Answer 11

Yah looks good so far :)

Answer 12

these tables might help jog a memory cell or two ;)

http://math2.org/math/integrals/tableof.htm

Answer 13

the e-able version is:

http://inst-mat.utalca.cl/~cdelpino/modelos/tecnologia/tablasdavid/integrals/more/sech.htm

Answer 14

... but then its squared :/

Answer 15

i see that the h makes no difference.

tan down to sec^2
tanh down to sech^2

Answer 16

ya, good times.

Answer 17

Sorry, I'm back. My Internet disconnected

Answer 18

Okay, so integrating that part that I wrote, I get
\[-\cot y = \tanh^2 x +c\]
Now what?

Answer 19

Oh we were given initial conditions? Hmm ok.\[\large \cot y=C-\tanh x\]

So plug the value in and let's see what we get.\[\cot\left(\frac{\pi}{2}\right)=C-\tanh 0\]

Answer 20

C = 0

so cot y = - tanh x
and y = arccot(-tanh x) ?

Answer 21

Yah that appears to be correct.

What a weird problem lol.

Answer 22

Yeah some of these problems are a little weird haha

Thanks you guys for the help!