# Function sin(2x)^(-1/2)

For a function sin(2x)^(-1/2) in ]0, pi/2[ counts:
D2(f) + f = 3*D(a) where Da stands for the a'd derivative
So is there any quick way to solve this? I also can't seem to find a formula for the n'd derivative.
Thanks!

Hello,

For a function sin(2x)^(-1/2) in ]0, pi/2[ counts:
D2(f) + f = 3*D(a) where Da stands for the a'd derivative
So is there any quick way to solve this? I also can't seem to find a formula for the n'd derivative.
Thanks!
What are you trying to do here? Are you trying to find the derivative of

sin(2x)^(-0.5) ?

What is a'd derivative?

Thanks
Matt

tiny-tim
Homework Helper
Hi JanClaesen!

(have a pi: π and a square-root: √ )
For a function sin(2x)^(-1/2) in ]0, pi/2[ counts:
D2(f) + f = 3*D(a) where Da stands for the a'd derivative
So is there any quick way to solve this? I also can't seem to find a formula for the n'd derivative.
Thanks!
(we say the nth derivative, not the n'd derivative )

Do you mean "what is the nth derivative of 1/√(sin2x)?"

And what is f?

Hello , I'm sorry if I was a little unclear:

So there is a certain relation for the function sin(2x)^(-1/2): the second derivative of this function plus the function itself gives the (n'th derivative)*3, the question is to determine n.

Rephrase:

If $y=1/\sqrt{sin(2x)} \text{ and }y''+y=3y^{(n)}$, solve for n. (Note $y^{(n)} = \frac{d^n y}{{dx}^n}$.)

I have hammered out the first 10 derivatives of y but none seem to be equal to (y'' + y)/3. The exponent of the denominator factor is of the order 1/2 + n, while the numerator is a trigonometric polynomial in cosine of order n. I do not forsee this collapsing into something nice. This exercise seems fishy.

N.B.: The expression on the left of the differential equation is

$$\frac{3}{(sin(2x))^{5/2}}.$$​

--Elucidus

tiny-tim
Homework Helper
If $y=1/\sqrt{sin(2x)} \text{ and }y''+y=3y^{(n)}$, solve for n. (Note $y^{(n)} = \frac{d^n y}{{dx}^n}$.)
Hi JanClaesen! Hi Elucidus!

I don't think there can be a solution …

just try to solve 3y(n) - y'' - y = 0 the usual way …

that gives you a characteristic equation, of which 1/√sin can't be an answer.

I think the exercise is correct, it's probably just me who misunderstood it, I scanned it so you guys can have a look at it: http://img212.imageshack.us/img212/2451/scan001001s.jpg [Broken]
It's exercice 17a, the answer should be 5, perhaps alpha is a power?

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tiny-tim
Homework Helper
I think the exercise is correct, it's probably just me who misunderstood it, I scanned it so you guys can have a look at it: http://img212.imageshack.us/img212/2451/scan001001s.jpg [Broken]
It's exercice 17a, the answer should be 5, perhaps alpha is a power?
D'oh!

Yes of course α is a power … Question 17a says so

f'' + f = 3fα

Why did you write D2(f) + f = 3*D(a) in your first post??

ok, since f = 1/√(sin(2x)), what is f'' + f?

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Because I only thought of that while I was writing the newest post
I'm sorry, but I never saw a power-variable being named alpha before, that's why I thought it was perhaps a way to note the n'th derivative, which seemed to me like a quite hard thing to solve.
Hope I didn't waste too much of your time!

Just curious ... what language is that question written in ... and are you from the Netherlands ?

Hah nice, you're right, it's Dutch , but I'm not from the Netherlands but from its little brother, Belgium. Most people here are American I guess?