Differential equation and substitution

[Pomico]

This is a step from my notes that I don't follow. I have sin$$\theta$$$$\frac{d}{d\theta}$$(sin$$\theta$$$$\frac{d\Theta}{d\theta}$$) and that, when substituting u=cos$$\theta$$ and writing that $$\Theta$$($$\theta$$)=P(u), $$\frac{d}{du}$$((1-u$$^{2}$$)$$\frac{dP}{du}$$) is obtained.

I can see $$\frac{d\Theta}{d\theta}$$=$$\frac{dP}{du}$$ for the far RHS but can't get the 1-u$$^{2}$$ to come out. I don't really remember how to do this though I'm sure I have been able to at some point so a starting tip would be much appreciated! The main trap I keep falling in is the temptation to try something with chain rule to get rid of all those d$$\theta$$ bits in the original equation. I know I'm not supposed to as I'm just supposed to be substituting, not solving, but I'm not sure how else to go about it.

 

[tiny-tim]

Hi Pomico!

(have a theta: θ )

This is a step from my notes that I don't follow. I have sin$$\theta$$$$\frac{d}{d\theta}$$(sin$$\theta$$$$\frac{d\Theta}{d\theta}$$) and that, when substituting u=cos$$\theta$$ and writing that $$\Theta$$($$\theta$$)=P(u), $$\frac{d}{du}$$((1-u$$^{2}$$)$$\frac{dP}{du}$$) is obtained.

I can see $$\frac{d\Theta}{d\theta}$$=$$\frac{dP}{du}$$ for the far RHS but can't get the 1-u$$^{2}$$ to come out.

Using the chain rule, sinθ d/dθ = sin²θ d/du.

(but I don't see how to get the whole thing … are you sure it isn't 1/sinθ on the left?)

 

[Mark44]

Using the chain rule, sinθ d/dθ = sin²θ d/du.

tiny-tim,
Maybe I'm missing something here, but the line above doesn't make sense to me.

Pomico,
Is there really a need to use uppercase and lowercase versions of the same letter? I.e., $\Theta$ and $\theta$.

 

[tiny-tim]

hmm … u = cosθ, so du/dθ = -sinθ,

so d/du = dθ/du d/dθ = -1/sinθ d/dθ,

so sin²θ d/du = -sinθ d/dθ

… ok, i missed out a minus sign ​

 

[Pomico]

(but I don't see how to get the whole thing … are you sure it isn't 1/sinθ on the left?)

Hi!
It's definitely not a reciprocal in the notes, that was one of the things bothering me

Pomico,
Is there really a need to use uppercase and lowercase versions of the same letter? I.e., $\Theta$ and $\theta$.

Yes there is, this is only part of the problem covered in the notes. Earlier on there was a function F(θ,$$\phi$$) which was split into variables as
F(θ,$$\phi$$) = $$\Theta$$(θ)$$\Phi$$($$\phi$$)

The equation in my first post isn't the full equation, just the first term. I don't think the rest will change anything but in case I've added an attachment of the lecture - the derivation I'm working from covers mostly pages 3 and 4 of the document. I don't know how to just include these two pages so sorry for adding the whole thing!

 

[Chrisas]

Try this. Consider the substitution $$\Theta(\theta)=P(u)$$

You want to find $$\frac{d\Theta}{d\theta}$$

$$\frac{d\Theta}{d\theta} = \frac{dP(u)}{d\theta} = \frac{dP(u)}{du} \frac{du}{d\theta}$$

You can get $$\frac{du}{d\theta}$$ from your substitution of u=cos($$\theta$$). Also sin($$\theta$$) = "something with u", using a well known trig identity.

 

[Mark44]

hmm … u = cosθ, so du/dθ = -sinθ,

so d/du = dθ/du d/dθ = -1/sinθ d/dθ,

so sin²θ d/du = -sinθ d/dθ

… ok, i missed out a minus sign ​

I didn't catch the minus sign, either. I guess what you were doing was equating operators. I think that's what threw me--having the operator without an operand for it to work on.

 

[tiny-tim]

The equation in my first post isn't the full equation, just the first term. I don't think the rest will change anything but in case I've added an attachment of the lecture - the derivation I'm working from covers mostly pages 3 and 4 of the document. I don't know how to just include these two pages so sorry for adding the whole thing!

D'oh! I was right!

You're referring to equations (2.26) and (2.27) …

but you can see that (2.26) is divided by sin²θ before the u-substitution, so then it does start with 1/sinθ !

 

[Pomico]

Aah yes I see, thanks for spotting that :p
I'll have another go!

Edit: Got it, thanks again :)