- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- Astrophysics

\[\frac{ d^2y }{ dx^2 }\] I'm trying to get this to equal \[\frac{ d^2y }{ dt^2 }t^2+\frac{ dy }{ dt }t\] so I have \[\frac{ d^2y }{ dx^2 } = \frac{ d }{ dx }\left( \frac{ dy }{ dt } t\right)\] which I got from the chain rule but how do I use the product rule here X_X

- Astrophysics

|dw:1444614858695:dw|

- Astrophysics

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- Kainui

Wait so are we saying y(x) and x(t) is that true?

- Astrophysics

Well I'm trying to figure out eulers equation \[t^2\frac{ d^2y }{ dt^2 }+\alpha t \frac{ dy }{ dt }+ \beta y=0~~~t>0\] and it says I need to calculate dy/dt and d^2y/dt^2 in terms of dy/dx and d^2y/dx^2

- Astrophysics

so I can use my substitution

- Kainui

Oh ok this makes more sense now.

- Kainui

wait are you not just allowed to substitute in \[y=At^n\] and solve?

- Astrophysics

No I have to use x = lnt

- Astrophysics

and it's telling me to do all that stuff lol

- Kainui

Oh alright then

- Astrophysics

Just want to know how to use the product rule in leibniz notation dont really care about the problem

- Astrophysics

f'g+g'f lol but I can't believe I don't know how to do this

- Kainui

\[y(x(t))\]
\[\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}\]
\[\frac{d}{dt} \left( \frac{dy}{dx} \frac{dx}{dt} \right) = \frac{d}{dt} \left( \frac{dy}{dx} \right)\frac{dx}{dt} +\frac{dy}{dx}\frac{d}{dt} \left( \frac{dx}{dt} \right) \]
Try to take it from here.

- Kainui

While you do that, I'll do it without logs to see how much faster it is.

- Astrophysics

It's the first part, \[\frac{ d }{ dt }\left( \frac{ dy }{ dx } \right)\frac{ dx }{ dt }\] there's no changeeee, I was expecting there to be d^2y/dx^2 or something

- Kainui

Ultimately they'll get the same answer and this method you're using is more general so it's not without its merits it's just this is a pretty common one to solve.

- Kainui

Here's only the derivative part of the first part (notice we'll have a squared dx/dt when we multiply this back in):
\[ \frac{d}{dt} \left( \frac{dy}{dx} \right) = \frac{d^2y}{dx^2} \frac{dx}{dt} \]

- Astrophysics

Ok should it not be \[\frac{ d^2y }{ dx^2 }\frac{ d }{ dt }\]

- Kainui

If this bothers you instead let:
\[\frac{dy}{dx} = u(x(t))\]
Then when you do the derivative you have:
\[\frac{d}{dt}(u) = \frac{du}{dx} \frac{dx}{dt}\]
Then you can look at the du/dx term on its own as the second derivative of y wrt x.

- Kainui

Explain your reasoning for this:
\[\frac{ d^2y }{ dx^2 }\frac{ d }{ dt }\]

- Astrophysics

Nvm I'm dumb, it's \[\frac{ d }{ dt }\left( \frac{ dy }{ dx } \right) = \frac{ dy }{ dt }\frac{ dt }{ dx }\frac{ d }{ dt } = \frac{ d^2y }{ dt^2 }\frac{ dt }{ dx }\] is this what you mean

- Astrophysics

and not dx^2

- Kainui

Nope, you're taking the derivatives in a strange order, and this hanging d/dt that's by itself is kind of meaningless here.

- Astrophysics

I just used the chain rule, the d/dt is what is confusing me

- Kainui

\[y(x(t))\]
\[\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}\]
\[\frac{d}{dt} \left( \frac{dy}{dx} \frac{dx}{dt} \right) = \color{red}{ \frac{d}{dt} \left( \frac{dy}{dx} \right)}\frac{dx}{dt} +\frac{dy}{dx}\frac{d}{dt} \left( \frac{dx}{dt} \right) \]
Looking only at the red part:
\[ \frac{d}{dt} \left( \frac{dy}{dx} \right) = \frac{d^2y}{dx^2} \frac{dx}{dt} \]
make the u substitution I described earlier to get through this part.

- Kainui

If there's anything about using the product and chain rule above that confuses you say something now before moving forward.

- Astrophysics

Nooo, it's not really the substitution part, I mean \[\frac{d}{dt} \left( \frac{dy}{dx} \frac{dx}{dt} \right) = \color{red}{ \frac{d}{dt} \left( \frac{dy}{dx} \right)}\frac{dx}{dt} +\frac{dy}{dx}\frac{d}{dt} \left( \frac{dx}{dt} \right)\] it was this in general, so you're just writing \[\frac{ d }{ dt }\left( \frac{ dy }{ dx } \frac{ dx }{ dt }\right)\] + this thing again, oh I'm overthinking this, I just realized how simple it actually is

- Astrophysics

\[\frac{ d }{ dx }\left( \frac{ dy }{ dt }t \right)=\frac{ d }{ dx }\left( \frac{ dy }{ dt } \right)t+\frac{ d }{ dx }\frac{ dy }{ dt } (t)\]

- Astrophysics

omggggggg kai

- Kainui

Where did this come from?

- Kainui

Explain why you're trying to use this: \[\frac{ d }{ dx }\left( \frac{ dy }{ dt }t \right)=\frac{ d }{ dx }\left( \frac{ dy }{ dt } \right)t+\frac{ d }{ dx }\frac{ dy }{ dt } (t)\]

- Astrophysics

That's my original q

- Kainui

Isn't this what you're trying to solve? \[t^2\frac{ d^2y }{ dt^2 }+\alpha t \frac{ dy }{ dt }+ \beta y=0~~~t>0\]

- Astrophysics

\[\frac{ d^2y }{ dx^2 } = \frac{ d }{ dx }\left( \frac{ dy }{ dt } t\right) = \frac{ d }{ dx }(\frac{ dy }{ dt })t+\frac{ d }{ dx }\left( t \right)\frac{ dy }{ dt }\]

- Astrophysics

I was trying to find d^y/dt^2 in terms of d^2y/dx^2 ahaha

- Astrophysics

I think I got it, I sort of just thought about this \[\frac{ d }{ dx }(u*v)=...\] I was just overthinking

- Kainui

No stop saying that lol, you can't claim that until you've finished solving the problem xD
Right now I'm trying to get us on the path of writing \(\frac{d^2y}{dt^2}\) in terms of x with the chain rule, but I'm not sure where you're going wrong so I'm having difficulty helping you out.

- Astrophysics

Oh I thought we were doing it in terms of d^2y/dt^2 the whole time

- Kainui

I don't know what you're doing so I don't know

- Astrophysics

lmao

- Kainui

\[\frac{ d^2y }{ dx^2 } = \frac{ d }{ dx }\left( \frac{ dy }{ dt } t\right)\] Explain where you're getting this from, cause I don't know. That doesn't mean it's wrong it just means you need to say words, I can't read your mind.

- Astrophysics

All I was doing is finding \[\frac{ d^2y }{ dt^2 }\] in terms of \[\frac{ d^2y }{ dx^2 }\] so I would get \[\frac{ d^2y }{ dt^2 }t^2+\frac{ dy }{ dt }t\]

- Kainui

Hmmm explain how you're supposed to solve this: " it says I need to calculate dy/dt and d^2y/dt^2 in terms of dy/dx and d^2y/dx^2"
So I am not seeing this here, so for example, dy/dt in terms of dy/dx would look like this, right:
\[\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}\]

- Astrophysics

\[\frac{ d^2y }{ dx^2 }\] and started using the chain rule, \[\frac{ d^2y }{ dx^2 } = \frac{ d }{ dx }\left( \frac{ dy }{ dx } \right)=\frac{ d }{ dx }\left( \frac{ dy }{ dt }\frac{ dt }{ dx } \right)=\frac{ d }{ dx }\left( \frac{ dy }{ dt } t\right)\] and then I was trying to apply chain rule to \[\frac{ d }{ dx }\left( \frac{ dy }{ dt } t\right)\]

- Astrophysics

I started with \[\frac{ d^2y }{ dx^2 }\]

- Astrophysics

product rule not chain rule* omg

- Kainui

So you're writing \[\frac{d^2y}{dx^2}\] in terms of \[\frac{d^2y}{dt^2}\] then? And you're plugging in \[\frac{dt}{dx}=t\] is that right?

- Astrophysics

Yes!

- Kainui

This is fine then I think, it's just weird cause instead of looking it as y(x(t)) you're looking at it as y(t(x)) which is kind of backwards but in this case works since \(x= \ln t\) which is surjective.

- Astrophysics

Haha, I thought that's how you were suppose to do it, I read it on pauls online notes or something, I really have no notes for this, so I was sort of coming up with stuff myself

- Astrophysics

I mean the point of getting this is so I can set up my equation as \[\frac{ d^2y }{ dx^2 }+(\alpha - 1)\frac{ dy }{ dx }+\beta y=0\]

- Kainui

Right of course, but doing it like this is awkward since time isn't generally thought of as a function of position. And it has no meaning in periodic motion.

- Astrophysics

true

- Astrophysics

I wasn't really thinking of it that way, haha just some math problem

- Astrophysics

And the question stated, let x = lnt and calculate dy/dt and d^2y/dt^2 in terms of dy/dx and d^2y/dx^2 ahah

- Kainui

Here, I'll just walk through how I did it:
\[\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}\]
\[\frac{d^2y}{dt^2} = \frac{d^2y}{dx^2} \left( \frac{dx}{dt} \right)^2 + \frac{dy}{dt} \frac{d^2x}{dt^2}\]
I guess we can do it now or later, so why not now:
\[\frac{dx}{dt} = \frac{1}{t}\]
\[\frac{d^2x}{dt^2} = \frac{-1}{t^2}\]
And then plugging all this in you get the same simplified differential equation since the \(t^2\) will disappear.

- Kainui

I messed up one piece but, oh well should have been a dy/dx on that last term I just noticed

- Kainui

At least this way is more straight forward to plug in (well in my opinion). However I wouldn't use this method at all lol. Best way is to notice you have a power next to the derivative which is basically compensating for a lost power by the power rule every derivative. Still pretty easy to spot.
\[t^2 y''+\alpha t y' + \beta y = 0\]
Now you can just pick \[y=At^n\] and plug it in:
\[n(n-1)+\alpha n + \beta = 0\]
Now you have a quadratic in n to solve for the two solutions, done!

- Astrophysics

Yo that's so simple

- Astrophysics

Yeah I already hate this method I have to use

- Kainui

Haha use the hate to fuel you through the tedious calculations and before you know it you'll be really fast. I feel like most of this discussion didn't have to happen I just had no idea what you were doing so we wasted a lot of time.

- Astrophysics

Yeah haha, well at least I know what I'm doing now X_X

- Astrophysics

I just realized how simple this was, I actually just misread something earlier, so was sort of confused haha xD

Looking for something else?

Not the answer you are looking for? Search for more explanations.