Second ODE - Using x = e^t show that the equation

[thomas49th]

Homework Statement

Show that the equation x= e^t converts the equation

$$ax^{2}\frac{d^{2}y}{dx^{2}} + bx\frac{dy}{dx} + cy = 0$$

in which a,b,c are coefficients

Homework Equations

The Attempt at a Solution

x = e^t and so does dx/dt. So you can write dx/dt = x

using the chain rule

dy/dx = dy/dt * dt/dx
=> dy/dt * 1/x

now here is the bit that is tricky.

$$ax^{2}\frac{d^{2}y}{dx^{2}} + b\frac{dx}{dt} + cy = 0$$

Apparently I cannot simply stick $$\frac{dy}{dt} * \frac{1}{x}$$
into the second order deravite to give $$\frac{d^{2}y}{d^{2}t} * \frac{1}{x^{2}}$$. Why not? Does $$\frac{d^{2}y}{dx^{2}} \neq \frac{dy}{dx^{2}}$$??

Also I thought maybe you differentiate the chain rule again
$$\frac{d^{2}y}{dx^{2}} = -\frac{dy}{dt}\frac{1}{x^{2}}$$

But that doesn't seem to help (have I differentiated it correctly as I didn't use the product rule because I don't think dy/dt is a function of x).

Any suggestions
Thomas

 

[lanedance]

so start with
$$y(x(t))$$

differntiate once to get
$$\frac{dy}{dt}= \frac{d}{dt}y(x(t)) = \frac{dy}{dx} \frac{dx}{dt}$$

then differntiate again using some product rule
$$\frac{d^2y}{dt^2} = \frac{d}{dt} ( \frac{dy(x(t))}{dx} \frac{dx(t)}{dt})$$

see if you can do the last step you'll need to use the product and chain rule, if it helps think of dy/dx as a function of x(t)
$$f(x(t)) = \frac{dy(x(t))}{dx}$$

then the chain rule becomes
$$\frac{d}{dt} f(x(t)) =\frac{df}{dx}\frac{dx}{dt} = \frac{d^2 y}{dx^2}\frac{dx}{dt}$$

 

[thomas49th]

I don't understand
$$f(x(t)) = \frac{dy(x(t))}{dx}$$

what exactly are you saying here? that the function of f that depends on x which in turns depends on t is equal to the dervative of x(t) with respect to x? How can I visualise this?

 

[lanedance]

I don't understand
$$f(x(t)) = \frac{dy(x(t))}{dx}$$

what exactly are you saying here? that the function of f that depends on x which in turns depends on t is equal to the dervative of x(t) with respect to x? How can I visualise this?

I was just trying to use a notation that would be easy to differentiate and help apply the chain rule, to finish off you get

$$\frac{d^2y}{dt^2} = \frac{d}{dt} ( \frac{dy(x(t))}{dx} \frac{dx(t)}{dt}) = ( \frac{d^2 y}{dx^2} (\frac{dx}{dt})^2 + \frac{dy}{dx} \frac{d^2 x}{dt^2}) )$$

 

[thomas49th]

Ahhh almost get it.

I get this part:

$$f(x(t)) = \frac{dy(x(t))}{dx}$$

and I see that you use the product rule

$$\frac{d^2y}{dt^2} = \frac{d}{dt} ( \frac{dy(x(t))}{dx} \frac{dx(t)}{dt})$$

on the RHS. I had a go and got the left part of it but the right most term is wrong

$$( \frac{d^2 y}{dx^2} (\frac{dx}{dt})^2 + \frac{dy}{dx} \frac{dx}{dt}) )$$

I don't get the squares. :( So this means the part I'm getting wrong is

$$\frac{d}{dt}(\frac{dx}{dt}x(t))$$

I'm having problems differentiating this. I like to think of it as
y = x(t)
for y' = x'(t)
dy/dt = x'(t) ? How do I write this is in dW/dZ form?

Thanks
Thomas

 

[thomas49th]

HANG ON I GOT IT! x'(t) is dx/dt derrrrr! so d/dt(dx/dt) = d^2x/dt^2

Now if I could only realize what we have just shown...

 

[lanedance]

so you know that if x = e^t, then
$$x = e^t = \frac{dx}{dt}= \frac{d^2x}{dt^2}$$


so the first derivative is
$$\frac{dy}{dt}= \frac{dy}{dx} e^t$$
or re-arranging
$$\frac{dy}{dx} = \frac{dy}{dt}e^{-t}$$


and the 2nd derivative is
$$\frac{d^2y}{dt^2} = ( \frac{d^2 y}{dx^2} (e^t)^2 + \frac{dy}{dt}e^{-t} e^t )$$
re-arranging
$$\frac{d^2 y}{dx^2} =(\frac{d^2y}{dt^2} - \frac{dy}{dt})e^{-2t}$$

now substitute into your original equation

 

[lanedance]

in the simpler notation if we let ' mean diff wrt t, while a dot is diff wrt x

differntiate once to get
$$y'= \dot{y} x'$$

differntiate again
$$y''= \ddot{y} (x')^2 + \dot{y} x''$$

 

[thomas49th]

so you know that if x = e^t, then
$$x = e^t = \frac{dx}{dt}= \frac{d^2x}{dt^2}$$so the first derivative is
$$\frac{dy}{dt}= \frac{dy}{dx} e^t$$
or re-arranging
$$\frac{dy}{dx} = \frac{dy}{dt}e^{-t}$$and the 2nd derivative is
$$\frac{d^2y}{dt^2} = ( \frac{d^2 y}{dx^2} (e^t)^2 + \frac{dy}{dt}e^{-t} e^t )$$
re-arranging
$$\frac{d^2 y}{dx^2} =(\frac{d^2y}{dt^2} - \frac{dy}{dt})e^{-2t}$$

now substitute into your original equation

I'm lost again It's the second derivative that's getting me. You've jumped to many steps for me

How do I work out

$$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dt}e^{-t})$$

Thanks

 

[lanedance]

we're looking at it 2 different ways, i was differentiating wrt t
$$\frac{d^2y}{dt^2} = ( \frac{d^2 y}{dx^2} (e^t)^2 + \frac{dy}{dt}e^{-t} e^t )$$

once you get here, re-arrange for $\frac{d^2 y}{dx^2}$

 

[HallsofIvy]

I'm lost again It's the second derivative that's getting me. You've jumped to many steps for me

How do I work out

$$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dt}e^{-t})$$

Thanks

If $x= e^t$, $t= ln(x)$, and
$$\frac{dy}{dx}= \frac{1}{x}\frac{dy}{dt}$$

So
$$\frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\right)$$
using the product rule, that is
$$-\frac{1}{x^2}\frac{dy}{dt}+ \frac{1}{x}\frac{d}{dx}\left(\frac{dy}{dt}\right)$$
$$= -\frac{1}{x^3}\frac{dy}{dt}+ \frac{1}{x}\left(\frac{1}{x}\frac{d^2y}{dt^2}\right)$$
$$= -\frac{1}{x^2}\frac{dy}{dt}+ \frac{1}{x^2}\frac{d^2y}{dt^2}$$

 

[hunt_mat]

In order to avoid all this nonsense about tranformations, just look for solutions of the form x^n of your original equation

 

[thomas49th]

I'm stupid, I know, but what is the derivatice of dy/dt. It must be a funciton of x right? Which we know as dy/dt = x * dy/dx . Otherwise we can treat it like a constant.

I can see what is going apart from the derivative of dy/dt with respect to x. It's implicit so you differentiate it as you would normally (d^2y/dt^2) then multiply it by dy/dx. So shouldn't term be

$$= -\frac{1}{x^{2}}\frac{dy}{dt}+ \frac{1}{x}\left(\frac{dy}{dx}\frac{d^2y}{dt^2}\right)$$

HOI - did you make a typo in your previous post ( shouldn't the left term be -1/x^2)??

Thanks

 

[thomas49th]

I'm still ending up with
$$= -\frac{1}{x^{2}}\frac{dy}{dt}+ \frac{1}{x}\left(\frac{dy}{dx}\frac{d^2y}{dt^2}\right)$$

from applying the product rule to

$$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}[\frac{dy}{dt}\frac{1}{x}]$$

$$= -\frac{1}{x^{2}}\frac{dy}{dt}+ \frac{1}{x}\frac{d}{dx}[\frac{dy}{dt}]$$

What the flup can I do. Want to be able to do this question before Christmas. What do I do!?

Thanks
Thomas

 

[lanedance]

Ok, so starting fresh, changing notation as I'm on mobile
x=e^t and consider

dy/dx=dy/dt.dt/dx=dy/dt.1/x

Differentiating again
d2y/dx2=d2y/dt2.1/x2-dy/dt.1/x2

Now sub these into original de and note all the x factors cancel

 

[lanedance]

The key to note is the original de has the second derivative multiplied by x^2, so the x's cancel with the 1/x^2 in the chain rule calculated derivative, and similarly with thenfirst derivative

 

[lanedance]

The key to note is the original de has the second derivative multiplied by x^2, so the x's cancel with the 1/x^2 in the chain rule calculated derivative, and similarly with the first derivative

 

[thomas49th]

Differentiating again
d2y/dx2=d2y/dt2.1/x2-dy/dt.1/x2

So, is $$\frac{d}{dx}\left \frac{dy}{dt} \right$$
just

$$\frac{d^{2}y}{dt^{2}}$$

I refuse to believe it. We're differentiating with respect to x. Doesn't it has to be implicit so you multiply d2y/dt2 by dy/dt?

If i can understand that differential I should be able to complete the question. It is this differential that has been stopping me complete the question.

Thanks
Thomas

 

[thomas49th]

Well is it? and how so?

 

[lanedance]

Use chain rule
(d/dx)dy/dt=(d/dt.(dy/dt)).(dt/dx)=(d2y/dt2).(dt/dx)

 

[thomas49th]

Use chain rule
(d/dx)dy/dt=(d/dt.(dy/dt)).(dt/dx)=(d2y/dt2).(dt/dx)

arghh! I thought so but how do I get

d2y/dx2=d2y/dt2.1/x2-dy/dt.1/x2
from

$$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}[\frac{dy}{dt}\frac{1}{x}]$$

When I try I get

$$= -\frac{1}{x^{2}}\frac{dy}{dt}+ \frac{1}{x}\left(\frac{dy}{dx}\frac{d^2y}{dt^2}\right)$$

Please. Going round in circles

oh and merry christmas!

 

[HallsofIvy]

arghh! I thought so but how do I get

d2y/dx2=d2y/dt2.1/x2-dy/dt.1/x2
from

$$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}[\frac{dy}{dt}\frac{1}{x}]$$

Yes, that was exactly what I said in post #11

When I try I get

$$= -\frac{1}{x^{2}}\frac{dy}{dt}+ \frac{1}{x}\left(\frac{dy}{dx}\frac{d^2y}{dt^2}\right)$$

No,
$$\frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\right)= \frac{d}{dx}(1/x)}\frac{dy}{dt}+ \frac{1}{x}\frac{d}{dx}\left(\frac{dy}{dt}\right)$$
$$= -\frac{1}{x^2}\frac{dy}{dt}+ \frac{1}{x}\left(\frac{1}{x}\frac{d}{dt}\left(\frac{dy}{dt}\right)\right)$$
$$= \frac{1}{x^2}\left(\frac{d^2y}{dt^2}- \frac{dy}{dt}\right)$$

Again, that was given in post #11.

Please. Going round in circles

oh and merry christmas!

 

[thomas49th]

So because d/dx is really dy/dx (this is where I'm going wrong and I still not that familiar with it)

dy/dx = dy.dt * 1/x
and so substitute that into the product rule expression

WHEY!
AT LAST. IT'S NOT THAT HARD!

Thank you!

 

[thomas49th]

I've come to look back over this and again I'm lost :(

Yes, that was exactly what I said in post #11 No,
$$\frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\right)= \frac{d}{dx}(1/x)}\frac{dy}{dt}+ \frac{1}{x}\textcolor{red}{\frac{d}{dx}\left(\frac{dy}{dt}\right)}$$
$$= -\frac{1}{x^2}\frac{dy}{dt}+ \frac{1}{x}\left(\textcolor{red}{\frac{1}{x}\frac{d}{dt}\left(\frac{dy}{dt}\right)}\right)$$
$$= \frac{1}{x^2}\left(\frac{d^2y}{dt^2}- \frac{dy}{dt}\right)$$

Again, that was given in post #11.

Can someone please show me at the simpliest of simpleist, with the chain rule how the part in red it done. I cannot see how to differentiate

$$\frac{d}{dx}\frac{dy}{dt}$$

I know $$\frac{dy}{dx} = \frac{1}{x}\frac{dy}{dt}$$ but I don't see how that is used.

Thank you

 

[hunt_mat]

The first line is just the product rule.

 

[thomas49th]

I have just had another attempt using implicit differentiation

$$\frac{d}{dx}[\frac{dy}{dt}] = \frac{1}{x}\frac{d^{2}y}{dt^2}\frac{dt}{dt}$$But you got a one term minus the other while I got the product

Hmmm :(

 

[hunt_mat]

It's like this:

$$\frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\right) =\frac{d}{dx}\left(\frac{1}{x}\right)\frac{dy}{dt}+\frac{1}{x}\frac{d}{dx}\frac{dy}{dt}$$

But:

$$\frac{d}{dx}=\frac{dt}{dx}\frac{d}{dt}=\frac{1}{\dot{x}}\frac{d}{dt}$$

So

$$\frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\right) =\frac{d}{dx}\left(\frac{1}{x}\right)\frac{dy}{dy}+\frac{1}{x\dot{x}}\frac{d^{2}y}{dt^{2}}$$

 

[thomas49th]

I know dt/dx = 1/x. Am I right in thinking

d/dx = dy/dx

How and why?$$\frac{d}{dx}=\frac{dt}{dx}\frac{d}{dt}=\frac{1}{\d ot{x}}\frac{d}{dt}$$also what is $$\frac{1}{x}\frac{\textcolor{red}{d}}{dt}$$

what does d mean on it's own. I thought it needs something to operate on?

Thanks
Thomas

 

[hunt_mat]

It's an operator, it acts on things to give something else, so if you like:

$$\frac{df}{dx}=\frac{dt}{dx}\frac{df}{dt}$$

via the chain rule..

 

[thomas49th]

Okay. But

$$\frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\right) =\frac{d}{dx}\left(\frac{1}{x}\right)\frac{dy}{dy} +\frac{1}{x\dot{x}}\frac{d^{2}y}{dt^{2}}$$

did you mean to write dy/dy or dy/dt? Please don't jump steps quickly :( Also please please please show me how you got the term $$\frac{1}{x\dot{x}}\frac{d^{2}y}{dt^{2}}$$

I see the 1/x must be there because of the product rule but how did the dt/dx (1/x dot) come in?

Thanks :D

 

[hunt_mat]

Standard terminology really, sorry if i didn't explain.

$$\dot{x}=\frac{dx}{dt}$$

You know the chain rule?

$$\frac{df}{dx}=\frac{dt}{dx}\frac{df}{dt}$$

Let:

$$f=\frac{dy}{dt}$$

and you have your answer.

 

[thomas49th]

Okay,

$$\frac{df}{dx}=\frac{dt}{dx}\frac{df}{dt}$$

You were using f but in our case we set y = dy/dt?

Thanks
I think I see ;)

 

[HallsofIvy]

In order to avoid all this nonsense about tranformations, just look for solutions of the form x^n of your original equation

In other words just don't do the problem you are given?

There are many good reasons for knowing that this substitution will change an Euler-type (or "equipotential") equation to an equation with constant coefficients having the same characteristic equation.

For example, how would you solve
$$\frac{d^2y}{dx^2}+ 3x\frac{dy}{dx}+ y= cos(ln(x))$$
by letting $y= x^r$?

 

[hunt_mat]

For this particular problem why not do as I suggested? It's a valid teachnique, it's a known solution.