# Homework Help: Change of variables in a differential equation

1. Jan 4, 2017

### jonjacson

1. The problem statement, all variables and given/known data

Transform the equation:

x2 * d2y/dx2 + 2 * x * dy/dx + (a2/x2)*y = 0

Using:

x=1/t

2. Relevant equations

The differential of a function of several variables, and the common rules of differentiation.

https://en.wikipedia.org/wiki/Derivative

3. The attempt at a solution

As I have x as function of t I can calculate dx as function of t and dt. So:

dx = (-1/t2)* dt , equation 1

d2x = (2/t3)*dt2, equation 2 (I considered d2t=0 because it is the independent variable)

To calculate dy/dx I symply change dx by its value at equation 1 , so I get:

dy/dx= dy/ (-1/t2)*dt = -t2 * (dy/dt)

(According to the book this is correct)

Now the problem is d2y/dx2

1.- First question, Why is it a mistake to simply substitute dx2 for its value calculated in equation 2?

2.- According to the book I have to differenciate dy/dx which was equal to:

-t2 dy/dt

I do it simply calculating the differential of a product:

(-2 * t * dt )* dy/dt -t2 * d2y/dt

(I didn't differenciate dt, I assume d2t =0)

Simply dividing this value by dt I get what I expected to be the right result:

(-2*t) * dy/dt -t2 * d2y/dt2

But according to the book, this is wrong, I am missing a term arising from dx/dt.

This is what the book does:

d2y/dx2 = step 1 = d/dx (dy/dx) = step 2 = (d/dt ( dy/dx))/ ( dx/dt ) to get this result:

- ( 2t dy/dt +t2 * d2y/dt2)(-t2)

I understand the step 1 it simply says that the second derivative is the derivative of the first derivative, but the step 2 is a mistery to me and it is the step that creates the term I am missing that is the dx/dt in the denominator. I try to differenciate thinking in terms of differentials, so I can manipulate them in the expressions like they were algebraic quantities, maybe I don't understand the operator version of the derivative.

The last term -t2 is what I didn't get in my calculation.

2.- Question two, in my previous calculation using differentials, Where I missed the dx/dt term?

2. Jan 4, 2017

### BvU

Differentiate to x, not to t !

3. Jan 4, 2017

### jonjacson

I don't get it, if I differenciate dy/dx to x I get d2y/dx2 right? But I am looking for d2y/dt2

4. Jan 4, 2017

### Ray Vickson

Chain rule!
$$\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt},$$
etc.

5. Jan 4, 2017

### vela

Staff Emeritus
It's probably best to avoid using differentials here.

For a test function $f$, the chain rule tells us that
$$\frac{df}{dx} = \frac{dt}{dx} \frac{df}{dt}.$$ In terms of operators, we'd say that
$$\frac{d}{dx} = \frac{dt}{dx} \frac{d}{dt}.$$
To calculate $dy/dx$ then, we'd have
$$\frac{d}{dx} y = \left(\frac{dt}{dx} \frac{d}{dt}\right) y = \frac{dt}{dx} \frac{dy}{dt},$$
and for the second derivative, we'd get
\begin{align*}
\frac{d}{dx} \left(\frac{d}{dx} y\right) &= \left(\frac{dt}{dx} \frac{d}{dt}\right) \left( \frac{dt}{dx} \frac{d}{dt} \right) y \\
&= \left(\frac{dt}{dx} \frac{d}{dt}\right) \left( \frac{dt}{dx} \frac{dy}{dt} \right) \\
&= \frac{dt}{dx} \left[\left(\frac{d}{dt}\frac{dt}{dx}\right) \frac{dy}{dt} + \frac{dt}{dx} \frac{d}{dt} \left( \frac{dy}{dt} \right) \right]
\end{align*}

6. Jan 4, 2017

### vela

Staff Emeritus
You calculated $\frac{d}{dt}\left(\frac{dy}{dx}\right)$, but you appear to want $\frac{d^2 y}{dt^2} = \frac{d}{dt}\left(\frac{dy}{dt}\right).$ What you actually want is $d^2 y/dx^2$ in terms of $y$, $dy/dt$, and $d^2y/dt^2$ to substitute into the original differential equation.

7. Jan 4, 2017

### jonjacson

Thanks folks, now I see it.

8. Jan 5, 2017

### jonjacson

Well, I have tried it hard but I don't get the right result.

d. will mean "the differential of"

So we have:

dy/dt = dy/dx * dx/dt

And I want to differenciate it, an infinitesimal change on t will change dy and dx, so let's calculate it:

d. (dy/dt) = d. (dy/dx * dx/dt)

I regard dt2 as equal to 0 since it is the independent variable.

d2y/dt = d.(dy/dx) * dx/dt + d. (dx/dt) * dy/dx

I apply the quotient rule for dy/dx so I get:

((dx * d2y - d2x * dy )/dx2)

So we have this expression:

d2y/dt= ((dx * d2y - d2x * dy )/dx2) * dx/dt +d2x/dt * dy/dx

d2y/dt= (d2y/dx2) * (dx/dt) * dx - (dx/dt)* (d2x/dx2) *dy + (dy/dx) * (d2x/dt)

Now I calculate d2x/dx2, which is a second order differential divided by the square of the first order differential. As we know that x=1/t I get for this quotient 2t.

Now I divide the whole equation by dt, so finally I get:

d2y/dt2= (d2y/dx2) * (dx/dt) * (dx/dt) - (dx/dt)* (2t) *(dy/dt) + (dy/dx) * (d2x/dt^2)

Now I need to substitute the derivatives of x over t, eliminate dy/dx which was known from the start, and get d2y/dx2 as function of derivatives respect to t.

I use:

dx/dt= -1/t2

d2x/dt2 = 2/t^3

dy/dx = -t2 * dy/dt

d2y/dt2= (d2y/dx2) * (1/t4) - (-1/t2)* (2t) *(dy/dt) + (-t2 * dy/dt) * (2/t3)

The last two terms cancel out and I only get this result:

d2y/dt2= (d2y/dx2) * (1/t4)

Which is partially correct, but I failed to get an extra term since according to the book the right result is:

d2y/dx2 = 2t3 * dy/dt + t4 * d2y/dt2

So folks, Where is my mistake? Is there a fundamental conceptual problem differentiating? Or is it a small mistake somewhere?

I apologize for my lack of understanding.

9. Jan 5, 2017

### BvU

You can't. You also cannot ignore dx2 in the original differential equation.

Did you read Vela's #5 and #6 ? What is the reason you ignore those ?

10. Jan 5, 2017

### BvU

dy/dx is not a quotient. It is a differential quotient.
You should get $\displaystyle {d^2y\over dx^2} dx$ because $\displaystyle{d\over dx}\left (\displaystyle{dy\over dx} \right ) = \displaystyle{d^2 y\over dx^2}$

11. Jan 5, 2017

### jonjacson

I didn't ignore, its just I read Euler Calculi differentialis and is the way I know to differenciate. I wanted to check if I got the right result but apparently I got it totally wrong and this was a disaster.

12. Jan 5, 2017

### Ray Vickson

Perhaps it would be better to eliminate the "differential" signs, and write the original DE as
$$x^2 y''(x) + 2 x y'(x) +\frac{a^2}{x^2} y(x) = 0$$.
You want to change variables to $Y(t) = \left. y(x) \right|_{x=1/t} = y(1/t)$.

The chain rule gives you
$$Y'(t) = \left. y'(x) \right|_{x=1/t} \: \cdot \frac{d}{dt} \frac{1}{t} = -\frac{1}{t^2} y'(1/t)$$.
In a similar way you can get $Y''(t) = (d/dt) Y'(t)$ in terms of $y'(x)$ and $y''(x)$, evaluated at $x = 1/t$.

13. Jan 5, 2017

### BvU

Not totally wrong, no disaster: the dx * d2y /dx2 is there alright !

So far, the step to get $\displaystyle {dy\over dx}$ in terms of $t$ has been successful. Next step is to get $\displaystyle {d^2 y\over dx^2 }$ using a similar procedure. As Ray says (and Vela and ..). Give it a try. $\displaystyle {d^2 y\over dt^2 }$ will pop up, don't worry.