- #1
jonjacson
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Homework Statement
Transform the equation:
x2 * d2y/dx2 + 2 * x * dy/dx + (a2/x2)*y = 0
Using:
x=1/t
Homework Equations
The differential of a function of several variables, and the common rules of differentiation.
https://en.wikipedia.org/wiki/Derivative
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?