Question

need help understanding reduction to constant coefficients by using x=ln(t)

Answer 1

t=ln(x)

Answer 2

if t=ln(x)

\[\frac{dy}{dx}=\frac{dy}{dt}*\frac{dt}{dx}\]

Answer 3

@mukushla

Answer 4

i understand where they get dt/dx as t=ln(X) so dt/dx = 1/x

Answer 5

but they leave dy/dt the same.

however when you take the second derivative

Answer 6

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

Answer 7

what is y in this case?

Answer 8

or would you assume that \[\frac{d}{dx}=\frac{dy}{dx}\]

Answer 9

y is the same but variable changes

Answer 10

the problem i'm having is where it gets the second derivative in respects of y

Answer 11

t* i mean

Answer 12

\[\frac{d}{dx}[\frac{dy}{dt}\]

Answer 13

i dont see what is going on here,

they end up with

\[\frac{d^2y}{dt^2}*\frac{1}{x}\]

Answer 14

well see this
\[\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx} ----> \frac{dy}{dt}=x \frac{dy}{dx}=e^{t} \frac{dy}{dx}\]

Answer 15

correct

Answer 16

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

Answer 17

am i right?

Answer 18

isnt this part of the chain rule of a second derivative? and yes tha is correct

Answer 19

with a parameter?

Answer 20

we want to change all x's to t's

Answer 21

yeah but this is parameters correct

Answer 22

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

Answer 23

where as you flip the denominator and you get exactly what you got above

Answer 24

u have the
\[\frac{dy}{dx} \]
in terms of t
u can use it for computing
\[\frac{d^2y}{dx^2}\]
in terms of t

Answer 25

and yes u r right

Answer 26

alright thanks