Undergrad Cauchy-Euler Diff. Eqn: Transformation & Solution

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The Cauchy-Euler differential equation can be transformed using the substitution x = e^t to facilitate its solution. The transformation requires converting derivatives, specifically transitioning from dy/dx to dy/dt, which involves applying the chain rule. The derivative dy/dx can be expressed as (dy/dt)(1/x), leading to the second derivative d²y/dx² being derived using the product rule. The discussion highlights confusion around the term df/dx and clarifies that it can be expressed as (df/dt)(dt/dx). Overall, the transformation process and derivative relationships are essential for solving the Cauchy-Euler equation effectively.
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Cauchy-Euler is a type of diff equation which is described by

$$a_0x^2(\frac {d^2y} {dx^2})+a_1x(\frac {dy} {dx})+a_2y=F(x)$$

The transformation of ##x=e^t## can solve the equation.

Now, in here I didnt understand how to transform ##\frac {dy} {dx}## to ##\frac {dy} {dt}##.

it goes like this ##\frac {dy} {dx}=\frac {dy} {dt} \frac {1} {x}## and then I am stuck I should take another derivative but I couldn't do it somehow.
 
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Let ##dy/dt = f## and apply the product rule for derivatives when differentiating ##f/x## wrt x.
 
Orodruin said:
Let ##dy/dt = f## and apply the product rule for derivatives when differentiating ##f/x## wrt x.
Okay hmm

##\frac {dy} {dx}=\frac {f} {x}##
##\frac {d^2y} {dx^2}=\frac {df} {dx} \frac {1} {x}-\frac {f} {x^2}##

now I understand until here, but I didnt understand this term ##\frac {df} {dx}=\frac {d^2y} {dt^2}\frac {1} {x}## which its done in the book.
 
What is ##df/dx## in terms of ##df/dt##?
 
Orodruin said:
What is ##df/dx## in terms of ##df/dt##?
##df/dx=(df/dt)(dt/dx)##

I understand it now, thanks :angel:
 

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