Proving 2nd Order Differential Eqns

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Homework Help Overview

The problem involves finding the value of x for the second derivative of a function, specifically when d²y/dx² = 0, where y is given as (3x)/e^(2x). The subject area pertains to differential equations and differentiation techniques.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to differentiate the function but encounters confusion regarding the implications of their results. Another participant provides a differentiation process, leading to a potential solution for x. There is also a light-hearted exchange regarding a misunderstanding of exponentiation.

Discussion Status

The discussion includes attempts to clarify differentiation methods and the implications of the results. While one participant suggests a value for x, there is no explicit consensus on the correctness of the approach or the interpretation of the results.

Contextual Notes

Participants express uncertainty about differentiation and the rules governing exponentiation, indicating a need for clarification on these concepts. The original poster mentions difficulties with differentiation, which may affect their understanding of the problem.

TannY
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Homework Statement


If y = (3x)/e^2x, find the value of x when d^2y/dx^2 = 0


Homework Equations


I'll just abbv d^2y/dx^2 as d2ydx2

The Attempt at a Solution


I kept getting stuck at:
d2ydx2 = 6/e^2
When d2ydx2 = 0,

6/e^2 = 0
6 = e^2

Then where's my x?? :confused: :confused:

P.S: i have serious issues with differentiation =.=
 
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y = 3x*e^(-2x)
y' = 3e^(-2x) - 6xe^(-2x)
y'' = -6e^(-2x) - 6e^(-2x) + 12xe^(-2x) = 0
-12e^(-2x) + 12xe^(-2x) = 0
12xe^(-2x) = 12e^(-2x)
x = 1

Differentiated using the chain rule (for exponentials) and product rule.
 
thanx! ^^
i thought that (e^2x)squared = (e^4x) lol :biggrin:
 
TannY said:
thanx! ^^
i thought that (e^2x)squared = (e^4x) lol :biggrin:

It is.

Filler Filler Filler.
 

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