Solve Integral Problem: dy/dx=7sec^2(pix/4)tan(pix/4), y'(1)=0, y(1)=5

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Homework Help Overview

The discussion revolves around solving a differential equation related to the second derivative of a function, specifically focusing on the equation d²y/dx² = 7sec²(πx/4)tan(πx/4) with initial conditions y'(1)=0 and y(1)=5.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the derivation of the first derivative and the integration process, noting discrepancies in their calculations. Some participants suggest that the first derivative should involve tan² and reference the identity 1 + tan²(...) = sec²(...).

Discussion Status

The discussion is ongoing, with participants attempting to verify their calculations and seeking confirmation on their approaches. There is a suggestion to simplify the expressions, indicating some progress in the exploration of the problem.

Contextual Notes

Participants express uncertainty about their calculations and seek clarification on the integration steps and the application of trigonometric identities. There is an acknowledgment of potential errors in the initial attempts.

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Homework Statement


d2y/dx2 = 7sec^2(pix/4)tan(pix/4)
y'(1)=0
y(1)=5


Homework Equations





The Attempt at a Solution



dy/dx = 14/pi sec^2(pix/4)-28/pi
y = 56tan(pix/4)/pi^2-28x/pi+5-56/pi^2+28/pi


Not sure where I went wrong, but I checked on my calc and it wasn't equal.
 
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The first derative should be something with tan^2(..).
 
1 + tan^2(...) = sec^2(...) but 1 is just part of c

EDIT:

dy/dx = 14/pi tan^2(pix/4) - 14/pi

y = 56tan(pix/4)/pi^2 - 56(pix/4)/pi^2 - 14x/pi + 5 - 56/pi^2 + 28/picould anyone check this please?
 
Last edited:
Chas3down said:
1 + tan^2(...) = sec^2(...) but 1 is just part of c

EDIT:

dy/dx = 14/pi tan^2(pix/4) - 14/pi

y = 56tan(pix/4)/pi^2 - 56(pix/4)/pi^2 - 14x/pi + 5 - 56/pi^2 + 28/pi


could anyone check this please?

Looks right, but you can simplify a little. Two terms are the same.
 

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