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

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SUMMARY

The integral problem presented involves the second derivative equation d²y/dx² = 7sec²(πx/4)tan(πx/4) with initial conditions y'(1)=0 and y(1)=5. The user derived the first derivative as dy/dx = 14/π tan²(πx/4) - 14/π and the function y as 56tan(πx/4)/π² - 56(πx/4)/π² - 14x/π + 5 - 56/π² + 28/π. A suggestion was made to simplify the expression as two terms were found to be identical. The discussion emphasizes the importance of correctly applying trigonometric identities in calculus.

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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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