Solve : 9 cosh9y dy = 4 sinh 4x dx

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Discussion Overview

The discussion revolves around the integration of hyperbolic functions, specifically the equation 9 cosh(9y) dy = 4 sinh(4x) dx. Participants are examining the integration process and the correctness of the resulting expressions.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents an integration attempt, suggesting that 9/9 sinh(9y) + C = -cosh(4x) is a valid result.
  • Another participant challenges this result, stating that the sign does not change when integrating sinh or cosh and implies that further simplification may be necessary, potentially requiring initial conditions.
  • A third participant confirms the integral of sinh x, stating that it equals cosh x + C.
  • A fourth participant provides a detailed breakdown of the integrals, showing that 9∫cosh(9y) dy results in sinh(9y) + C and 4∫sinh(4x) dx results in cosh(4x) + C.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of the initial integration attempt. There are competing views on the integration process and the implications of signs in the results.

Contextual Notes

Some assumptions about the integration constants and initial conditions are not explicitly stated, which may affect the interpretation of the results.

Naeem
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I did this:

9 Integral cosh9ydy = 4 Integral sinh4xdx

9/9 sinh9y + C = -cosh4x
C = - cosh4x -sinh9y

Is this right , or wrong, or is there more to it.
 
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This is wrong, you will find the sign does not change when integrating sinh or cosh.

Also if you wanted an explicit equation, there are ways of simplifying it (actually you'd probabily need some initial conditions).
 
Last edited:
[tex]\int \sinh x \ dx =\cosh x +C[/tex] !

Daniel.
 
[tex]9\int\cosh{9y}dy = \frac{9\sinh{9y}}{9} + C = {\sinh{9y}} + C[/tex]

[tex]4\int\sinh{4x}dx = \frac{4\cosh{4x}}{4} + C = {\cosh{4x} + C[/tex]
 
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