Solve : 9 cosh9y dy = 4 sinh 4x dx

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The discussion focuses on solving the equation 9 cosh(9y) dy = 4 sinh(4x) dx through integration. The initial integration steps presented are incorrect due to a misunderstanding of the signs when integrating hyperbolic functions. The correct integrals yield sinh(9y) + C for the left side and cosh(4x) + C for the right side. The conversation also notes that to find an explicit equation, initial conditions would be necessary for simplification. Overall, the importance of correctly applying integration rules for hyperbolic functions is emphasized.
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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:
\int \sinh x \ dx =\cosh x +C !

Daniel.
 
9\int\cosh{9y}dy = \frac{9\sinh{9y}}{9} + C = {\sinh{9y}} + C

4\int\sinh{4x}dx = \frac{4\cosh{4x}}{4} + C = {\cosh{4x} + C
 
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