Solve the hyperbolic equation: 2cosh 2x +10sinh2x=5
Answer (1)
Rewrite the hyperbolic functions in terms of exponentials.
Simplify the equation and transform it into a quadratic equation by substitution.
Solve the quadratic equation and choose the positive solution.
Take the natural logarithm to solve for x .
The solution is x = 2 1 ln ( 3 4 ) ≈ 0.1438 .
x = 2 1 ln ( 3 4 )
Explanation
Problem Statement We are given the hyperbolic equation: 2 cosh 2 x + 10 sinh 2 x = 5
Definitions of Hyperbolic Functions We will use the definitions of hyperbolic cosine and hyperbolic sine: cosh u = 2 e u + e − u sinh u = 2 e u − e − u
Substitution and Simplification Substituting these definitions into the given equation, we get: 2 ( 2 e 2 x + e − 2 x ) + 10 ( 2 e 2 x − e − 2 x ) = 5 Simplifying, we have: e 2 x + e − 2 x + 5 e 2 x − 5 e − 2 x = 5 6 e 2 x − 4 e − 2 x = 5
Quadratic Transformation To eliminate the negative exponent, we multiply both sides of the equation by e 2 x :
6 e 4 x − 4 = 5 e 2 x Rearranging the terms, we obtain a quadratic equation in e 2 x :
6 e 4 x − 5 e 2 x − 4 = 0
Substitution for Quadratic Form Let y = e 2 x . Then the equation becomes: 6 y 2 − 5 y − 4 = 0
Solving the Quadratic Equation We can solve this quadratic equation for y using the quadratic formula: y = 2 a − b ± b 2 − 4 a c where a = 6 , b = − 5 , and c = − 4 . Thus, y = 2 ( 6 ) 5 ± ( − 5 ) 2 − 4 ( 6 ) ( − 4 ) = 12 5 ± 25 + 96 = 12 5 ± 121 = 12 5 ± 11
Possible Values for y The two possible values for y are: y 1 = 12 5 + 11 = 12 16 = 3 4 y 2 = 12 5 − 11 = 12 − 6 = − 2 1
Choosing the Positive Solution Since y = e 2 x , y must be positive. Therefore, we discard the negative solution y 2 = − 2 1 .
So, we have: e 2 x = 3 4
Solving for x Taking the natural logarithm of both sides, we get: 2 x = ln ( 3 4 ) x = 2 1 ln ( 3 4 ) x ≈ 2 1 × 0.2877 ≈ 0.1438
Final Answer Therefore, the solution to the hyperbolic equation is: x = 2 1 ln ( 3 4 ) x ≈ 0.1438
Examples
Hyperbolic functions are used in various fields such as physics and engineering. For example, they appear in the study of catenary curves, which describe the shape of a hanging cable or chain suspended between two points. They are also used in special relativity to describe the velocity addition formula. Solving hyperbolic equations can help determine parameters in models involving these functions, such as finding the sag of a power line or analyzing the trajectory of a relativistic particle.
