Solving Hyperbolic Equations

This lesson covers techniques for solving equations involving hyperbolic functions: using exponential definitions, using identities to simplify, and recognising quadratic forms.

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1 Using Exponential Definitions

The most direct method: replace sinh, cosh, tanh with their exponential definitions, then solve.

Worked Example 1

Solve sinh x = 2.

(e^x - e^(-x))/2 = 2 leads to e^(2x) - 4e^x - 1 = 0.

Let u = e^x: u = 2 + sqrt(5) (rejecting the negative root).

x = ln(2 + sqrt(5))

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2 Worked Example 2

Solve cosh x = 3.

e^(2x) - 6e^x + 1 = 0 gives u = 3 +/- 2 sqrt(2).

Both roots are positive, so: x = +/- ln(3 + 2 sqrt(2))

(cosh is even, so two solutions when k > 1)

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3 Worked Example 3

Solve tanh x = 3/5.

(e^x - e^(-x))/(e^x + e^(-x)) = 3/5

5e^x - 5e^(-x) = 3e^x + 3e^(-x)

2e^x = 8e^(-x), so e^(2x) = 4.

x = ln 2

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4 Using Identities

For equations with mixed or squared terms, use identities to reduce to one function.

Worked Example 4

Solve 2 cosh^2 x - 5 sinh x = 5.

Using cosh^2 x = 1 + sinh^2 x:

2 + 2 sinh^2 x - 5 sinh x = 5

2 sinh^2 x - 5 sinh x - 3 = 0

(2 sinh x + 1)(sinh x - 3) = 0

sinh x = -1/2: x = ln((-1 + sqrt(5))/2)

sinh x = 3: x = ln(3 + sqrt(10))

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5 Worked Example 5

Solve cosh x - 2 sinh x = 3.

Substituting definitions:

(e^x + e^(-x))/2 - 2(e^x - e^(-x))/2 = 3

(-e^x + 3e^(-x))/2 = 3

Multiply by e^x: -e^(2x) + 3 = 6e^x, so e^(2x) + 6e^x - 3 = 0.

u = -3 + 2 sqrt(3) (taking the positive root).

x = ln(-3 + 2 sqrt(3))

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6 Worked Example 6

Solve cosh^2 x - sinh x = 1.

Using cosh^2 x = 1 + sinh^2 x:

sinh^2 x - sinh x = 0

sinh x(sinh x - 1) = 0

x = 0 or x = arsinh(1) = ln(1 + sqrt(2))

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7 Worked Example 7

Given cosh x + sinh x = 5, find x.