What is the general solution for solvefor y,x=sinh(y) ?

The general solution for solvefor y,x=sinh(y) is y=ln(x+sqrt(x^2+1)),y=ln(x-sqrt(x^2+1))

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solvefor y,x=sinh(y)

Solution

solve for

y, x = sinh (y)

Solution

y = ln (x + x^{2} + 1), y = ln (x - x^{2} + 1)

Solution steps

x = sinh (y)

Switch sides

sinh (y) = x

Rewrite using trig identities

sinh (y) = x

Use the Hyperbolic identity:

sinh (x) = \frac{e ^{x} - e ^{- x}}{2}

\frac{e ^{y} - e ^{- y}}{2} = x

\frac{e ^{y} - e ^{- y}}{2} = x

\frac{e ^{y} - e ^{- y}}{2} = x : y = ln (x + x^{2} + 1), y = ln (x - x^{2} + 1)

\frac{e ^{y} - e ^{- y}}{2} = x

Multiply both sides by

2

\frac{e ^{y} - e ^{- y}}{2} \cdot 2 = x \cdot 2

Simplify

e^{y} - e^{- y} = x \cdot 2

Apply exponent rules

e^{y} - e^{- y} = x \cdot 2

Apply exponent rule:

a^{b c} = (a^{b})^{c}

e^{- y} = (e^{y})^{- 1}

e^{y} - (e^{y})^{- 1} = x \cdot 2

e^{y} - (e^{y})^{- 1} = x \cdot 2

Rewrite the equation with

e^{y} = u

u - u^{- 1} = x \cdot 2

Solve

u - u^{- 1} = x \cdot 2 : u = x + x^{2} + 1, u = x - x^{2} + 1

u - u^{- 1} = x \cdot 2

Refine

u - \frac{1}{u} = x \cdot 2

Multiply both sides by

u

u - \frac{1}{u} = x \cdot 2

Multiply both sides by

u

uu - \frac{1}{u} u = x \cdot 2 u

Simplify

uu - \frac{1}{u} u = x \cdot 2 u

Simplify

uu : u^{2}

uu

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= u^{1 + 1}

Add the numbers:

1 + 1 = 2

= u^{2}

Simplify

- \frac{1}{u} u : - 1

- \frac{1}{u} u

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{1 \cdot u}{u}

Cancel the common factor:

u

= - 1

u^{2} - 1 = x \cdot 2 u

u^{2} - 1 = x \cdot 2 u

u^{2} - 1 = x \cdot 2 u

Solve

u^{2} - 1 = x \cdot 2 u : u = x + x^{2} + 1, u = x - x^{2} + 1

u^{2} - 1 = x \cdot 2 u

Move

x 2 u

to the left side

u^{2} - 1 = x \cdot 2 u

Subtract

x 2 u

from both sides

u^{2} - 1 - x \cdot 2 u = x \cdot 2 u - x \cdot 2 u

Simplify

u^{2} - 1 - x \cdot 2 u = 0

u^{2} - 1 - x \cdot 2 u = 0

Write in the standard form

a x^{2} + b x + c = 0

u^{2} - x \cdot 2 u - 1 = 0

Solve with the quadratic formula

u^{2} - x \cdot 2 u - 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - x 2, c = - 1

u_{1, 2} = \frac{- ( - x \cdot 2 ) \pm ( - x \cdot 2 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - x \cdot 2 ) \pm ( - x \cdot 2 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

Simplify

(- x \cdot 2)^{2} - 4 \cdot 1 \cdot (- 1) : 2 x^{2} + 1

(- x \cdot 2)^{2} - 4 \cdot 1 \cdot (- 1)

Apply rule

- (- a) = a

= (- x \cdot 2)^{2} + 4 \cdot 1 \cdot 1

(- x \cdot 2)^{2} = 2^{2} x^{2}

(- x \cdot 2)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2 x)^{2} = (2 x)^{2}

= (2 x)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} x^{2}

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4

= 2^{2} x^{2} + 4

Factor

x^{2} 2^{2} + 4 : 4 (x^{2} + 1)

x^{2} \cdot 2^{2} + 4

Rewrite as

= 4 x^{2} + 4 \cdot 1

Factor out common term

4

= 4 (x^{2} + 1)

= 4 (x^{2} + 1)

Apply radical rule:

assuming

a \geq 0, b \geq 0

= 4 x^{2} + 1

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= 2 x^{2} + 1

u_{1, 2} = \frac{- ( - x \cdot 2 ) \pm 2 x ^{2} + 1}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - x \cdot 2 ) + 2 x ^{2} + 1}{2 \cdot 1}, u_{2} = \frac{- ( - x \cdot 2 ) - 2 x ^{2} + 1}{2 \cdot 1}

u = \frac{- ( - x 2 ) + 2 x ^{2} + 1}{2 \cdot 1} : x + x^{2} + 1

\frac{- ( - x \cdot 2 ) + 2 x ^{2} + 1}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{x \cdot 2 + 2 x ^{2} + 1}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2 x + 2 x ^{2} + 1}{2}

Factor out common term

2

= \frac{2 ( x + x ^{2} + 1 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= x + x^{2} + 1

u = \frac{- ( - x 2 ) - 2 x ^{2} + 1}{2 \cdot 1} : x - x^{2} + 1

\frac{- ( - x \cdot 2 ) - 2 x ^{2} + 1}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{x \cdot 2 - 2 x ^{2} + 1}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2 x - 2 x ^{2} + 1}{2}

Factor out common term

2

= \frac{2 ( x - x ^{2} + 1 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= x - x^{2} + 1

The solutions to the quadratic equation are:

u = x + x^{2} + 1, u = x - x^{2} + 1

u = x + x^{2} + 1, u = x - x^{2} + 1

u = x + x^{2} + 1, u = x - x^{2} + 1

Substitute back

u = e^{y},

solve for

y

Solve

e^{y} = x + x^{2} + 1 : y = ln (x + x^{2} + 1)

e^{y} = x + x^{2} + 1

Apply exponent rules

e^{y} = x + x^{2} + 1

f (x) = g (x)

, then

ln (f (x)) = ln (g (x))

ln (e^{y}) = ln (x + x^{2} + 1)

Apply log rule:

ln (e^{a}) = a

ln (e^{y}) = y

y = ln (x + x^{2} + 1)

y = ln (x + x^{2} + 1)

Solve

e^{y} = x - x^{2} + 1 : y = ln (x - x^{2} + 1)

e^{y} = x - x^{2} + 1

Apply exponent rules

e^{y} = x - x^{2} + 1

f (x) = g (x)

, then

ln (f (x)) = ln (g (x))

ln (e^{y}) = ln (x - x^{2} + 1)

Apply log rule:

ln (e^{a}) = a

ln (e^{y}) = y

y = ln (x - x^{2} + 1)

y = ln (x - x^{2} + 1)

y = ln (x + x^{2} + 1), y = ln (x - x^{2} + 1)

y = ln (x + x^{2} + 1), y = ln (x - x^{2} + 1)

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor y,x=sinh(y) ?
The general solution for solvefor y,x=sinh(y) is y=ln(x+sqrt(x^2+1)),y=ln(x-sqrt(x^2+1))