# Homework Help: Huperbolic Equation Help

1. Oct 24, 2009

### fan_103

1.Solve the equation sinh2x+ 4coshx = 6sinhx+ 12 giving your answers in terms of natural logarithms

2. Sinh2x= 2SinhxCoshx
Cosh2x=2Cosh^2x-1

3. 2SinhxCoshx+4Coshx =6Sinhx+12
I don't know what to do next plz help me!

Last edited: Oct 24, 2009
2. Oct 24, 2009

### fan_103

I ve got the answer! ;)

3. Oct 24, 2009

### Staff: Mentor

Can you share it with us? I filled in most of a page last night, but stopped before getting solutions for x. There are two other identities that you didn't list: cosh x = (1/2)(ex + e-x) and sinh x = (1/2)(ex - e-x).

I converted the equation using these identities and eventually got to
e2x - 2ex - e-2x - 10e-x = 24.
I completed the square in the first pair of terms and the last pair of terms on the left side, and got to this equation:
(ex - 1)2 = (e-x + 5)2
or
ex - 1 = +/-(e-x + 5)

It was getting past my bedtime, so I quit there.

After thinking about it a bit more this morning, I can see that the equation above can be split into two cases. Multiplying by ex yields these two equations:
e2x - 6ex - 1 = 0
e2x + 4ex + 1 = 0

Each of these is a quadratic in ex, so we can solve for ex in each of them, and finally take the natural log to get x.

Is that about what you did?

4. Oct 25, 2009

### fan_103

Re: Hyperbolic Equation Help

Its very easy indeed mate.No need to use $$e^-x$$
Sinh$$2x$$+4Cosh$$x$$=6Sinh$$x$$+12
2Sinh$$x$$Cosh$$x$$+4Cosh$$x$$-6Sinh$$x$$+12=0
2Cosh$$x$$(Sinh$$x$$+2)-6(Sinh$$x$$+2)=0
(2Cosh$$x$$-6)(Sinh$$x$$+2)=0

Either 2Cosh$$x$$-6=0
Coshx=3
$$x$$=Cosh^-1(3)
=ln(3+$$\sqrt{9-1}$$)
=ln(3+$$\sqrt{6}$$)
Or Sinh$$x$$=-2
$$x$$=Sinh^-1(-2)
=ln (-2+$$\sqrt{4+1}$$)
=ln($$\sqrt{5}$$)-2)
=ln(-2+$$\sqrt{5}$$)

5. Oct 25, 2009

### Staff: Mentor

Re: Hyperbolic Equation Help

Your way is simpler than mine, but after correcting a sign error I made, I get three solutions to your two. BTW, there is an error in the first solution you show. sqrt(9 - 1) = sqrt(8), not sqrt(6). So your two solutions are x = ln(3 + 2sqrt(2)) and x = ln(-2 + sqrt(5)).

My equation (ex - 1)2 = (e-x + 5)2 should have been (ex - 1)2 = (e-x - 5)2

so that ex - 1 = +/-(e-x - 5)

e2x - 6ex + 1 = 0 and
e2x + 4ex - 1 = 0

Substituting u = ex

We get u2 -6u + 1 = 0, and
u2 + 4u - 1 = 0

Using the quadratic formula on the first, we get
u = 3 +/- 2sqrt(2), so x = ln(3 +/- 2sqrt(2))

Using the quadratic formula on the second, we get
u = -2 + sqrt(5), so x = ln(-2 + sqrt(5))

The solution I have but you don't is x = ln(3 - 2sqrt(2)).

It's possible that your missing solution came from using cosh-1, but I don't know this for sure.