Question

The function f(x)=Acosh(Cx)+Bsinh(Cx). I need help determining the value of constant C>0 when the second derivative f''(x)=25(f(x)).

Answer 1

C = 5.

Explanation:

First, you need to remember that:

For the function:

h(x) = Sinh(k*x)

We have:

h'(x) = k*Cosh(k*x)

and for the Cosh function:

g(x) = Cosh(k*x)

g'(x) = k*Cosh(k*x).

Now let's go to our problem:

We have f(x) = A*cosh(C*x) + B*Sinh(C*x)

We want to find the value of C such that:

f''(x) = 25*f(x)

So let's derive f(x):

f'(x) = A*C*Sinh(C*x) + B*C*Cosh(C*x)

and again:

f''(x) = A*C*C*Cosh(C*x) + B*C*C*Sinh(C*x)

f''(x) = C^2*(A*cosh(C*x) + B*Sinh(C*x)) = C^2*f(x)

And we wanted to get:

f''(x) = 25*f(x) = C^2*f(x)

then:

25 = C^2

√25 = C

And because we know that C > 0, we take the positive solution of the square root, then:

C = 5