Question

Show that for any constants M, k, and a, the function

y(t) = 1/2 M (1+ tanh (k(t – a) /2)

satishes the logistic equation: y'/y = k(1 - y/M)

Answer 1

The question involves a mathematical proof in calculus, demonstrating that a given function involving a hyperbolic tangent satisfies a logistic differential equation. By differentiating the function and manipulating it to match the form of the equation, this proof is verified step-by-step using calculus techniques.

Step-by-step explanation:

The student has posed a mathematics question that involves showing that the function y(t) = 1/2 M (1+ tanh (k(t – a) /2)) satisfies the logistic differential equation y'/y = k(1 - y/M). We will start by finding the derivative of the given function with respect to t (denoted as y'), apply the properties of tanh, and then compare the derivative to the original function multiplied by the expression on the right side of the logistic equation.

First, we find the derivative of y(t):

1. Differentiate the tanh function using the chain rule, recalling that the derivative of tanh(u) with respect to u is sech2(u), which is (1-tanh2(u)).
2. As we find the derivative, we account for the constants and the chain rule as it applies to k(t-a)/2.
3. After simplifying, we obtain the expression y'(t) = 1/4 M k sech2(k(t-a)/2).

Next, we verify that this derivative satisfies the logistic equation:

1. Rewrite sech2(u) as (1 - tanh2(u)) according to its definition.
2. Substitute y(t) into the expression (1 - y/M).
3. Compare the resulting expression with y'/y and show that they are equivalent.

After completing these steps, you will find that y'/y = k(1 - y/M) is satisfied, proving that y(t) is a solution to the logistic equation.