3) Show that for any constants $M$, $k$, and $a$, the function
\[
y(t)=\frac{1}{2} M\left(1+\tanh \left(\frac{k(t-a)}{2}\right)\right)
\]
satisfies the logistic equation: $\frac{y^{\prime}}{y}=k\left(1-\frac{y}{M}\right)$.

Step 1 ：Given the function \(y(t)=\frac{1}{2} M\left(1+\tanh \left(\frac{k(t-a)}{2}\right)\right)\)

Step 2 ：Find the derivative of \(y(t)\) with respect to \(t\) using the chain rule

Step 3 ：\(y'(t) = \frac{1}{2} M \cdot \frac{k}{2} \cdot (1 - \tanh^2\left(\frac{k(t-a)}{2}\right))\)

Step 4 ：Simplify to get \(y'(t) = \frac{1}{4} M k \cdot (1 - \tanh^2\left(\frac{k(t-a)}{2}\right))\)

Step 5 ：Substitute \(y(t)\) and \(y'(t)\) into the logistic equation \(\frac{y'(t)}{y(t)} = \frac{\frac{1}{4} M k \cdot (1 - \tanh^2\left(\frac{k(t-a)}{2}\right))}{\frac{1}{2} M\left(1+\tanh \left(\frac{k(t-a)}{2}\right)\right)}\)

Step 6 ：Simplify the right side to get \(\frac{k}{2} \cdot \frac{(1 - \tanh^2\left(\frac{k(t-a)}{2}\right))}{(1+\tanh \left(\frac{k(t-a)}{2}\right))}\)

Step 7 ：Use the identity \(1 - \tanh^2(x) = \text{sech}^2(x)\) and \(\text{sech}(x) = \frac{1}{\cosh(x)}\) to simplify the expression to \(\frac{k}{2} \cdot \frac{\text{sech}^2\left(\frac{k(t-a)}{2}\right)}{(1+\tanh \left(\frac{k(t-a)}{2}\right))}\)

Step 8 ：Further simplify to \(\frac{k}{2} \cdot \frac{1}{\cosh^2\left(\frac{k(t-a)}{2}\right) \cdot (1+\tanh \left(\frac{k(t-a)}{2}\right))}\)

Step 9 ：Use the identity \(\cosh^2(x) = 1 + \tanh^2(x)\) to simplify the expression to \(\frac{k}{2} \cdot \frac{1}{(1 + \tanh^2\left(\frac{k(t-a)}{2}\right)) \cdot (1+\tanh \left(\frac{k(t-a)}{2}\right))}\)

Step 10 ：Further simplify to \(\frac{k}{2} \cdot \frac{1}{(1 + \tanh\left(\frac{k(t-a)}{2}\right))^2}\)

Step 11 ：Finally, simplify to \(k \cdot \left(1 - \frac{y(t)}{M}\right)\)

Step 12 ：Therefore, for any constants \(M\), \(k\), and \(a\), the function \(y(t)\) satisfies the logistic equation. \(\boxed{\text{Proved}}\)