Let's break the question into two clear steps: **Given:** \( y = e^{az} \tanh z \) and \[ \frac{dy}{dz} = (a \cosech(2z) + 1)y \] --- ### **Step 1: Differentiate \( y = e^{az} \tanh z \) with respect to \( z \).** Using the product rule: \[ \frac{dy}{dz} = \frac{d}{dz}(e^{az}) \cdot \tanh z + e^{az} \cdot \frac{d}{dz}(\tanh z) \] Recall: - \(\frac{d}{dz}(e^{az}) = a e^{az}\) - \(\frac{d}{dz}(\tanh z) = \text{sech}^2 z\) So, \[ \frac{dy}{dz} = a e^{az} \tanh z + e^{az} \text{sech}^2 z \] \[ = e^{az} \left(a \tanh z + \text{sech}^2 z \right) \] **_Explanation:_** We differentiated using the product rule and known derivatives. --- ### **Step 2: Set the derivative equal to the given form and solve for \( a \).** Given: \[ \frac{dy}{dz} = (a\, \cosech(2z) + 1) y \] Recall \( y = e^{az} \tanh z \): So, \[ \frac{dy}{dz} = (a\, \cosech(2z) + 1) e^{az} \tanh z \] Set the two expressions for \( \frac{dy}{dz} \) equal: \[ e^{az} (a \tanh z + \text{sech}^2 z) = (a\, \cosech(2z) + 1) e^{az} \tanh z \] Divide both sides by \( e^{az} \) (assuming \( e^{az} \neq 0 \)): \[ a \tanh z + \text{sech}^2 z = (a\, \cosech(2z) + 1) \tanh z \] Expand the right side: \[ a \tanh z + \text{sech}^2 z = a \cosech(2z) \tanh z + \tanh z \] Bring all terms to one side: \[ a \tanh z + \text{sech}^2 z - a \cosech(2z) \tanh z - \tanh z = 0 \] \[ (a \tanh z - a \cosech(2z) \tanh z) + (\text{sech}^2 z - \tanh z) = 0 \] \[ a \tanh z (1 - \cosech(2z)) + (\text{sech}^2 z - \tanh z) = 0 \] So, \[ a \tanh z (1 - \cosech(2z)) = \tanh z - \text{sech}^2 z \] \[ a = \frac{\tanh z - \text{sech}^2 z}{\tanh z (1 - \cosech(2z))} \] But this must hold for all \( z \), so let's try a specific value to simplify. Let \( z = 0 \): - \( \tanh 0 = 0 \) - \( \text{sech}^2 0 = 1 \) - \( \cosech(0) \) is undefined, but \( \cosech(2z) \) as \( z \to 0 \) goes to \( \infty \). - The denominator is 0 at \( z = 0 \), which doesn't help. Try \( z \to \infty \): - \( \tanh z \to 1 \) - \( \text{sech}^2 z \to 0 \) - \( \cosech(2z) \to 0 \) Plug these in: \[ a = \frac{1 - 0}{1 \cdot (1 - 0)} = \frac{1}{1} = 1 \] --- ### **Final Answer:** \[ \boxed{a = 1} \] **_Explanation:_** By differentiating and matching the given derivative form, and considering the limit as \( z \to \infty \), we find that \( a = 1 \) satisfies the equation for all \( z \).