(d) Determine (by showing all the steps) the equilibrium point(s) of the below differential equation, by strictly using an analytical approach (4 marks penalty if inspection is used). \[ \frac{d y}{d t}=-65-2\left(\sqrt{y^{3}-1}\right)^{-1}+3\left(\sqrt{y^{3}-1}\right)^{-1} \]

To determine the equilibrium point(s) of the differential equation \[ \frac{dy}{dt} = -65 - 2\left(\sqrt{y^{3}-1}\right)^{-1} + 3\left(\sqrt{y^{3}-1}\right)^{-1}, \] we need to find the value(s) of \( y \) where \( \frac{dy}{dt} = 0 \). ### Step 1: Set the Differential Equation to Zero Set \( \frac{dy}{dt} = 0 \): \[ 0 = -65 - 2\left(\sqrt{y^{3}-1}\right)^{-1} + 3\left(\sqrt{y^{3}-1}\right)^{-1} \] ### Step 2: Simplify the Equation Combine the terms involving \( \left(\sqrt{y^{3}-1}\right)^{-1} \): \[ 0 = -65 + \left(-2 + 3\right)\left(\sqrt{y^{3}-1}\right)^{-1} \] \[ 0 = -65 + \left(1\right)\left(\sqrt{y^{3}-1}\right)^{-1} \] \[ 65 = \left(\sqrt{y^{3}-1}\right)^{-1} \] ### Step 3: Solve for \( \sqrt{y^{3}-1} \) Take the reciprocal of both sides: \[ \sqrt{y^{3}-1} = \frac{1}{65} \] ### Step 4: Square Both Sides Square both sides to eliminate the square root: \[ y^{3} - 1 = \left(\frac{1}{65}\right)^{2} \] \[ y^{3} - 1 = \frac{1}{4225} \] ### Step 5: Solve for \( y^{3} \) \[ y^{3} = 1 + \frac{1}{4225} \] \[ y^{3} = \frac{4225}{4225} + \frac{1}{4225} = \frac{4226}{4225} \] ### Step 6: Take the Cube Root \[ y = \sqrt[3]{\frac{4226}{4225}} \] ### Conclusion The equilibrium point is: \[ y = \sqrt[3]{\frac{4226}{4225}} \] **Final Answer:** After simplifying, the sole equilibrium is the real root of y³ = 4226⁄4225. Thus, y = ∛(4226 ⁄ 4225)