Solve the initial value problem: \( \frac{x^{2}}{y^{2}-4} \frac{d y}{d x}=\frac{1}{2 y}, \quad y(1)=\sqrt{5} \) \( y(x)= \)

We start with the differential equation \[ \frac{x^2}{y^2-4}\,\frac{dy}{dx}=\frac{1}{2y} \] with the initial condition \[ y(1)=\sqrt{5}. \] **Step 1. Rewrite the Differential Equation** Divide both sides by \( \frac{x^2}{y^2-4} \) (or equivalently multiply by its reciprocal) to obtain \[ \frac{dy}{dx} = \frac{y^2-4}{2y\,x^2}. \] **Step 2. Separate the Variables** Multiply both sides by \( \frac{2y}{y^2-4} \) and by \( dx \) to separate the \( y \) and \( x \) variables: \[ \frac{2y}{y^2-4}\,dy = \frac{dx}{x^2}. \] **Step 3. Integrate Both Sides** Integrate the left side with respect to \( y \) and the right side with respect to \( x \): \[ \int \frac{2y}{y^2-4}\,dy = \int \frac{dx}{x^2}. \] **Left Side Integration:** Let \[ u = y^2-4 \quad \text{so that} \quad du = 2y\,dy. \] Then, \[ \int \frac{2y}{y^2-4}\,dy = \int \frac{du}{u} = \ln|u| + C_1 = \ln|y^2-4| + C_1. \] **Right Side Integration:** Rewrite the integral as \[ \int \frac{dx}{x^2} = \int x^{-2}\,dx = -\frac{1}{x} + C_2. \] **Step 4. Combine the Results** Equate the two sides (the constants of integration can be combined into one arbitrary constant, \( C \)): \[ \ln|y^2-4| = -\frac{1}{x} + C. \] Exponentiate both sides to solve for \( |y^2-4| \): \[ |y^2-4| = e^{-\frac{1}{x}+C} = Ae^{-\frac{1}{x}}, \] where \( A = e^C \) is a positive constant. Since we expect (from the initial condition) that \( y^2-4 \) is positive, we can drop the absolute value: \[ y^2-4 = Ae^{-\frac{1}{x}}. \] **Step 5. Determine the Constant Using the Initial Condition** Substitute \( x = 1 \) and \( y(1) = \sqrt{5} \) into the equation: \[ (\sqrt{5})^2-4 = A e^{-1}. \] This simplifies to \[ 5-4 = A e^{-1} \quad \Rightarrow \quad 1 = A e^{-1}. \] Solving for \( A \) gives \[ A = e. \] **Step 6. Write the Final Solution** Substitute \( A = e \) back into the equation: \[ y^2-4 = e e^{-\frac{1}{x}} = e^{1-\frac{1}{x}}. \] Thus, \[ y^2 = 4+e^{1-\frac{1}{x}}, \] and taking the positive square root (since \( y(1)=\sqrt{5}>0 \)) we obtain \[ y(x) = \sqrt{4+e^{1-\frac{1}{x}}}. \]