Question:

**Step 1: Solve the linear differential equation** Divide by $(x^2 - 5)$ to get it in standard form: $\frac{dy}{dx} - \frac{2x}{x^2 - 5}y = -2x(x^2 - 5)$.\nThe integrating factor is $I.F. = e^{\int \frac{-2x}{x^2 - 5} dx} = e^{-\ln|x^2 - 5|} = \frac{1}{|x^2 - 5|}$.\nWe can use $\frac{1}{x^2 - 5}$ as the IF. Multiplying the original equation by $\frac{1}{(x^2 - 5)^2}$ gives:\n$\frac{1}{x^2 - 5} \frac{dy}{dx} - \frac{2x}{(x^2 - 5)^2} y = -2x$.\nThis is $\frac{d}{dx} \left( \frac{y}{x^2 - 5} \right) = -2x$. **Step 2: Integrate both sides** Integrating with respect to $x$ yields:\n$\frac{y}{x^2 - 5} = -x^2 + C$.\n$y = (-x^2 + C)(x^2 - 5)$. **Step 3: Apply initial condition** We are given $y(2) = 7$. Substitute $x = 2$ and $y = 7$:\n$7 = (-4 + C)(4 - 5) \implies -7 = -4 + C \implies C = -3$.\nSo the function is $y(x) = (-x^2 - 3)(x^2 - 5)$. **Step 4: Find the maximum value** Let $u = x^2$. Since $x \in \mathbb{R}$, $u \ge 0$. The function is $Y(u) = (-u - 3)(u - 5) = -u^2 + 2u + 15$.\nTo find the maximum, set the derivative to 0: $Y'(u) = -2u + 2 = 0 \implies u = 1$. Since $u=1 \ge 0$, this is a valid value for $x^2$.\nAt $u = 1$, the maximum value is $-(1)^2 + 2(1) + 15 = 16$.\nThus, the maximum value of $y(x)$ is $16$.