Step 1: Identify and rewrite the differential equation

The given differential equation is: $\left(x^2-5\right) \frac{d y}{d x}-2 x y=-2 x\left(x^2-5\right)^2$ This equation is defined for $x^2-5 \neq 0$, i.e., $x \neq \pm\sqrt{5}$. To solve it, we can rewrite it in the standard form of a linear first-order differential equation, $\frac{dy}{dx} + P(x)y = Q(x)$. Dividing the entire equation by $(x^2-5)$, we get: $\frac{d y}{d x} - \frac{2x}{x^2-5} y = -2x(x^2-5)$ Here, $P(x) = -\frac{2x}{x^2-5}$ and $Q(x) = -2x(x^2-5)$.

Step 2: Find the integrating factor (I.F.)

The integrating factor is given by the formula $I.F. = e^{\int P(x) dx}$. $I.F. = e^{\int -\frac{2x}{x^2-5} dx}$ To evaluate the integral, we use the substitution $u = x^2-5$, so $du = 2x dx$. $\int -\frac{2x}{x^2-5} dx = \int -\frac{du}{u} = -\ln|u| = -\ln|x^2-5| = \ln\left(\frac{1}{|x^2-5|}\right)$ So, the integrating factor is: $I.F. = e^{\ln\left(\frac{1}{|x^2-5|}\right)} = \frac{1}{|x^2-5|}$ The initial condition is given at $x=2$. For $x=2$, we have $x^2-5 = 4-5 = -1 < 0$. The solution is therefore defined in the interval $(-\sqrt{5}, \sqrt{5})$, where $x^2-5$ is negative. In this interval, $|x^2-5| = -(x^2-5) = 5-x^2$. We can choose the integrating factor as $I.F. = \frac{1}{x^2-5}$ for simplicity, as the sign will be consistent throughout the calculation.

Step 3: Find the general solution

The solution of the linear differential equation is given by $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$. $y \cdot \frac{1}{x^2-5} = \int -2x(x^2-5) \cdot \frac{1}{x^2-5} dx + C$ $\frac{y}{x^2-5} = \int -2x dx + C$ $\frac{y}{x^2-5} = -x^2 + C$ Thus, the general solution is: $y(x) = (C-x^2)(x^2-5)$

Step 4: Apply the initial condition

We are given the condition $y(2)=7$. Substituting $x=2$ and $y=7$ into the general solution: $7 = (C - 2^2)(2^2-5)$ $7 = (C-4)(4-5)$ $7 = (C-4)(-1)$ $-7 = C-4$ $C = 4-7 = -3$ So, the particular solution is: $y(x) = (-3-x^2)(x^2-5) = -(x^2+3)(x^2-5)$ Expanding this expression, we get: $y(x) = -(x^4 - 5x^2 + 3x^2 - 15) = -(x^4 - 2x^2 - 15)$ $y(x) = -x^4 + 2x^2 + 15$

Step 5: Find the maximum value of the function y(x)

To find the maximum value of $y(x)$, we need to find its critical points by taking the first derivative and setting it to zero. $\frac{dy}{dx} = -4x^3 + 4x$ Set $\frac{dy}{dx} = 0$: $-4x^3 + 4x = 0$ $-4x(x^2-1) = 0$ $-4x(x-1)(x+1) = 0$ The critical points are $x=0$, $x=1$, and $x=-1$. All these points lie within the interval $(-\sqrt{5}, \sqrt{5})$.

Now, we evaluate the function $y(x)$ at these critical points:

• At $x=0$: $y(0) = -(0^2+3)(0^2-5) = (-3)(-5) = 15$.
• At $x=1$: $y(1) = -(1^2+3)(1^2-5) = -(4)(-4) = 16$.
• At $x=-1$: $y(-1) = -((-1)^2+3)((-1)^2-5) = -(1+3)(1-5) = -(4)(-4) = 16$.

To determine if these points are maxima or minima, we can use the second derivative test: $\frac{d^2y}{dx^2} = -12x^2 + 4$

• At $x=0$: $\frac{d^2y}{dx^2} = 4 > 0$, which indicates a local minimum.
• At $x=1$: $\frac{d^2y}{dx^2} = -12(1)^2 + 4 = -8 < 0$, which indicates a local maximum.
• At $x=-1$: $\frac{d^2y}{dx^2} = -12(-1)^2 + 4 = -8 < 0$, which indicates a local maximum.

The local maximum values are both 16. As $x \to \pm\sqrt{5}$, $y(x) = -(x^2+3)(x^2-5) \to 0$. Since 16 is greater than the other values calculated (15 and 0), the absolute maximum value of the function $y(x)$ is 16.