Maximum Value of Solution to First Order Linear Differential Equation

To find the maximum value of the function $y(x)$, we first solve the given differential equation:

$(x^2 - 5)\frac{dy}{dx} - 2xy = -2x(x^2 - 5)^2$

Dividing both sides of the equation by $(x^2 - 5)^2$, we obtain:

$\frac{(x^2 - 5)\frac{dy}{dx} - 2xy}{(x^2 - 5)^2} = -2x$

Recognizing the left-hand side as the derivative of a quotient, we can rewrite the equation as:

$\frac{d}{dx} \left( \frac{y}{x^2 - 5} \right) = -2x$

Integrating both sides with respect to $x$:

$\frac{y}{x^2 - 5} = \int -2x \, dx$

$\frac{y}{x^2 - 5} = -x^2 + C$

$y(x) = (x^2 - 5)(C - x^2)$

We are given the initial condition $y(2) = 7$. Substituting $x = 2$ and $y = 7$ into the solution:

$7 = (2^2 - 5)(C - 2^2)$ $7 = (-1)(C - 4)$ $7 = 4 - C \implies C = -3$

Substituting $C = -3$ back into the solution for $y(x)$:

$y(x) = (x^2 - 5)(-3 - x^2) = -(x^2 - 5)(x^2 + 3)$ $y(x) = -(x^4 - 2x^2 - 15) = -x^4 + 2x^2 + 15$

To find the maximum value of $y(x)$, we can rewrite the expression by completing the square:

$y(x) = -(x^4 - 2x^2 + 1) + 1 + 15$ $y(x) = -(x^2 - 1)^2 + 16$

Since $(x^2 - 1)^2 \ge 0$ for all $x \in \mathbb{R}$, the maximum value of $y(x)$ occurs when $x^2 - 1 = 0$ (i.e., $x = \pm 1$).

Thus, the maximum value of the function $y(x)$ is:

$y_{\text{max}} = 16$