For $$x \in \mathbb{R}$$, let $$y(x)$$ be a solution of the differential equation $$(x^2 - 5)\frac{dy}{dx} - 2xy = -2x(x^2 - 5)^2$$ such that $$y(2) = 7$$. Then the maximum value of the function $$y(x)$$ is

The given differential equation is$$(x^{2}-5)\frac{dy}{dx}-2xy=-2x(x^{2}-5)^{2}$$

Divide by $$(x^{2}-5)$$ (which is non-zero everywhere except at $$x=\pm\sqrt5$$, the points where the original equation is not defined):
$$\frac{dy}{dx}-\frac{2x}{x^{2}-5}\,y=-2x(x^{2}-5)$$

This is a linear first-order ODE of the form $$\dfrac{dy}{dx}+P(x)\,y=Q(x)$$ with
$$P(x)=-\frac{2x}{x^{2}-5},\qquad Q(x)=-2x(x^{2}-5).$$

Integrating factor (I.F.):
$$\text{I.F.}=e^{\int P(x)\,dx}=e^{\int-\frac{2x}{x^{2}-5}\,dx}.$$

Put $$u=x^{2}-5\;\Rightarrow\;du=2x\,dx\;,$$ so
$$\int-\frac{2x}{x^{2}-5}\,dx=-\int\frac{du}{u}=-\ln|u|=\ln|u|^{-1}.$$
Thus $$\text{I.F.}=|x^{2}-5|^{-1}$$. Ignoring the absolute value (it affects the sign only in regions where $$x^{2}-5\lt 0$$), we take
$$\text{I.F.}=\frac1{x^{2}-5}.$$

Multiplying the differential equation by the integrating factor:
$$\frac1{x^{2}-5}\frac{dy}{dx}-\frac{2x}{(x^{2}-5)^{2}}\,y=-2x.$$

The left side is the derivative of $$\dfrac{y}{x^{2}-5}$$ because
$$\frac{d}{dx}\!\left(\frac{y}{x^{2}-5}\right)= \frac{(x^{2}-5)dy/dx-2xy}{(x^{2}-5)^{2}} =\frac1{x^{2}-5}\frac{dy}{dx}-\frac{2x}{(x^{2}-5)^{2}}\,y.$$

So we have
$$\frac{d}{dx}\!\left(\frac{y}{x^{2}-5}\right)=-2x.$$

Integrate with respect to $$x$$:
$$\frac{y}{x^{2}-5}=\int-2x\,dx=-x^{2}+C,$$
where $$C$$ is the constant of integration.

Therefore
$$y=( -x^{2}+C)(x^{2}-5).$$

Use the initial condition $$y(2)=7$$:
For $$x=2\ (\;x^{2}=4\;),\qquad y(2)=(C-4)(-1)=7\; \Rightarrow\;C-4=-7\; \Rightarrow\;C=-3.$$

Hence the particular solution is
$$y(x)=(-x^{2}-3)(x^{2}-5).$$

Simplify by letting $$t=x^{2}\;(\,t\ge0\,):$$
$$y=-(t+3)(t-5)=-(t^{2}-2t-15)=-t^{2}+2t+15.$$

The quadratic $$y(t)=-t^{2}+2t+15$$ opens downwards (coefficient of $$t^{2}$$ is $$-1$$), so its maximum occurs at the vertex
$$t=\frac{-b}{2a}=\frac{-2}{2(-1)}=1.$$(That means $$x^{2}=1\Rightarrow x=\pm1$$.)

Maximum value:
$$y_{\max}=-(1)^{2}+2(1)+15=-1+2+15=16.$$

Therefore, the maximum value of $$y(x)$$ is 16.