If the solution curve $$y =f (x)$$ of the differential equation
$$(x^{2}-4)y^{'}-2xy+2x(4-x^{2})^{2}=0,x>2,$$
passes through the point (3, 15), then the local maximum value of $$f$$ is __________

The given differential equation is$$(x^{2}-4)\,y' \;-\;2x\,y \;+\;2x\,(4-x^{2})^{2}=0,\qquad x\gt 2$$

Rewrite it in the standard linear form $$y'+P(x)\,y=Q(x)$$.
Divide the whole equation by $$x^{2}-4$$ (which is positive for $$x\gt 2$$):

$$y' \;-\;\frac{2x}{x^{2}-4}\,y \;+\;\frac{2x(4-x^{2})^{2}}{x^{2}-4}=0$$

Because $$(4-x^{2})^{2}=(x^{2}-4)^{2}$$, the last term simplifies to $$2x(x^{2}-4)$$. Thus

$$y' \;-\;\frac{2x}{x^{2}-4}\,y \;+\;2x(x^{2}-4)=0$$

Move the non-homogeneous term to the right:

$$y' \;-\;\frac{2x}{x^{2}-4}\,y = -\,2x(x^{2}-4)\qquad -(1)$$

Here $$P(x)=-\dfrac{2x}{x^{2}-4}$$ and $$Q(x)=-\,2x(x^{2}-4)$$.

Integrating factor
The integrating factor (I.F.) is $$\displaystyle \mu(x)=\exp\!\bigl(\int P(x)\,dx\bigr)$$.

$$\int P(x)\,dx=\int\!-\frac{2x}{x^{2}-4}\,dx$$
Put $$u=x^{2}-4\;\Longrightarrow\;du=2x\,dx$$, giving

$$\int\!-\frac{du}{u}=-\ln|u|=-\ln|x^{2}-4|$$

Hence $$\mu(x)=e^{-\ln|x^{2}-4|}=(x^{2}-4)^{-1},\qquad x\gt 2$$

Using the integrating factor
Multiplying equation $$(1)$$ by $$(x^{2}-4)^{-1}$$ gives

$$\frac{1}{x^{2}-4}\,y' -\frac{2x}{(x^{2}-4)^{2}}\,y = -\,2x$$

The left side is the derivative of $$y\,(x^{2}-4)^{-1}$$, so

$$\frac{d}{dx}\Bigl[\frac{y}{x^{2}-4}\Bigr]= -\,2x$$

Integrate with respect to $$x$$:

$$\frac{y}{x^{2}-4}= -\int 2x\,dx + C = -x^{2}+C$$

Therefore $$y=(x^{2}-4)\,(-x^{2}+C)= -x^{4}+4x^{2}+C\,x^{2}-4C$$

Simplify:

$$y=-x^{4}+(4+C)\,x^{2}-4C\qquad -(2)$$

Finding the constant using the point (3, 15)
Substitute $$x=3,\;y=15$$ into $$(2)$$:

$$15=-81+(4+C)\,9-4C=-81+36+9C-4C=-45+5C$$

$$5C=60\;\Longrightarrow\;C=12$$

Hence the explicit solution is

$$y=-x^{4}+16x^{2}-48\qquad -(3)$$

Locating critical points
Differentiate $$(3)$$:

$$y'=\frac{dy}{dx}=-4x^{3}+32x=-4x\,(x^{2}-8)$$

Set $$y'=0$$:

$$-4x\,(x^{2}-8)=0\;\;\Longrightarrow\;\;x=0,\;\;x=\pm\sqrt{8}= \pm2\sqrt{2}$$

Because the domain is $$x\gt 2$$, the only relevant critical point is$$x=2\sqrt{2}\ (\approx2.828).$$

Second-derivative test
Compute $$y''$$:

$$y''=\frac{d}{dx}(-4x^{3}+32x)=-12x^{2}+32$$

At $$x=2\sqrt{2}$$, $$x^{2}=8$$, hence

$$y''=-12(8)+32=-96+32=-64\lt 0$$

Since $$y''\lt 0$$, $$x=2\sqrt{2}$$ gives a local maximum.

Maximum value of $$y$$
Substitute $$x=2\sqrt{2},\;x^{2}=8,\;x^{4}=64$$ into $$(3)$$:

$$y_{\text{max}}=-64+16\!\times\!8-48=-64+128-48=16$$

Thus, the local maximum value of $$f$$ is $$16$$.