For x ∈ ℝ , let y x be a solution of the differential equation x 2 - 5 d y d x - 2 x y = - 2 x x…

Given,

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

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

Which is a linear differential equation,

So, integrating factor,$\text{ I.F.}=e^{\int \frac{-2x}{x^2-5}dx}=\frac{1}{\left|x^2-5\right|}$

Now solution of differential equation is given by,

$y\cdot \frac{1}{\left|x^2-5\right|}=\int -2x\cdot \frac{x^2-5}{\left|x^2-5\right|}dx$

$\Rightarrow \frac{y}{\left|x^2-5\right|}=\frac{x^2-5}{\left|x^2-5\right|}\left(-x^2\right)+C$

Now using the given value, $y\left(2\right)=7\Rightarrow C=3$

$\Rightarrow y=-x^2\left(x^2-5\right)+3\left|x^2-5\right|$

$\Rightarrow y=f\left(x\right)$ is even function

If $0<x<\sqrt{5}, y=-x^4+5x^2-3x^2+15$

$\Rightarrow y=-x^4+2x^2+15$

For increasing function $\frac{dy}{dx}>0$

$\Rightarrow \frac{dy}{dx}=-4x^3+4x>0$

$\Rightarrow x<1$
If $x>\sqrt{5}, y=-x^4+5x^2+3x^2-15$

$\Rightarrow y=-x^4+8x^2-15$
For increasing function $\frac{dy}{dx}>0$

$\Rightarrow \frac{dy}{dx}=-4x^3+16x>0$

$\Rightarrow -4x\left(x^2-4\right)>0$

$\Rightarrow 4x\left(x^2-4\right)<0$

$\Rightarrow x\in \left(-\infty ,-2\right)\cup \left(0,2\right)$ but $x>\sqrt{5}$

$\Rightarrow x=\varphi$

$\Rightarrow y\left(x\right)$ is increasing over $(0,1)$

Now plotting the graph we get,

Hence, maximum value of the function is given by,

$\Rightarrow f(x{)}_{max}=16$