For x ∈ ℝ , let y(x) be a solution of the differential equation ( x...

Given,

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

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

Which is a linear differential equation,

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

Now solution of differential equation is given by,

$y ⋅ \dfrac{1}{\left|{x}^{2} - 5\right|} = ∫ - 2x ⋅ \dfrac{{x}^{2} - 5}{\left|{x}^{2} - 5\right|}dx$

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

Now using the given value, $y ( 2 ) = 7 ⇒ C = 3$

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

$⇒ y = f ( x )$ is even function

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

$⇒ y = - {x}^{4} + 2{x}^{2} + 15$

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

$⇒ \dfrac{dy}{dx} = - 4{x}^{3} + 4x > 0$

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

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

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

$⇒ - 4x\left({x}^{2} - 4\right) > 0$

$⇒ 4x\left({x}^{2} - 4\right) < 0$

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

$⇒ x = ϕ$

$⇒ y ( x )$ is increasing over $( 0, 1 )$

Now plotting the graph we get,

Hence, maximum value of the function is given by,

⇒ f ( x{)}_\max = 16