For x ∈ 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 correct answer is 16.

Step 1: Express the given differential equation in standard first-order linear form.

The given differential equation is:

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

Dividing the entire equation by $x^2-5$ (assuming $x^2-5\ne 0$), we get:

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

This is a linear differential equation of the form $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$, where:

$P\left(x\right)=\frac{-2x}{x^2-5}$

and

$Q\left(x\right)=-2x(x^2-5)$

Step 2: Find the Integrating Factor (I.F.).

The integrating factor is given by:

$\text{I.F.}=e^{\int P\left(x\right)dx}$

Evaluating the integral:

$\int \frac{-2x}{x^2-5}dx=-\ln |x^2-5|=\ln |x^2-5{|}^{-1}$

Therefore:

$\text{I.F.}=e^{\ln |x^2-5{|}^{-1}}=\frac{1}{x^2-5}$

Step 3: Solve the differential equation.

The general solution is given by:

$y\cdot \left(\text{I.F.}\right)=\int Q\left(x\right)\cdot \left(\text{I.F.}\right)dx+C$

Substitute the values of $\text{I.F.}$ and $Q\left(x\right)$:$y\cdot \frac{1}{x^2-5}=\int -2x(x^2-5)\cdot \frac{1}{x^2-5}dx+C$

Simplifying the integrand:

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

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

Thus, the general solution for $y\left(x\right)$ is:

$y\left(x\right)=(x^2-5)(C-x^2)$

Step 4: Use the initial condition to find the constant C.

We are given that $y\left(2\right)=7$. Substituting $x=2$ and $y=7$:

$7=(2^2-5)(C-2^2)$

$7=(4-5)(C-4)$

$7=-1\cdot (C-4)$

$-7=C-4\Rightarrow C=-3$

Step 5: Write the explicit function y(x) and find its maximum value.

Substituting $C=-3$ back into the expression for $y\left(x\right)$:

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

Expanding this polynomial:

$y\left(x\right)=-(x^4-2x^2-15)=-x^4+2x^2+15$

To maximize $y\left(x\right)$, complete the square with respect to $x^2$:

$y\left(x\right)=-(x^4-2x^2+1-1)+15$

$y\left(x\right)=-{(x^2-1)}^2+1+15$

$y\left(x\right)=16-{(x^2-1)}^2$

Since ${(x^2-1)}^2\ge 0$ for all real $x$, the maximum value occurs when $x^2-1=0$ (i.e., at $x=\pm 1$).

Thus, the maximum value of $y\left(x\right)$ is $16$.