Question

For \(x ∈ R\), let \(y(x)\) be the 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

Answer 1

Correct Answer: 16

The answer is 16.

Answer 2

Given,
\((x^2-5)\frac{dy}{dx}-2xy=-2x(x^2-5)\)
\(\frac{dy}{dx}+(\frac{-2x}{x^2-5})y=-2x(x^2-5)\)
Now Both side \(\frac{1}{|x^2-5|}\)

The solution for the Derivative equation is 
\(y.\frac{1}{|x^2-5|}=\int-2x.\frac{x^2-5}{|x^2-5|}dx\)

\(\frac{y}{|x^2-5|}=\frac{x^2-5}{|x^2-5|}(-x^2)+C\)
y(2)=7
C=3
Now put the value
\(y=-x^2(x^2-5)+3|x^2-5|\)
y=f(x) is even function
\(If 0 < x < \sqrt(5), y = - x ^ 4 + 5x ^ 2 - 3x ^ 2 + 15 = - x ^ 4 + 2x ^ 2 + 15\)
For increasing function\(\frac{dy}{dx}\gt0=\gt x\lt1\)

\(If\ x>√5, y = -x^4 + 5x^2 + 3x²-15\)
For increasing function\(\frac{dy}{dx}\gt0=\gt x=\phi\)
y(x) is increasing over (0, 1)

f(x)max = 16

So, the answer is 16.