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

Given: The differential equation is:

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

Step 1: Rearranging the Equation

We start by rearranging the equation as follows:

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

The integrating factor is:

\(I.F. = \frac{1}{|x^2 - 5|}\)

Step 2: Solving the Differential Equation

Now, we solve the differential equation:

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

Substituting \( y(2) = 7 \), we find:

\(C = 3\)

Step 3: Final Solution

The solution to the differential equation is:

\(y = -x^2(x^2 - 5) + 3 |x^2 - 5|\)

Step 4: Function Behavior

Given the form of the solution, \( y = f(x) \) is an even function. The solution is split into two parts:

• If \( 0 < x < \sqrt{5} \), then \( y = -x^4 + 5x^2 - 3x^2 + 15 \) or \( y = -x^4 + 2x^2 + 15 \).
• If \( x > \sqrt{5} \), then \( y = -x^4 + 5x^2 + 3x^2 - 15 \) or \( y = -x^4 + 5x^2 + 3x^2 - 15 \).

Step 5: Determining the Increasing Function

For an increasing function, we check when the derivative \( \frac{dy}{dx} > 0 \). Solving this gives:

\(x = 1 \quad \text{(for increasing function)}\)

The function \( y(x) \) is increasing over the interval \( (0, 1) \).

Step 6: Graph and Maximum Value

We plot the graph of \( y(x) \) and observe the following points:

• At \( (-1, 16) \) and \( (1, 16) \), the value is 16.
• At \( (0, 15) \), the value is 15.

The maximum value of \( f(x) \) is 16, and the function \( y(x) \) reaches its maximum at \( x = 1 \).

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.