Differential Equation Solution and Properties of Real Valued Function

To determine which of the given statements are correct, we first solve the given differential equation for the function $y = f(x)$.

Step 1: Solving the Differential Equation

The given differential equation is: $x \frac{dy}{dx} = y - x^3 \quad \text{for } x \in (0, \infty)$

Rearranging the terms: $x \frac{dy}{dx} - y = -x^3$

Dividing both sides by $x^2$ (since $x > 0$): $\frac{x \frac{dy}{dx} - y}{x^2} = -x$

The left-hand side is the derivative of $\frac{y}{x}$: $\frac{d}{dx}\left(\frac{y}{x}\right) = -x$

Integrating both sides with respect to $x$: $\frac{y}{x} = -\frac{x^2}{2} + C$ $y = -\frac{x^3}{2} + Cx$

Using the initial condition $y(1) = 0$: $0 = -\frac{1^3}{2} + C(1) \implies C = \frac{1}{2}$

Thus, the function $f(x)$ is given by: $f(x) = \frac{x - x^3}{2}$

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Step 2: Analyzing Maxima and Minima (Options A and B)

To find the local extrema, we differentiate $f(x)$ with respect to $x$: $f'(x) = \frac{d}{dx}\left(\frac{x - x^3}{2}\right) = \frac{1 - 3x^2}{2}$

Setting $f'(x) = 0$ for $x \in (0, \infty)$: $1 - 3x^2 = 0 \implies x = \frac{1}{\sqrt{3}}$

Now, computing the second derivative: $f''(x) = \frac{d}{dx}\left(\frac{1 - 3x^2}{2}\right) = -3x$

Evaluating $f''(x)$ at the critical point $x = \frac{1}{\sqrt{3}}$: $f''\left(\frac{1}{\sqrt{3}}\right) = -3\left(\frac{1}{\sqrt{3}}\right) = -\sqrt{3} < 0$

Since $f''\left(\frac{1}{\sqrt{3}}\right) < 0$, the function $f(x)$ has a local maximum at $x = \frac{1}{\sqrt{3}}$.

• Therefore, Option A is FALSE and Option B is TRUE.

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Step 3: Analyzing Monotonicity in the Interval $(1, 2)$ (Option C)

For any $x \in (1, 2)$: $x > 1 \implies x^2 > 1 \implies 3x^2 > 3 \implies 1 - 3x^2 < -2 < 0$

Since $f'(x) = \frac{1 - 3x^2}{2} < 0$ for all $x \in (1, 2)$, the function $f(x)$ is strictly decreasing in the interval $(1, 2)$.

• Therefore, Option C is FALSE.

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Step 4: Number of Solutions for $f(x) = g(x)$ (Option D)

Given: $g(x) = 4x^3 - 5x^2 + \frac{3}{2}x \quad \text{for } x > 0$

Setting $f(x) = g(x)$: $\frac{x - x^3}{2} = 4x^3 - 5x^2 + \frac{3}{2}x$

Since $x > 0$, we can divide the entire equation by $x$: $\frac{1 - x^2}{2} = 4x^2 - 5x + \frac{3}{2}$

Multiplying both sides by 2: $1 - x^2 = 8x^2 - 10x + 3$

Rearranging into a quadratic equation: $9x^2 - 10x + 2 = 0$

Using the quadratic formula, the roots are: $x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(9)(2)}}{2(9)} = \frac{10 \pm \sqrt{100 - 72}}{18} = \frac{10 \pm \sqrt{28}}{18} = \frac{5 \pm \sqrt{7}}{9}$

Both roots are positive real numbers: $x_1 = \frac{5 + \sqrt{7}}{9} > 0 \quad \text{and} \quad x_2 = \frac{5 - \sqrt{7}}{9} > 0 \quad (\text{since } \sqrt{7} \approx 2.65 < 5)$

Thus, there are exactly 2 elements in the set $\{x \in (0, \infty) : f(x) = g(x)\}$.

• Therefore, Option D is TRUE.

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Conclusion

The correct options are B and D.