Let $y=f\left(x\right)$ be the real valued function defined on the interval $(0,\infty )$, satisfying $f\left(1\right)=0$ and the differential equation $x\frac{dy}{dx}=y-x^3$

Then which of the following statements is (are) TRUE?

The correct statements are:

1. The function $f$ has a local maximum at $x=\frac{1}{\sqrt{3}}$

2. If $g\left(x\right)=4x^3-5x^2+\frac{3}{2}x$ for $x>0$, then the number of elements in the set $\{x\in (0,\infty ):f(x)=g(x\left)\right\}$ is $2$

Step 1: Solve the Differential Equation

We are given the linear differential equation:

$x\frac{dy}{dx}=y-x^3$

Rearranging the equation gives:

$x\frac{dy}{dx}-y=-x^3$

Dividing both sides by $x^2$ (since $x>0$):

$\frac{x\frac{dy}{dx}-y}{x^2}=-x$

Notice that the left-hand side is the quotient rule derivative of $\frac{y}{x}$:

$\frac{d}{dx}\left(\frac{y}{x}\right)=-x$

Integrating both sides with respect to $x$:

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

Multiplying by $x$, we get the general solution:

$f\left(x\right)=y=-\frac{x^3}{2}+Cx$

Step 2: Apply the Initial Condition

We are given that $f\left(1\right)=0$. Substituting $x=1$ into the general solution:

$0=-\frac{1}{2}+C\Rightarrow C=\frac{1}{2}$

Therefore, the function $f\left(x\right)$ is:

$f\left(x\right)=\frac{x-x^3}{2}$

Step 3: Analyze Critical Points and Extrema of $f\left(x\right)$

To find local extrema, differentiate $f\left(x\right)$ with respect to $x$:

$f^{\prime }\left(x\right)=\frac{1-3x^2}{2}$

Set $f^{\prime }\left(x\right)=0$ for $x>0$:

$1-3x^2=0\Rightarrow x=\frac{1}{\sqrt{3}}$

Taking the second derivative:

$f^{\prime \prime }\left(x\right)=-3x$

At $x=\frac{1}{\sqrt{3}}$:

$f^{\prime \prime }\left(\frac{1}{\sqrt{3}}\right)=-\frac{3}{\sqrt{3}}<0$

Since the second derivative is negative, $f\left(x\right)$ has a local maximum at $x=\frac{1}{\sqrt{3}}$. Thus, the second option statement is TRUE.

Step 4: Check the Number of Solutions for $f\left(x\right)=g\left(x\right)$

Equate $f\left(x\right)$ and $g\left(x\right)$ for $x>0$:

$\frac{x-x^3}{2}=4x^3-5x^2+\frac{3}{2}x$

Multiply both sides by 2:

$x-x^3=8x^3-10x^2+3x$

Combine like terms:

$9x^3-10x^2+2x=0$

Since we are given that $x\in (0,\infty )$, $x\ne 0$. Dividing by $x$:

$9x^2-10x+2=0$

Using the quadratic formula to solve for $x$:

$x=\frac{10\pm \sqrt{{(-10)}^2-4\left(9\right)\left(2\right)}}{2\left(9\right)}=\frac{10\pm \sqrt{100-72}}{18}=\frac{10\pm \sqrt{28}}{18}=\frac{5\pm \sqrt{7}}{9}$

Since $7\approx 2.65$, both roots:

$x_1=\frac{5+\sqrt{7}}{9}>0$

$x_2=\frac{5-\sqrt{7}}{9}>0$

are real and strictly positive. Therefore, there are exactly 2 elements in the set $\{x\in (0,\infty ):f(x)=g(x\left)\right\}$, making the fourth option statement TRUE.