\text{An ordinary differential equation is given below.}

\text{<br>}

$$\begin{gathered} \\ {x}^2\frac{d^{2}y}{d{x}^2}=6y \end{gathered}$$

\mathrm{<br>}

$$\begin{gathered} \mathrm{<br>} \\ \text{Considering a and b as arbitrary constants, the general solution of the equation is} \end{gathered}$$

The correct option is: y(x) = ax³ + b/x²

Step-by-step Explanation:

The given ordinary differential equation is a second-order Euler-Cauchy (or Cauchy-Euler) equation:
$$x^2\frac{d^{2}y}{d{x}^2}=6y$$
We can rewrite this in standard Euler-Cauchy form as:
$$x^2\frac{d^{2}y}{d{x}^2}-6y=0$$

To solve this differential equation, we assume a solution of the form:
$$y=x^m$$
Differentiating $y$ twice with respect to $x$ gives:
$$\frac{dy}{dx}=m{x}^{m-1}$$
$$\frac{d^{2}y}{d{x}^2}=m(m-1)x^{m-2}$$

Substituting these expressions back into the original differential equation, we get:
$$x^2\left[m\right(m-1\left)x^{m-2}\right]-6x^m=0$$
Simplifying the term $x^2\cdot x^{m-2}=x^m$, we obtain:
$$m(m-1)x^m-6x^m=0$$
Factoring out $x^m$ (since $x^m\ne 0$ for non-trivial solutions):
$$\left[m\right(m-1)-6]x^m=0$$
This yields the characteristic (or auxiliary) equation:
$$m^2-m-6=0$$

To find the roots of this quadratic equation, we factor it:
$$(m-3)(m+2)=0$$
Solving for $m$ gives:
$$m_1=3\text{and}{m}_2=-2$$

Since the roots are real and distinct, the general solution of the differential equation is a linear combination of the two independent solutions, $x^3$ and $x^{-2}$:
$$y\left(x\right)=a{x}^3+b{x}^{-2}$$
which can be written as:
$$y\left(x\right)=a{x}^3+\frac{b}{x^2}$$
where $a$ and $b$ are arbitrary constants.