Let 𝑦 be the solution of the differential equation with the initial conditions given below. If 𝑦(𝑥=2)= 𝐴 ln2, then the value of 𝐴 is ____________ (rounded off to 2 decimal places).

$x^2\frac{d^{2}y}{d{x}^2}+3x\frac{dy}{dx}+y=0$ , $\mathrm{ \; y}(x=1)=0, \frac{dy}{dx}(x=1)=1$

The correct answer is 0.55.

Consider the given second-order Cauchy-Euler differential equation:
$x^2\frac{d^{2}y}{d{x}^2}+3x\frac{dy}{dx}+y=0$

To solve this, we use the substitution $x=e^t$, which implies $t=\ln x$.
Using this substitution, the differential terms transform as:
$x\frac{dy}{dx}=\frac{dy}{dt}=Dy$
and
$x^2\frac{d^{2}y}{d{x}^2}=D(D-1)y$
where $D=\frac{d}{dt}$.

Substituting these relations into the original differential equation yields:
$\left[D\right(D-1)+3D+1]y=0$
Simplifying the operator:
$(D^2+2D+1)y=0$

The auxiliary characteristic equation is:
$m^2+2m+1=0$
Factoring gives:
${(m+1)}^2=0$
Thus, we have repeated roots: $m=-1,-1$.

The general solution in terms of $t$ is:
$y=(C_1+C_{2}t)e^{-t}$
Substituting back $t=\ln x$ and $e^{-t}=\frac{1}{x}$:
$y\left(x\right)=\frac{C_1+C_2\ln x}{x}$

Now we apply the initial conditions:
1. Using $y\left(1\right)=0$:
$0=\frac{C_1+C_2\ln \left(1\right)}{1}=C_1$
Therefore, $C_1=0$. The solution simplifies to:
$y\left(x\right)=\frac{C_2\ln x}{x}$

2. Next, we differentiate