$$\mbox{Use the Euler equation method to solve the given equation for t $>$ 0}$$
$$t^2{y}''-4t{y}'+6y=0$$
$$\frac{d^2y}{dx^2}+(\alpha-1)\frac{dy}{dx}+\beta y=0$$
$$\frac{d^2y}{dx^2}+(-4-1)\frac{dy}{dx}+6y=0$$
$$\frac{d^2y}{dx^2}-5\frac{dy}{dx}+6y=0$$
$$r^2-5r+6=0$$
$$(r-2)(r-3)=0$$
$$r_{1}=2 \quad r_{2}=3$$
$$\mbox{Thus, } y(x)=c_{1}e^2x+c_{2}e^3x$$
$$\mbox{Substituting }x=ln(t)$$
$$y(t)=c_{1}e^2ln(t)+c_{2}e^3ln(t)$$
$$y(t)=c_{1}e^{ln(t)^2}+c_{2}e^{ln(t)^3}$$
$$\mbox{Therefore, the solution is } y(t)=c_{1}t^2+c_{2}t^3$$
