= Solution
For $c\ne0$ and on an interval where $x\ne0$, the <logarithmic derivative> substitution gives
$$
y'=\frac{x''}{cx}-\frac{x'^2}{cx^2},\qquad cy^2=\frac{x'^2}{cx^2}.
$$
The quadratic terms cancel. Multiplying by $cx$ gives the <linearization of a Riccati equation>:
$$
\boxed{x''+a(t)x'+cb(t)x=0.}
$$
Multiplying $x$ by a nonzero constant leaves $y$ unchanged. If $c=0$, that substitution is undefined, but the original equation is already a first-order <linear ordinary differential equation>, solvable by an <integrating factor>.
For the particular equation, $c=1$ and the linear equation is $x''+x'/t-\lambda^2x/t^2=0$. Use the PDF's logarithmic coordinate $\tau=\log t$ and write $X(\tau)=x(e^\tau)$. The <chain rule> gives $x'=X'/t$ and $x''=(X''-X')/t^2$, leaving $X''-\lambda^2X=0$. Thus $x=A t^\lambda+B t^{-\lambda}$, as also follows from the <Euler-Cauchy equation>. Recovering the <Riccati equation> solution yields
$$
y(t)=\frac\lambda t\frac{A t^{2\lambda}-B}{A t^{2\lambda}+B}.
$$
The initial value imposes $A=-3B$; choose $A=3,B=-1$. Therefore
$$
\boxed{y(t)=\frac\lambda t\frac{3t^{2\lambda}+1}{3t^{2\lambda}-1},\qquad t_*=3^{-1/(2\lambda)}.}
$$
The denominator vanishes only at $t_*>0$, and its numerator is then nonzero. This is a simple pole: $y(t)\sim1/(t-t_*)$. The maximal interval containing $t=1$ is $(t_*,\infty)$; the same expression on $(0,t_*)$ is a separate solution branch. These are <poles of a Riccati solution from zeros of its linearizing solution>.
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