Show, by using the substitution $w=\dfrac{y}{t^{2}}$, that the general solution of the differential equation $t\dfrac{dw}{dt}=w\!\,^{2}t^{2}+2wt-2w$ is $w=\dfrac{1}{t^{2}}\left(\dfrac{2Ae^{2t}}{1-Ae^{2t}}\right)$. Sketch, on a single diagram, two members of the family of solution curves corresponding to $A=0$ and to $A>0$ respectively.

1. Find the curve for A = 0

Putting $A=0$ into $w=\dfrac{1}{t^{2}}\left(\dfrac{2Ae^{2t}}{1-Ae^{2t}}\right)$ gives $w=\dfrac{1}{t^{2}}\left(\dfrac{0}{1}\right)=0$.

So this solution curve is simply the horizontal line $w=0$ (the $t$-axis).

2. Identify features of the A > 0 curve and sketch both

For $A>0$, the denominator vanishes when $1-Ae^{2t}=0$, i.e. at $t=-\tfrac{1}{2}\ln A$, giving a vertical asymptote there.

Limiting behaviour: as $t\to-\infty$, $e^{2t}\to0$ so $w\to0^{+}$; as $t\to+\infty$, $\dfrac{2Ae^{2t}}{1-Ae^{2t}}\to-1$ so $w\to0^{-}$. Hence $w=0$ is a horizontal asymptote at both ends.

Sign: $w>0$ for $t<-\tfrac12\ln A$ (rising steeply to $+\infty$ at the asymptote) and $w<0$ for $t>-\tfrac12\ln A$ (rising from $-\infty$ back up towards $0^{-}$).

Sketch both on one $w$–$t$ diagram: the straight line $w=0$ for $A=0$, and the two-branch curve with the vertical asymptote at $t=-\tfrac12\ln A$ for $A>0$.