More on the Wronskian — Question 9

Question 9 – Solution

Strategy. The singular leading coefficient changes which interval hypotheses are available; explicitly match derivatives to determine global solutions.

Step 1: Verify and compute. The residuals are $-2t+2t=0$ and $2t^2-4t^2+2t^2=0$. Also $$\boxed{W[t,t^2]=t(2t)-1(t^2)=t^2.}$$ It is zero only at $t=0$. The functions are independent on $\mathbb R$, since $At+Bt^2\equiv 0$ forces $A=B=0$.

Step 2: Respect the regular intervals. Division by $t^2$ gives $p=-2/t$, $q=2/t^2$, continuous on $(-\infty,0)$ and $(0,\infty)$. On either interval Abel’s identity gives $W=Ct^2$, with the constant fixed within that interval. For this pair it is $C=1$ on both sides. There is no regular interval containing zero. Abel’s conclusion applies separately on the two intervals and makes no assertion across zero.

Step 3: Match all global solutions. The pair is fundamental on either side, so write $A_-t+B_-t^2$ for $t<0$ and $A_+t+B_+t^2$ for $t>0$. Continuity forces $y(0)=0$. Matching first derivatives forces $A_-=A_+$; matching second derivatives forces $B_-=B_+$. Conversely every resulting polynomial solves the equation globally. Thus $$\boxed{y=At+Bt^2\quad\text{on }\mathbb R.}$$

Step 4: Inspect the singular data. The data give $A=1$ but leave $B$ arbitrary, so all solutions are $\boxed{y=t+Bt^2}$. Uniqueness fails despite this two-dimensional global family. No regular normalized equation through zero could include $v=t^2$: at zero its value and slope vanish while its second derivative is $2$, making the normalized equation impossible there.