More on the Wronskian — Question 4

Question 4 – Solution

Strategy. Abel’s identity determines the first-derivative coefficient; the known nonzero solution then determines the remaining coefficient.

Step 1: Recover the equation. Since $W$ never vanishes, $$p=-W'/W=2t.$$ Substituting $u=e^t$ gives $1+p+q=0$, hence $$\boxed{y''+2ty'-(1+2t)y=0.}$$

Step 2: Recover the companion. The identity $W=e^t(v'-v)$ gives $v'-v=2e^{-t^2-t}$. Multiplying by $e^{-t}$ and integrating from zero yields $$\boxed{v(t)=e^t\int_0^t2e^{-s^2-2s}\,ds.}$$ The integral is smooth on all of $\mathbb R$. It gives $v(0)=0$ and $v'(0)=2$.

Step 3: Verify by differentiation. Write $J$ for the integral, so $J'=2e^{-t^2-2t}$. Then $v'=v+2e^{-t^2-t}$ and $e^t(v'-v)=2e^{-t^2}$, as prescribed. If $L[y]=y''+py'+qy$, direct expansion gives $$uL[v]-vL[u]=W'+pW.$$ Here $L[u]=0$ and $W'+2tW=0$. Since $u=e^t\ne 0$, this proves $L[v]=0$ everywhere.

Step 4: Identify what the Wronskian omits. With $p=2t$ and any continuous choice of $q$, regular existence supplies solutions with data $(1,0)$ and $(0,2)$ at zero. Their initial Wronskian is $2$, so Abel’s identity gives the same $2e^{-t^2}$ for every such $q$. Thus the Wronskian alone fixes $p$ but not $q$; the supplied nonzero seed was essential for the latter.