More on the Wronskian — Question 7

Question 7 – Solution

Strategy. A common nonzero multiplier changes the Wronskian by its square and can remove the first-derivative coefficient.

Step 1: Differentiate the substitution. Put $h=e^{-P/2}$, so $h'/h=-p/2$ and $h''/h=-p'/2+p^2/4$. Substitution of $y=hz$ gives $$z''+(2h'/h+p)z'+(h''/h+ph'/h+q)z=0,$$ and therefore $$\boxed{z''+[q-p'/2-p^2/4]z=0.}$$

Step 2: Transform the determinant. Expanding derivatives cancels the terms containing $hh'$: $$\boxed{W[hz_1,hz_2]=h^2W[z_1,z_2]=e^{-P}W[z_1,z_2].}$$ The transformed equation has no first-derivative term, so its Wronskian has derivative zero. The nonzero multiplier preserves independence.

Step 3: Solve the transformed example. Here $P=t^2$ and $q-p'/2-p^2/4=t^2+3-1-t^2=2$. Thus $z''+2z=0$, and the normalized pair is $$\boxed{\phi=e^{-t^2/2}\cos(\sqrt 2t),\qquad \psi=e^{-t^2/2}\frac{\sin(\sqrt 2t)}{\sqrt 2}.}$$ At zero, $h=1$ and $h'=0$, so the data remain $(1,0)$ and $(0,1)$. The transformed Wronskian is $1$, giving $W[\phi,\psi]=e^{-t^2}>0$ everywhere.

Step 4: Match and interpret. The normalized data give $$\boxed{y=\phi+2\psi=e^{-t^2/2}[\cos(\sqrt 2t)+\sqrt 2\sin(\sqrt 2t)].}$$ Its value and slope at zero are $1,2$. Both basis functions decay to zero because their trigonometric factors are bounded and the Gaussian factor decays. Dependence would require an exact constant relation on the interval; the strictly positive finite-time Wronskian rules that out.