More on the Wronskian — Question 2

Question 2 – Solution

Strategy. Separate the constant determinant of a basis change from the time evolution supplied by Abel’s identity.

Step 1: Expand the determinant. Terms containing $uu'$ or $vv'$ cancel, leaving $$\boxed{W[au+bv,cu+dv]=(ad-bc)W[u,v].}$$ The constants may be factored through differentiation; this step would require extra terms for time-dependent coefficients.

Step 2: Apply the two determinants. Here $W[u,v](0)=1$ and $p=2$, so $W[u,v]=e^{-2t}$. The coefficient determinant is $2\cdot 4-(-1)\cdot 3=11$, hence $$\boxed{W[f,g]=11e^{-2t}>0.}$$ Both combinations solve the equation and are independent, so they are fundamental. Reversing their order negates the Wronskian but leaves independence unchanged.

Step 3: Compute the area. The columns $(f,f')$, $(g,g')$ have signed area $W[f,g]$ and ordinary area $|W[f,g]|$. At zero they are $(2,-1)$ and $(3,4)$, with area $11$. The half-area equation gives $$11e^{-2t}=11/2,\qquad\boxed{t=\tfrac 12\ln 2.}$$ The area is strictly decreasing for $t\ge 0$, so this is the first such time.

Step 4: Distinguish collapse from a limit. The determinant never vanishes at finite $t$, so the two state vectors never become parallel there. The area tends to zero as $t\to\infty$; this does not destroy finite-time independence. Area alone also does not specify the lengths of the two vectors or their angle separately.