Basic Concepts for nth Order Linear Equations — Question 4

Question 4 – Solution

Strategy. The coefficient of the next-to-highest derivative controls the entire Wronskian, not just its value at one point.

Step 1: Differentiate the determinant. The Wronskian rows contain derivative orders $0,1,2,3$. Differentiating any of the first three rows duplicates the following row, giving determinant zero. In the last-row term, substitute $y_j^{(4)}=-p_3y_j'''-p_2y_j''-p_1y_j'-p_0y_j$ for every column. All terms except the multiple of the original last row again duplicate a row. Therefore $$\boxed{W'=-p_3W.}$$

Step 2: Determine the Wronskian globally. Here $p_3=-2x/(1+x^2)$, so $$W(x)=-6\exp\!\left(\int_0^x\frac{2s}{1+s^2}\,ds\right) =\boxed{-6(1+x^2)}.$$ It never vanishes. The four solutions are linearly independent, and the regular fourth-order homogeneous solution space has dimension four; hence they form a fundamental set on $\mathbb R$.

Step 3: Interpret invertibility of the derivative matrix. At any finite $x_0$, the matrix with columns $(y_j,y_j',y_j'',y_j''')(x_0)$ has determinant $W(x_0)\ne 0$. It therefore maps one and only one coefficient vector to any prescribed initial derivative vector. The resulting combination solves the equation; the linear IVP theorem shows it is the unique solution for those data on $\mathbb R$.

Step 4: Reject a plausible-looking candidate. First, $\widetilde W(1)=0$ contradicts the nonvanishing formula forced by $W(0)\ne 0$ on this regular connected interval. Second, it fails the differential identity even at the initial point: $\widetilde W'(0)=6$, whereas $-p_3(0)\widetilde W(0)=0$. Matching one determinant value is insufficient; a valid Wronskian must obey the identity throughout the interval.