Systems of Differential Equations — Question 6

Question 6 – Solution

Strategy. Keep the factors in their stated order and regard intermediate derivatives as state coordinates.

Step 1: Form the ordered cascade. The definitions give $$\boxed{q_1'=q_1+q_2,\qquad q_2'=tq_2+q_3,\qquad q_3'=-q_3,\qquad q(0)=(0,0,1)^T.}$$ In particular, $q_3=y''-(t+1)y'+ty$; using only $y''$ as this coordinate would lose the variable-coefficient terms.

Step 2: Expand and invert the coordinate map. Differentiating the expression for $q_3$ and adding $q_3$ yields $$\boxed{y'''-t y''-2y'+(t+1)y=0.}$$ The coordinate matrix is $$q=\begin{pmatrix}1&0&0\\-1&1&0\\t&-(t+1)&1\end{pmatrix} \begin{pmatrix}y\\y'\\y''\end{pmatrix},\qquad \det=1.$$ Its inverse relations are $y=q_1$, $y'=q_1+q_2$ and $y''=q_1+(t+1)q_2+q_3$, so the conversion is globally reversible.

Step 3: Integrate in triangular order. Successive integrating factors give $$q_3=e^{-t},\qquad q_2=e^{t^2/2}\int_0^t e^{-v-v^2/2}\,dv,$$ $$\boxed{y(t)=e^t\int_0^t e^{-u+u^2/2} \left(\int_0^u e^{-v-v^2/2}\,dv\right)du.}$$ These oriented integrals apply to every real $t$. Their derivatives give the three cascade equations. At zero the state is $(0,0,1)$, and the inverse relations give $(y,y',y'')=(0,0,1)$. Finally $q_3'+q_3=0$ verifies the original ordered product directly.

Step 4: Quantify the order error. The commutator $(D+1)(D-t)-(D-t)(D+1)$ is multiplication by $-1$. Thus the original operator minus the reordered one equals $-(D-1)$. The reordered operator is $D^3-tD^2-D+t$, whereas the original has the extra $-D+1$. For $y=e^{-t}$ the reordered product vanishes, but the original gives $2e^{-t}\ne 0$. Reordering these factors changes the solution space.