Systems of Differential Equations — Question 1

Question 1 – Solution

Strategy. Normalize first, and distinguish the evolution of a family of states from that of one trajectory.

Step 1: Preserve the derivative chain. Since $1+t^2>0$ for every real $t$, $$\boxed{A(t)=\begin{pmatrix}0&1&0\\0&0&1\\0&-2/(1+t^2)&2t/(1+t^2)\end{pmatrix}, \qquad X(0)=\begin{pmatrix}1\\0\\2\end{pmatrix}.}$$ The conversion is regular on $\mathbb R$, and its first two rows enforce $X_2=X_1'$, $X_3=X_1''$, so it is reversible.

Step 2: Verify the complete state. The initial data select $y=1+t^2$ among quadratic polynomials. Its derivatives are $2t,2,0$, and the scalar residual is $-4t+4t=0$. Thus $$\boxed{X(t)=(1+t^2,\,2t,\,2)^T.}$$ The system’s last row gives $(-4t+4t)/(1+t^2)=0=X_3'$; the other rows and initial vector also agree. Continuity of $A(t)$ gives uniqueness on $\mathbb R$.

Step 3: Compute the state-volume factor. Liouville’s formula gives $$\det\Phi(t)=\exp\!\left(\int_0^t\frac{2v}{1+v^2}\,dv\right)=\boxed{1+t^2}.$$ An initial parallelepiped of states has its oriented volume multiplied by this positive factor. The factor increases with $|t|$, not with $t$ on the entire line.

Step 4: Test the individual-growth claim. For any constant $c$, $X=(c,0,0)^T$ is a constant solution. Hence expansion of state volume does not force every trajectory to grow. The prescribed components are shown below; their different behaviors also caution against that inference.