Systems of Differential Equations — Question 1

Question 1 – Solution

Strategy. Store successive derivatives as independent state coordinates, then recover their required relations from the first rows.

Step 1: Build the derivative chain. Set $x_1=y$, $x_2=y'$, $x_3=y''$. Solving for the highest derivative gives $$\boxed{X'=\begin{pmatrix}0&1&0\\0&0&1\\-1&2&-1/t\end{pmatrix}X +\begin{pmatrix}0\\0\\e^t\end{pmatrix},\quad X(1)=(2,-1,0)^T.}$$ In particular the last row is $x_3'=-x_1+2x_2-x_3/t+e^t$.

Step 2: Separate the classifications. The three-dimensional system is linear: coefficients depend on $t$, not on the unknown state. It is nonhomogeneous because the forcing is nonzero, and nonautonomous because $1/t$ and $e^t$ depend explicitly on time. The largest open coefficient-continuity interval containing $1$ is $(0,\infty)$.

Step 3: Reorder the coordinates consistently. Here $z_1=x_3$, $z_2=x_1$, $z_3=x_2$, so $$\boxed{Z'=\begin{pmatrix}-1/t&-1&2\\0&0&1\\1&0&0\end{pmatrix}Z +\begin{pmatrix}e^t\\0\\0\end{pmatrix},\quad Z(1)=(0,2,-1)^T.}$$ A fixed permutation is invertible. It relabels coordinates and initial data without adding or removing information; there are still three initial values.

Step 4: Check equivalence and the domain limitation. Any scalar solution supplies a state satisfying all three rows. Conversely, for a continuously differentiable state solution, the first two rows imply $x_2=x_1'$ and $x_3=x_1''$; the third then makes $x_1$ three times differentiable and gives the original equation and initial data. The standard linear-system theorem gives a unique solution on $(0,\infty)$. It does not apply through $0$, where $1/t$ is undefined. This alone does not prove that a particular solution cannot have a smooth extension; any extension would require a separately specified equation at the singular point.