#6,pg.389 (part d): Let $\vec{y}^{(1)} = \begin{bmatrix} t \\ 1 \end{bmatrix}$ and $\vec{y}^{(2)} = \begin{bmatrix} t^2 \\ 2t \end{bmatrix}$. Find a system $\vec{y}' = P(t)\vec{y} + \vec{g}$.
Solution: We will assume this to be homogeneous so we will pick $\vec{g}=\vec{0}$ (note: nonhomogeneous ones could exist but they will be harder to find). We write $P(t) = \begin{bmatrix} p_{11}(t) & p_{12}(t) \\ p_{21}(t) & p_{22}(t) \end{bmatrix}$ and then plug in $\vec{y}^{(1)}$ into the equation $\vec{y}' = P\vec{y}$ to see
$$\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} p_{11}(t) & p_{12}(t) \\ p_{21}(t)
& p_{22}(t) \end{bmatrix} \begin{bmatrix} t \\ 1 \end{bmatrix}$$
which after multiplying the right side out and simplifying yields
$$\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} tp_{11} + p_{12} \\ t p_{21} + p_{22} \end{bmatrix}.$$
Similarly, plug $\vec{y}^{(2)}$ into $\vec{y}'=Py$ and simplify to get
$$\begin{bmatrix} 2t \\ 2 \end{bmatrix} = \begin{bmatrix} t^2 p_{11} + 2t p_{12} \\ t^2 p_{21} + 2tp_{22} \end{bmatrix}$$
These two equations encode the following system of equations:
$$\left\{ \begin{array}{ll}
tp_{11} + p_{12} &= 1 \\
tp_{21}+p_{22} &= 0 \\
t^2p_{11}+2tp_{12} &= 2t \\
t^2 p_{21}+2tp_{22} &= 2.
\end{array} \right.$$
From the first equation we see that $p_{12} = 1-tp_{11}$. Plug this into the third equation to get
$$t^2 p_{11} + 2t (1-tp_{11}) = 2t,$$
and upon simplifying we get
$$p_{11} = 0,$$
and hence
$$p_{12}=1.$$
From the second equation we get $p_{22}=-tp_{21}$. Plug this into the fourth equation to get
$$t^2 p_{21} -2t^2 p_{21} = 2$$
and simplifying for $p_{21}$ yields
$$p_{21} = -\dfrac{2}{t^2}$$
and hence
$$p_{22} = -t p_{21} = -\dfrac{2}{t}.$$
Thus our functions are solutions to the system
$$\vec{y}' = \begin{bmatrix} 0 & 1 \\ -\dfrac{2}{t^2} & -\dfrac{2}{t}.\end{bmatrix}y,$$
which matches the answer in the back of the book.
