#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.