How to find g(t) of an initial value problem

Question

Consider the initial value problem $$y′=\begin{bmatrix} 4 & 3t \\ 4t^2 & -3 \\ \end{bmatrix} y +g(t)$$

$$y(1)=\begin{bmatrix} -1 \\ 3\\ \end{bmatrix}$$

We have system of equations : Suppose we know that $$y(t)=\begin{bmatrix} t+α \\ -3t^2+ β\\ \end{bmatrix}$$

is the unique solution to this initial value problem. Find g(t) and the constants α and β.

Picture of Question

I was already able to calculate α and β as -2 and 6 respectively, however I'm not sure what to do next to find g(t).

Sorry I am so bad at rendering these equations, but i hope the attached image will help make them clear! Thank You!

Answer 1

Hint

$$y(t)=\begin{bmatrix} t+α \\ -3t^2+ β\\ \end{bmatrix} \implies y(1)=\begin{bmatrix} 1+α \\ -3+ β\\ \end{bmatrix} \implies \begin{bmatrix} α +1\\ -3+ β\\ \end{bmatrix}=\begin{bmatrix} -1 \\ 3\\ \end{bmatrix} \implies \begin{bmatrix} \alpha= -2 \\ \beta =6\\ \end{bmatrix}$$

$$g(t)=y′-\begin{bmatrix} 4 & 3t \\ 4t^2 & -3 \\ \end{bmatrix}y \implies g(t)=\begin{bmatrix} 1 \\ -6t\\ \end{bmatrix}-\begin{bmatrix} 4 & 3t \\ 4t^2 & -3 \\ \end{bmatrix}\begin{bmatrix} t-2 \\ -3t^2+ 6\\ \end{bmatrix}$$

Answer 2

Just rearrange: $$g(t)=y'-\begin{bmatrix} 4 & 3t \\ 4t^2 & -3 \\ \end{bmatrix} y.$$