Determining interval of stability for multistep method

Question

I am attaching a solved example for solving an IVP for converting a second order IVP to first order IVP. I Know it may appear simple but I am unable to understand the circled steps. Any help towards explaining the same would be really appreciated

Answer 1

We are given:

$$y'' = y' + 1\\y(0) = 1, y'(0) = 1$$

We can proceed as follows

• Let $y_1 = y \implies y_1' = y' = y_2$
• Let $y_2 = y' \implies y_2' = y'' = y'+ 1 = y_2 + 1$

Note that the given initial conditions would similarly follow as

• $y(0) = 1 \implies y_1(0) = 1$
• $y'(0) = 1 \implies y_2(0) = 1$

The new system can be written as

$$\tag 1 Y' = \begin{bmatrix} y_1' \\ y_2' \end{bmatrix} = \begin{bmatrix} y_2 \\ 1 + y_2 \end{bmatrix}$$

Taking the derivative of $(1)$

$$\tag 2 Y'' = \begin{bmatrix} y_1'' \\ y_2'' \end{bmatrix} = \begin{bmatrix} y_2' \\ y_2' \end{bmatrix}$$

From our original substitutions, we have that $y_2' = 1 + y_2$, so we can rewrite this as

$$\tag 3 Y'' = \begin{bmatrix} y_1'' \\ y_2'' \end{bmatrix} = \begin{bmatrix} 1 + y_2 \\ 1 + y_2 \end{bmatrix}$$