Problem: I want to use Laplace transforms to solve the following: $$\def\b{\begin{pmatrix}}\def\e{\end{pmatrix}}\def\d{\dot}$$ $$ \b\d y_1 \\ \d y_2\e = \b 2y_1 + y_2 \\ -y_1 \e$$
where $y_1(0)=0$ and $y_2(0)=1$
Firstly I would like to say: I have never done a question like this. Secondly I am following what would appear to be a very badly explained method in my Kreyszig - Advanced Engineering Mathematics textbook(which is actually usually clear and pleasna
My attempt: We convert everything into the form the textbook uses: $$A= \b 2 & 1\\-1 & 0\e,G=\b0\\0\e,Y=\mathfrak{L}(y)$$ We jump straight into the form: $$(A-sI)Y = -y(0) - G$$ $$\b 2-s & 1\\-1 & -s\e\b Y_1 \\ Y_2\e=\b0\\-1\e$$ $$(2-s)Y_1 + Y_2 = 0$$ $$-Y_1-sY_2 + 1=0$$ $$Y_2=\frac{1-Y_1}s$$ $$(2-s)Y_1 + \frac{1-Y_1}s = 0$$ $$Y_1+\frac{1-Y_1}{s(2-s)}=0$$ $$\frac{(1-s)Y_1}{(2-s)}=0$$ $$(1-s)Y_1=0$$ $$Y_1=sY_1$$
Now I don't know what is going on.
We have $$y_1' = 2y_1 + y_2 \\ y_2' = -y_1 \\ y_1(0) = 0, y_2(0) = 1$$
Taking Laplace transforms, we have:
$$sy_1(s) - y_1(0) = 2y_1(s) + y_2(s) \\ s y_2(s) - y_2(0) = -y_1(s)$$
Substituting ICs yields:
$$sy_1(s) = 2y_1(s) + y_2(s) \\ s y_2(s) - 1 = -y_1(s)$$
You can put this into matrix form to solve if desired (but it is instructive to see where things went wrong). We would get:
$$\b s-2 & -1\\1 & s\e\b y_1 \\ y_2\e=\b0\\1\e$$
From the first equation, we have:
$$y_2(s) = (s-2)y_1(s)$$
Substituting that into the second equation and simplifying yields:
$$y_1(s) = \dfrac{1}{(s-1)^2}$$
The inverse Laplace yields:
$$y_1(t) = t e^t$$
I am sure you can take it from here.
You should end up with:
$$y_2(t) = -e^t(t-1)$$