# Wronskian and differential equations

Given the fact that solutions $$Y_1(t) = (x_1(t), y_1(t))$$ and $$Y_2(t) = (x_2(t), y_2(t))$$ for the system of $$\frac{dy}{dt} = AY$$, where

$$A=\begin{pmatrix} a & b\\ c&d \end{pmatrix}$$ and then clearly the Wronskian is defined as $$W(t)= x_1(t)y_2(t)-x_2(t)y_1(t)~,$$then

1. When asked to compute $$\frac{dW}{dt}$$, does it equal $$AW$$, i.e. the matrix about multiplied by $$x_1(t)y_2(t)-x_2(t)y_1(t)~?$$
2. If $$Y_1(t)$$ and $$Y_2(t)$$ are solutions to a linear system, how do I show that $$\frac{dW}{dt} = (a+d)(W(t))~?$$
3. How do I find the general solution of the above differential equation? i.e. how do I go about figuring the determinant of this linear system?

Thanks!

details: $$W(t)=\begin{vmatrix} x_1(t) & x_2(t)\\ y_1(t)& y_2(t) \end{vmatrix}\\ W'(t)=\begin{vmatrix} x'_1(t) & x_2(t)\\ y'_1(t)& y_2(t) \end{vmatrix}+\begin{vmatrix} x_1(t) & x'_2(t)\\ y_1(t)& y'_2(t) \end{vmatrix} \\= \begin{vmatrix} ax_1(t) + bx_2(t) & x_2(t)\\ ay_1(t) + by_2(t) & y_2(t) \end{vmatrix} + \begin{vmatrix} x_1(t) & cx_1(t) + dx_2(t)\\ y_1(t)& cy_1(t) + dy_2(t) \end{vmatrix} \\= \begin{vmatrix} ax_1(t) & x_2(t)\\ ay_1(t) & y_2(t) \end{vmatrix} + \begin{vmatrix} x_1(t) & dx_2(t)\\ y_1(t)& dy_2(t) \end{vmatrix}=(a+d)W(t)$$
You conclude that $$W(t) = [x_1(0)y_2(0) - x_2(0)y_1(0)]\exp((a+d)t)$$