Add page 3 of notes up to phase portrait.

src/differential_equations_1.md

@@ -161,6 +161,81 @@ $$\Rightarrow F(x(t)) = G(t) + c $$
 
 Given this form of general solution, knowledge of specific functions $f, g$ would
 make it possible to solve for $x(t)$. 
+
+!!! check "Example: First order linear differential equation with coefficient t"
+    Let us apply the above strategy to the following equation,
+    
+    $$\dot{x}= t x^2 .$$
+    
+    Comparison to the strategy indicates that we should define $f(x)=x^2$ and 
+    $g(t)=t$. As before, we can re-arrange the equation
+    
+    $$\frac{\dot{x}}{x^2} = t. $$
+    
+    It is then necessary to find $F(x)$, the anti-derivative of $\frac{1}{f(x)}$,
+    or the left hand side of the above equation, as well as $G(t)$, the 
+    anti-derivative of $g(t)$, or the right hand side of the previous equation.
+    
+    Integrating, one finds
+    
+    $$F(x) = - \frac{1}{x} $$
+    $$G(t)=\frac{1}{2}t^2 + c. $$
+    
+    Accordingly then, the equation we have is
+    
+    $$- \frac{1}{x} = \frac{1}{2} t^2 + c. $$
+    
+    At this point, it is possible to solve for $x(t)$ by re-arrangement
+    
+    $$x(t)= \frac{-2}{t^2 + c_0}, $$
+    
+    where in the last line we have defined $c_0 = 2c$. Once again, specification
+    of an initial condition would allow determination of $c_0$ directly. To see 
+    this, suppose $x(0) = 2$. Inserting this into the equation for $x(t)$ we have
+    
+    $$2 = \frac{-2}{c_0} $$
+    $$ \Rightarrow c_0 = -1.$$
+    
+    Having solved for $c_0$, with the choice of initial condition $x(0)=2$, the 
+    full equation for $x(t)$ is 
+    
+    $$x(t)=\frac{-2}{t^2 -1}. $$
+    
+!!! check "Example: First order linear differential equation with general 
+    non-constant coefficient function"
+
+    Let us apply the strategy for dealing with non-constant coefficient functions
+    to the more general equation
+
+    $$\dot{x}= g(t) \cdot x. $$
+
+    This equation suggests that we first define $f(x)=x$ and then find $F(x)$ and 
+    $G(t)$, the anti-derivatives of $\frac{1}{f(x)}$ and $g(t)$, respectively. Doing
+    so, we determine 
+
+    $$F(x) = log(x) $$
+
+    Continuing to follow the protocol, we arrive at the equation
+
+    $$log(x) = G(t) + c.$$
+
+    Exponentiating and defining $c_0:=e^c$, we obtain an equation for $x(t)$,
+
+    $$x(t)= c_0 e^{G(t)} .$$
+    
+So far we have only considered first order differential equations. If we consider
+extending the strategies we have developed to higher order equations such as
+
+$$x^{(2)}(t)=f(x), $$
+
+with f(x) a linear function, then our work will swiftly become tedious. Later on
+we will develop the general theory for linear equations which will allow us to 
+tackle such higher order equations. For now, we move on to considering systems 
+of coupled first order linear DE's. 
+
+
+    
+