Add first page and first example of second page from first set of lecture notes on DE's.

src/differential_equations_1.md

@@ -7,12 +7,119 @@ title: Differential Equations
 A differential equation is any equation which involves both a function and some 
 derivative of that function. In this course we will be focusing on 
 *Ordinary Differential Equations*, meaning that our equations will involve 
-functions of one dependent variable and hence any derivatives will be full 
-derivatives. Equations which involve a function of several dependent variables
+functions of one independent variable and hence any derivatives will be full 
+derivatives. Equations which involve a function of several independent variables
 and their partial derivatives are handled in courses on 
 *Partial Differential Equations*. 
 
-We consider functions $x(t)$ and define $\dot{x}(t)=\frac{dx}{dt}$, $x^{(n)}(t)=\frac{d^{n}x}{dt^{n}}$
+We consider functions $x(t)$ and define $\dot{x}(t)=\frac{dx}{dt}$, 
+$x^{(n)}(t)=\frac{d^{n}x}{dt^{n}}$. An $n$*-th* order differential equation is 
+an equation of the form 
 
+$$x^{(n)}(t) = f(x^{(n-1)}(t), \cdots, x(t), t).$$ 
 
+Typically, $n \leq 2$. Such an equation will usually be presented with a set of 
+initial conditions,
+
+$$x^{(n-1)}(t_{0}) = x^{(n-1)}_{0}, \cdots, x(t_0)=x_0. $$
+
+This is because to fully specify the solution to an $n$*-th* order differential 
+equation, $n-1$ initial conditions are necessary. To understand why we need 
+initial conditions, look at the following example.
+
+!!! check "Example: Initial conditions"
+    Consider the following calculus problem,
+    
+    $$\dot{f}(x)=x. $$ 
+
+    By integrating, one finds that the solution to this equation is 
+
+    $$\frac{1}{2}x^2 + c,$$
+
+    where $c$ is an integration constant. In order to specify the integration 
+    constant, an initial condition is needed. For instance, if we know that when 
+    $x=2$ then $f(2)=4$, we can plug this into the equation to get 
+
+    $$\frac{1}{2}*4 + c = 4, $$
+
+    which implies that $c=2$. 
+    
+Essentially initial conditions are needed when solving differential equations so
+that unknowns resulting from integration may be determined.
+
+!!! info Terminology for Differential Equations
+    1. If a differential equation does not explicitly contain the 
+        independent variable $t$, it is called an *autonomous equation*.
+    2. If the largest derivative in a differential equation is of first order, 
+        i.e. $n=1$, then the equation is called a first order differential 
+        equation.
+    3. Often you will see differential equation presented using $y(x)$ 
+        instead of $x(t)$. This is just a different nomenclature. 
+            
+In this course we will be focusing on *Linear Differential Equations*, meaning 
+that we consider differential equations $x^{(n)}(t) = f(x^{(n-1)}(t), \cdots, x(t), t)$
+where the function $f$ is a linear ploynomial function of the unknown function
+$x(t)$. A simple way to spot a non-linear differential euation is to look for 
+non-linear terms, such as $x(t)*\dot{x}(t)$ or $x^{(n)}(t)*x^{(2)}(t)$. 
+
+Often, we will be dealing with several coupled differential equations. In this 
+situation we can write the entire system of differential equations as a vector 
+equation, involving a linear operator. For a system of $m$ equations, denote 
+
+$$**x(t)** = \begin{bmatrix}
+x_1(t) \\
+\vdots \\
+x_{m}(t) \\
+\end{bmatrix}.$$
+
+A system of first order linear equations is then written as 
+
+$$\dot{**x(t)**} = **f**(**x(t)**,t) $$
+
+with initial condition $**x(t_0)** = **x_0**$.
+
+# Basic examples and strategies
+
+The simplest type of differential equation is the type learned about in the 
+integration portion of a calculus course. Such equations have the form,
+
+$$\dot{x}(t) = f(t). $$
+
+When $F(t)$ is an anti-derivative of $f(t)$ i.e. $\dot{F}=f$, then the solutions
+to this type of equation are 
+
+$$x(t) = F(t) + c. $$
+
+!!! check "Example: Differential equations from calculus"
+    Given the equation
+    
+    $$\dot{x}(t)=t, $$
+    
+    one finds by integrating that the solution is $\frac{1}{2}t^2 + c$. But we 
+    can also re-write this equation in the form we have been discussing, 
+    
+    $$\dot{x}(t)=f(x(t)).$$ 
+    
+    This implies that $\frac{f(x(t))}{\dot{x}(t)} = 1$. Let $F(x)$ be the 
+    anti-derivative of $f(x)$. Then, making use of the chain rule
+    
+    $$\frac{f(x)}{dot{x}(t)} = \frac{\frac{dF}{dx}}{\frac{dx}{dt}} = \frac{d}{dt}F(x(t)) = 1$$
+    
+    $$\Leftrightarrow F(x(t)) = t + c.$$
+    
+    From this we notice that if we can solve for $x(t)$ then we have the 
+    solution!
+
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