diff --git a/src/differential_equations_1.md b/src/differential_equations_1.md
index 21545ba4f70056d2efe3ede6e7b7c545d5997987..261674a8229cb7e4836781754961d08606e37216 100644
--- a/src/differential_equations_1.md
+++ b/src/differential_equations_1.md
@@ -369,7 +369,6 @@ $$**\Phi**(t) \dot{**c**}(t) = **b**(t). $$
 
 In order to cancel terms in the previous line we made use of the fact that 
 $**\Phi**(t)$ solves the homogeneous equation $\dot{**\Phi**} = **A** **\Phi**$.
-
 By way of inverting and integrating, we can write an equation for the coefficient
 vector $**c**(t)$
 
@@ -380,9 +379,94 @@ the particular solution,
 
 $$**\psi**(t)= **\Phi**(t) \cdot \int **\Phi**^{-1}(t) **b**(t) dt .$$
 
-
-
-
+!!! check "Example: Inhomogeneous first order linear differential equation"
+
+    The technique for solving a system of inhomogeneous equations also works for a 
+    single inhomogeneous equation. Let us apply the technique to the equation
+
+    $$ \dot{x} = \lambda x + a. $$
+
+    In this particular inhomogenous equaiton, the function $g(t)=a$. As discussed in
+    an earlier example, the solution to the homogenous equation is
+    $c e^{\lambda t}$. Hence we define $\phi(t)=e^{\lambda t}$ and make the ansatz
+
+    $$\psi(t) = c(t) e^{\lambda t}. $$
+
+    Solving for $c(t)$ results in 
+
+    $$c(t) = \int e^{- \lambda t} a  dt$$
+    $$c(t) = \frac{- a }{\lambda} e^{- \lambda t} $$
+
+    Overall then, the solution (which can be easily verified by substitution) is 
+
+    $$\psi(t) = - \frac{a}{\lambda}.  $$ 
+    
+## Homogeneous linear system with constant coefficients ##
+
+The type of equaiton under consideration in this section looks like 
+
+$$ \dot{**x**}(t) = **A** **x**(t),$$
+
+where throughout the section $**A**$ will be a constant matrix. It is possible 
+to define a formal solution using the *matrix exponential*, 
+$**x**(t) = e^{**A** t}$.
+
+!!! info "Definition: Matrix Exponential"
+
+    Before defining the matrix exponential, recall the definition of the regular 
+    exponential function in terms of Taylor series,
+    
+    $$e^{x} = \overset{\infty}{\underset{n=0}{\Sigma}} \frac{x^n}{n!},$$
+    
+    in which it is agreed that $0!=1$. The matrix exponential is defined in 
+    exactly the same way, only now instead of taking powers of a number or 
+    function, powers of a matrix are calculated 
+    
+     $$e^{**A**} = \overset{\infty}{\underset{n=0}{\Sigma}} \frac{{**A**}^n}{n!}.$$
+     
+     It is important to use caution when translating the properties of the normal
+     exponential function over to the matrix exponential, because not all of the
+     regular properties hold generally. In particular, 
+     
+     $$e^{**X** + **Y**} \neq e^{**X**} e^{**Y**},$$
+     
+     unless it happens that 
+     
+     $$**XY** = **YX**.$$
+     
+     The necessary condition for this property to hold, stated on the previous 
+     line, is called *commutativity*. Recall that in general, matrices are not
+     commutative so such a condition is only met for particular choises of 
+     matrices. The property of *non-commutativity* (what happens when the 
+     condition is not met) is of central importance in the mathematical 
+     structure of quantum mechanics. For example, mathematically, 
+     non-commutativity is responsible for the Heisenberg uncertainty relations.
+     
+     On the other hand, one property that does hold, is that $e^{- **A** t}$ is 
+     the inverse of the matric exponential of $**A**$. 
+     
+     Furthermore, it is possible to derive the derivative of the matrix 
+     exponential by making use of the Taylor series formulation,
+     
+     $$\frac{d}{dt} e^{**A** t} = \frac{d}{dt} \overset{\infty}{\underset{n=0}{\Sigma}} \frac{(**A** t)^n}{n!} $$
+     $$\frac{d}{dt} e^{**A** t} = \overset{\infty}{\underset{n=0}{\Sigma}} \frac{1}{n!} \frac{d}{dt} (**A** t)^n$$
+     $$\frac{d}{dt} e^{**A** t} = \overset{\infty}{\underset{n=0}{\Sigma}} \frac{n **A**}{n!}(**A** t)^{n-1}$$
+     $$\frac{d}{dt} e^{**A** t} = \overset{\infty}{\underset{n=1}{\Sigma}} \frac{**A**}{(n-1)!}(**A** t)^{n-1}$$
+     $$\frac{d}{dt} e^{**A** t} = \overset{\infty}{\underset{n=0}{\Sigma}} \frac{**A**}{n!}(**A** t)^n$$
+     $$\frac{d}{dt} e^{**A** t} = **A** e^{**A** t}.$$
+
+Armed with the matrix exponential and it's derivative, 
+$\frac{d}{dt} e^{**A** t} = **A** e^{**A** t}$, it is simple to verify that 
+the matrix exponential solves the differential equation. Properties of this 
+solution are:
+
+1. The columns of $e^{**A** t}$ form a basis for the solution space.
+2. Accounting for initial conditions, the full solution of the equation is 
+   $\dot{**x**}(t) = e^{**A** t} {**x**}_{0}$, with initial condition 
+    $**x**(0) = e^{**A** 0}{**x**}_0 = \mathbbm{1} {**x**}_{0} = {**x**}_{0}$.
+
+Next we will discuss how to determine a solution in general, going beyond the 
+formal solution just presented.