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The Applications of Ordinary Differential Equation

Term Paper Instructions:

no specific instruction from professor

just a paper that solve a problem use ode. For example use ode to predict the population or predict a simple economic model or anything use ode to solve

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                                    The Applications of Ordinary Differential Equation Author Name University Name                                   Linear Systems of Ordinary Differential Equations   For the linear systems of ordinary differential equations, we will have to consider the effects of some unknown functions on one another. The two by two system of ordinary differential equations will look something like this:  
  dx = ax + by                    
dt   d x   a b   x  
         
    or =    
dy       y c d   y  
       
  = cx + dy dt         ·  
dt            
                       
Let’s say that Y has been set to an unknown vector and A has been set as coefficient matrix, then we will have the following matrix equation   dY (1) dt  
  1. Source and Sink Equilibrium
  First Theorem: If A consists of two distinctive, real values λ1 and λ2 with corresponding eigenvectors as v1 and v2, then the main solution will be as following: Y(t) = k1 exp ( λ1 t ) v1  + k2 exp ( λ2 t ) v2.   Example: The unstable system of ordinary differential equations
  dx = x   d x 1 0 x  
  dt  
    or         =      
  dy                  
    = 2y dt y0 2 ·  y  
dt  
It should be noticed that the coefficient matrix A consists of two values, λ+ = 2 and λ− = 1, along with eigenvectors v+ = (0, 1) and v− = (1, 0). In that case, the solution to this problem will be something like this:
Y(t) = k1et 1   + k2e2t 0 .  
0 1  


  1. Periodic Equilibrium
  Second Theorem: Let’s suppose that the values of coefficient matrix A are in the form of λ = α ± iβ, then the solution will be as following: Y(t) = eαt  cos(βt) + i sin(βt) v We should not forget that here v is the eigenvector of A.   Example: Periodic solution will be something like this:
    dx   =   y                              
  dt                                
  −             d x   0 1 x      
                           
            or             =   −0        
  dy = x               y 1 y      
                dt ·    
                                         
dt                                    
Here we will get Char(A) = λ2 + 1 = 0. Hence λ = ±i.     i ,  
Only one vector is needed in that case, so let’s take v =  
1  
corresponding to λ = i and coming fro...
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