Example 1

We solved the system u'={{5,2},{2,2}}u in class. The solution is c1{1,-2}Exp[t]+c2{2,1}Exp[6t] which looks like:

In[55]:=

  uold={{c1 Exp[t]+2c2 Exp[6t]},
  {-2c1 Exp[t]+c2 Exp[6t]}};
  MatrixForm[uold]

Out[55]=

      t         6 t
  c1 E  + 2 c2 E
  
         t       6 t
  -2 c1 E  + c2 E

Let's use our "new" method

In[56]:=

  Clear[a,m,s,d,evalues]

In[57]:=

  a={{5,2},{2,2}};
  {evalues, m}=Eigensystem[a];
  s=Transpose[m];
  d=DiagonalMatrix[evalues];
  c={c1,c2};
  MatrixForm[a]
  MatrixForm[s]
  MatrixForm[d]

Out[57]=

  5   2
  
  2   2

Out[58]=

  -1   2
  
  2    1

Out[59]=

  1   0
  
  0   6

We want u=Exp[tA]C. To find Exp[tA] we will find S.Exp[tD].Inv[S].

In[60]:=

  expta=s.MatrixExp[t d].Inverse[s];
  MatrixForm[expta]

Out[60]=

   t      6 t          t      6 t
  E    4 E         -2 E    2 E
  -- + ------      ----- + ------
  5      5           5       5
  
      t      6 t      t    6 t
  -2 E    2 E      4 E    E
  ----- + ------   ---- + ----
    5       5       5      5

In[61]:=

  unew=expta.c;
  MatrixForm[unew]

Out[61]=

          t      6 t         t      6 t
      -2 E    2 E           E    4 E
  c2 (----- + ------) + c1 (-- + ------)
        5       5           5      5
  
         t    6 t            t      6 t
      4 E    E           -2 E    2 E
  c2 (---- + ----) + c1 (----- + ------)
       5      5            5       5

It will be difficult to see if uold and unew are the same solution without initial conditions. Let u(0)={3,4}. Then c1=-1, c2=2 in uold (you can verify this!). So oldsol is:

In[62]:=

  oldsol=uold /. {c1->-1,c2->2};
  MatrixForm[%]

Out[62]=

    t      6 t
  -E  + 4 E
  
     t      6 t
  2 E  + 2 E

To use the new technique we need Exp[tA].{3,4}

In[63]:=

  expta.{3,4}

Out[63]=

          t      6 t        t      6 t
      -2 E    2 E          E    4 E
  {4 (----- + ------) + 3 (-- + ------), 
        5       5          5      5
   
          t    6 t           t      6 t
       4 E    E          -2 E    2 E
    4 (---- + ----) + 3 (----- + ------)}
        5      5           5       5

In[64]:=

  Simplify[%];
  MatrixForm[%]

Out[64]=

    t      6 t
  -E  + 4 E
  
     t      6 t
  2 E  + 2 E

And there you have it!

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