<p> Solving Linear Systems of Differential Equations:</p><p>You are given a linear system of differential equations: </p><p> d a b  Y  Y  AY, Y (0)  Y0 . dt c d </p><p>The type of behavior depends upon the eigenvalues of matrix A. The procedure is to determine the eigenvalues and eigenvectors and use them to construct the general solution. </p><p>If you have an initial condition, you can determine your two arbitrary constants in the general solution in order to obtain the particular solution. Thus, if Y1 (t ) and Y2 (t ) are two linearly independent solutions, then the general solution is given as Y(t ) c1 Y 1 ( t )  c 2 Y 2 ( t ). </p><p>Then, setting t  0, you get two linear equations for c1 and c2 : c1Y 1(0) c 2 Y 2 (0)  Y 0 .</p><p>The major work is in finding the linearly independent solutions. This depends upon the different types of eigenvalues that you obtain from solving the characteristic equation, det(Y I )  0.</p><p>I. Two real, distinct roots. a. Solve the eigenvalue problem AV  V for each eigenvalue obtaining two </p><p> eigenvectors V1 ,V 2 .</p><p>1t  2 t b. Write the general solution as a linear combination Y(t )  c1 e V 1  c 2 e V2 II. Two complex conjugate roots. a. Solve the eigenvalue problem AV  V for one eigenvalue,    i  , obtaining one eigenvector V. Note that this eigenvector may have complex entries. b. Write the vector Y(t ) et V  e  t (cos t  i sin  t ) V . c. Construct two linearly independents solutions to the problem by letting</p><p>V1  Re( Y (t )) and V2  Im( Y (t )) .</p><p>Note that since the original system of equations does not have any i's, then d we have [Re(Y (t )) i Im( Y ( t ))]  A [Re( Y ( t ))  i Im( Y ( t ))]. Differentiating the dt sum and splitting the real and imaginary parts of the equation, gives d d Re(Y (t )) i Im( Y ( t ))  A [Re( Y ( t ))]  i A [Im( Y ( t ))]. Setting the real and dt dt d imaginary parts equal, we have Re(Y (t )) A [Re( Y ( t ))], and dt d Im(Y (t )) A [Im( Y ( t ))]. Therefore, the real and imaginary parts each are dt solution of the system!</p><p> d. Write the general solution as a linear combination Y(t )  c1 V 1  c 2 V2 I. One Repeated Root e. Solve the eigenvalue problem AV  V for the one eigenvalue obtaining </p><p> the first eigenvector V1.</p><p> f. Solve the eigenvalue problem AV2 V 2  V 1 for V2 . t  t g. The general solution is then given by Y(t ) c1 e V 1  c 2 e ( V2  t V 1 ) . II. Examples 4 2  h. A   . 3 3 </p><p>4  2 0  3 3  Eigenvalues: 0 (4  )(3   )  6 Thus,   1,6. 02  7   6 0 (  1)(   6)</p><p>Eigenvectors:   1:   6 :</p><p>4 2 v1   v 1  4 2 v1   v 1          6   3 3 v2   v 2  3 3 v2   v 2 </p><p>3 2 v1   0  2 2 v1   0              3 2 v2   0  3 3 v2   0 </p><p>v1  2  v1  1  3v1 2 v 2  0,      . 2v1  2 v 2  0,      . v2  3  v2  1 </p><p>1t  2 t Y(t )  c1 e V 1  c 2 e V2</p><p> t2 6 t  1  Y(t )  c1 e   c 2 e   General Solution: 3   1  t6 t 2c1 e c 2 e   t6 t . 3c1 e  c 2 e </p><p>3 5  i. A   . 1 1 </p><p>3  5 0  1 1   0 (3  )(  1   )  6 Eigenvalues: Thus,  1 i ,1  i . 02  2   2 (  2)  4  4(1)(2)   1  i . 2 Eigenvectors:  1  i :</p><p>3 5 v1   v 1    (1  i )   1 1 v2   v 2 </p><p>2i  5 v1   0        1 2  i v2   0 </p><p>v1  2  i  (2i ) v1  5 v 2  0,      . v2  1 </p><p>Complex Solution:</p><p>t2i (1 i ) t  2  i  e  e   1   1 </p><p> t 2  i  e(cos t  i sin t )  1 </p><p> t (2i )(cos t  i sin t )   e   cost i sin t </p><p> t (2cost sin t )  i (cos t  2sin t )   e   cost i sin t </p><p> t2cost sin t  t  cos t  2sin t  e   ie  . cost   sin t </p><p> t2cost sin t  t  cos t  2sin t  Y(t )  c1 e   c 2 e   cost   sin t  General Solution: t c1(2cos t sin t )  c 2 (cos t  2sin t )   e  . c1cos t c 2 sin t </p><p> t2c1 c 2  t  2 c 2  c 1  Note: This can be rewritten as Y(t ) e cos t   e sin t   . c1   c 2 </p><p>7 1  j. A   . 9 1 </p><p>7  1 0  9 1  0 (7  )(1   )  9 Eigenvalues: Thus,   4. 02  8   16 0 (  4)2 .</p><p>Eigenvectors:   4 :</p><p>7 1 v1   v 1     4   9 1 v2   v 2 </p><p>3 1 v1   0        9 3 v2   0 </p><p>v1  1  3v1 v 2  0,      . v2  3 </p><p>Second Linearly Independent Solution:</p><p>Solve AV2 V 2  V 1</p><p>7 1 u1   u 1   1    4      9 1 u2   u 2   3 </p><p>3 1 u1   1        9 3 u2   3 </p><p>3u1 u 2  1   u 1  1       . 9u1 3 u 2  3  u 2  2 </p><p>General Solution: </p><p>t  t Y(t ) c1 e V 1  c 2 e ( V2  t V 1 )</p><p>4t1  4 t   1   1   Y(t )  c1 e   c 2 e    t    3   2   3  </p><p>4t c1 c 2 (1  t )   e  . 3c1 c 2 (2  3 t ) </p>