Consider a 3rd order linear homogeneous DE of the form L u = u ‴ + a 2 ( x ) u ″ + a 1 ( x ) u ′ + a 0 ( x ) u = f ( x ) ( 1 )and for which u 1 = e − x and u 2 = e − 2 x are solutions to the homogeneous form of (1).Let f ( x ) = 10 e − 2 x . Give an example of a form of a 2 , a 1 and a 0 such that (1) has a stable equilibrium point and an example such that (1) has no stable equilibrium point.My attempt:When I think of stability, I immediately think of eigenvalues (nodes etc). Hence I reduced (1) into a system of linear equations: d u d t = y , d y d t = z , d z d t = − a 2 z − a 1 y − a 0 u . This gives a corresponding matrix A = ( 0 1 0 0 0 1 − a 0 − a 1 − a 2 ) .But after working with this, I feel as if I'm not on the right track. Any advice would be greatly appreciated.

Question

Consider a 3rd order linear homogeneous DE of the form  L  u  =      u    ‴    +      a    2    (  x  )      u    ″    +      a    1    (  x  )      u    ′    +      a    0    (  x  )  u  =  f  (  x  )              (  1  )and for which       u    1    =      e          −      x       and       u    2    =      e          −      2      x       are solutions to the homogeneous form of (1).Let   f  (  x  )  =  10      e          −      2      x      . Give an example of a form of       a    2  ,      a    1   and       a    0   such that (1) has a stable equilibrium point and an example such that (1) has no stable equilibrium point.My attempt:When I think of stability, I immediately think of eigenvalues (nodes etc). Hence I reduced (1) into a system of linear equations:                                          d            u                                d            t                                              =        y        ,                                                  d            y                                d            t                                              =        z        ,                                                  d            z                                d            t                                              =        −                  a          2                z        −                  a          1                y        −                  a          0                u        .            This gives a corresponding matrix  A  =      (                            0                          1                          0                                      0                          0                          1                                      −                      a            0                                    −                      a            1                                    −                      a            2                                )    .But after working with this, I feel as if I&#039;m not on the right track. Any advice would be greatly appreciated.

popman14ee · Accepted Answer

In particular,       u    1   and       u    2   are homogeneous solutions:                    −        1        +                  a          2                −                  a          1                +                  a          0                                    =        0                            −        8        +        4                  a          2                −        2                  a          1                +                  a          0                                    =        0            i.e.       a    1    =  2  +      3    2        a    0   and       a    2    =  3  +      1    2        a    0  . Now, it remains to examine the stability of the equilibrium points, which are the constants       u    ∗   such that       a    0        u    ∗    =  f. The eigenvalues of   A are   −  2,   −  1 and   −      1    2        a    0  . Therefore, if       a    0    =  α      u    2   with   α  &amp;gt;  0, the system has a single stable equilibrium point       u    ∗    =  10      /    α. However, if for instance       a    0    =  β      u    1  , the system has no equilibrium point.

Consider a 3rd order linear homogeneous DE of the form L u = u ‴

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