Let's say we have the general solution to X ′ = A ( t ) X, where X = ( x 1 , x 2 ) T . How do you find the general solution to the system X ′ = A ( t ) X + b ( t ) where b ( t ) is a 2 × 1 matrix with two polynomials as entries. How do you find the particular solution?

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Let&#039;s say we have the general solution to       X    ′    =  A  (  t  )  X, where   X  =  (      x    1    ,      x    2        )    T  . How do you find the general solution to the system       X    ′    =  A  (  t  )  X  +  b  (  t  ) where   b  (  t  ) is a   2  ×  1 matrix with two polynomials as entries. How do you find the particular solution?

postotnojeyf · Accepted Answer

Let&#039;s call the two solutions differently, and clean up notation. Lowercase boldface will denote vectors (  2  ×  1). Plain uppercase will denote matrices. Plain lowercase will be scalars. All explicit dependence on   t will be dropped. Now the one without any forcing term on the right hand side is denoted with       x  :            x        ′    =  A      x  The one with forcing we&#039;ll denote by       y  ,            y        ′    =  A      y    +      b    .Let&#039;s guess a form for       y  . Let&#039;s guess it is the product of       x   and some unknown scalar function   u,      y    =  u      x  Then your equation becomes  u            x        ′    +      u    ′        x    =  u  A      x    +      b  But we already know that             x        ′    =  A      x   so substitute that  u  A      x    +      u    ′        x    =  u  A      x    +      b  Canceling on both sides we are left with      u    ′        x    =      b  Multiplying by             x        T        u    ′              x        T        x    =            x        T        b  But             x        T        x    =      |        |        x        |              |        2  , so dividing by that scalar gives      u    ′    =                              x                T                    b                            |                    |                    x                    |                              |                2              ,and integrating and putting back all the explicit   t dependence gives  u  (  t  )  =  ∫                              x                T            (      t      )              b            (      t      )                      |                    |                    x            (      t      )              |                              |                2              d  tIf you take that and multiply this by your original known solution, then I think you have a particular solution   y.

Let's say we have the general solution to X′=A(t)X, where X=(x)1,x)2)^T. How do you find the general solution to the system X′=A(t)X+b(t) where b(t) is a 2*1 matrix with two polynomials as entries. How do you find the particular solution?

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