Determinants and Differential Equations Problem"[Here] we explore a relationship between determinants and solutions to a differential equation. The 3 × 3 matrix consisting of solutions to a differential equation and their derivatives is called the Wronskian and, as we will see in later chapters, plays a pivotal role in the theory of differential equations."This is the question following the description:"Verify that y 1 ( x ) = cos ⁡ ( 2 x ) , y 2 ( x ) = sin ⁡ ( 2 x ) , y 3 ( x ) = e x are solutions to the differential equation: y ‴ − y ″ + 4 y ′ − 4 y = 0, and show that { { y 1 , y 2 , y 3 } , { y 1 ′ , y 2 ′ , y 3 ′ } , { y 1 ″ , y 2 ″ , y 3 ″ } } is nonzero on any interval."Now I'm really just looking for an explanation on how I would show that it is nonzero on any interval. I completed the first part, I believe, by just taking up to the third derivative for each and plugging each set in to verify an identity of 0=0 at the end. I would then put the terms into the matrix which would give me: { { cos ⁡ ( 2 x ) , sin ⁡ ( 2 x ) , e x } , { − 2 sin ⁡ ( 2 x ) , 2 cos ⁡ ( 2 x ) , e x } , { − 4 cos ⁡ ( 2 x ) , − 4 sin ⁡ ( 2 x ) , e x } }So I've gotten this far and if anyone could point me in the right direction on where to go from here it would be greatly appreciated.

Question

Determinants and Differential Equations Problem&quot;[Here] we explore a relationship between determinants and solutions to a differential equation. The   3  ×  3 matrix consisting of solutions to a differential equation and their derivatives is called the Wronskian and, as we will see in later chapters, plays a pivotal role in the theory of differential equations.&quot;This is the question following the description:&quot;Verify that       y    1    (  x  )  =  cos  ⁡  (  2  x  )  ,      y    2    (  x  )  =  sin  ⁡  (  2  x  )  ,      y    3    (  x  )  =      e    x   are solutions to the differential equation:       y    ‴    −      y    ″    +  4      y    ′    −  4  y  =  0, and show that   {  {      y    1    ,      y    2    ,      y    3    }  ,  {      y    1    ′    ,      y    2    ′    ,      y    3    ′    }  ,  {      y    1    ″    ,      y    2    ″    ,      y    3    ″    }  } is nonzero on any interval.&quot;Now I&#039;m really just looking for an explanation on how I would show that it is nonzero on any interval. I completed the first part, I believe, by just taking up to the third derivative for each and plugging each set in to verify an identity of 0=0 at the end. I would then put the terms into the matrix which would give me:  {  {  cos  ⁡  (  2  x  )  ,  sin  ⁡  (  2  x  )  ,      e    x    }  ,  {  −  2  sin  ⁡  (  2  x  )  ,  2  cos  ⁡  (  2  x  )  ,      e    x    }  ,  {  −  4  cos  ⁡  (  2  x  )  ,  −  4  sin  ⁡  (  2  x  )  ,      e    x    }  }So I&#039;ve gotten this far and if anyone could point me in the right direction on where to go from here it would be greatly appreciated.

alinearjb · Accepted Answer

Step 1Calculating the Wronskian involves taking the determinant of the matrix you&#039;ve calculated. The typical formula given looks like      |                            a                          b                          c                                      d                          e                          f                                      g                          h                          i                      |    =  a      |                            e                          f                                      h                          i                      |    −  b      |                            d                          f                                      g                          i                      |    +  c      |                            d                          e                                      g                          h                      |    =  a  e  i  −  a  f  h  −  b  d  i  +  b  f  g  +  c  d  h  −  c  e  g  .Step 2When you calculate this for your matrix you&#039;ll get some function in x (since each entry is a function in x). This function is the Wronskian, and you simply examine it to make sure it is never zero.

cubanwongux · Accepted Answer

Step 1  {  {  cos  ⁡  (  2  x  )  ,  sin  ⁡  (  2  x  )  ,      e    x    }  ,  {  −  2  sin  ⁡  (  2  x  )  ,  2  cos  ⁡  (  2  x  )  ,      e    x    }  ,  {  −  4  cos  ⁡  (  2  x  )  ,  −  4  sin  ⁡  (  2  x  )  ,      e    x    }  }Step 2This is a matrix. You&#039;re asked to prove there is no x which makes the determinant of this matrix zero.  det      [                            cos          ⁡          (          2          x          )                          sin          ⁡          (          2          x          )                                      e            x                                                −          2          sin          ⁡          (          2          x          )                          2          cos          ⁡          (          2          x          )                                      e            x                                                −          4          cos          ⁡          (          2          x          )                          −          4          sin          ⁡          (          2          x          )                                      e            x                                ]    =  0

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