The function y ( x ) satisfies the linear equation y ″ + p ( x ) y ′ + q ( x ) = 0.The Wronskian W ( x ) of two independant solutions, y 1 ( x ) and y 2 ( x ) is defined as | y 1 y 2 y 1 ′ y 2 ′ | . Let y 1 ( x ) be given, use the Wronskian to determine a first-order inhomogeneous differential equation for y 2 ( x ). Hence show that y 2 ( x ) = y 1 ( x ) ∫ x 0 x W ( t ) y 1 ( t ) 2 d t ( ∗ )The only first-order inhomogeneous differential equation I can get for y 2 ( x ) is simply y 1 ( x ) y 2 ′ ( x ) − y 1 ′ ( x ) y 2 ( x ) = W ( x ). Which is trivial and simply the definition of W ( x ), so I don't think this is what it wants. I can easily get to the second result by differentiating y 2 ( x ) y 1 ( x ) and then integrating, but the wording suggests I'm supposed to get to this result from the differential equation that I can't determine.So I would ask please that someone simply complete the above question, showing the differential equation we are to determine and how to get to the second result from this.Edit: The question later goes on to give the differential equation x y ″ − ( 1 − x 2 ) y ′ − ( 1 + x ) y = 0 call this equation ( † ), we have already confirmed that y 1 ( x ) = 1 − x is a solutionHence, using ( ∗ ) with x 0 = 0 and expanding the integrand in powers of t to order t 3 , find the first three non-zero terms in the power series expansion for a solution, y 2 , of ( † ) that is independent of y 1 and satisfies y 2 ( 0 ) = 0 , y ′ ′ 2 ( 0 ) = 1.I'm stuck here too as I don't see how we could expand W ( t ) as it is surely a function of the unknown y 2 . Apologies for my ignorance, this is independant study, and aside from a brief look at some online notes that simply define W ( x ), I've never worked on these problems before.

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The function   y  (  x  ) satisfies the linear equation       y    ″    +  p  (  x  )      y    ′    +  q  (  x  )  =  0.The Wronskian   W  (  x  ) of two independant solutions,       y    1    (  x  ) and       y    2    (  x  ) is defined as       |                                        y            1                                                y            2                                                            y            1            ′                                                y            2            ′                                |  . Let       y    1    (  x  ) be given, use the Wronskian to determine a first-order inhomogeneous differential equation for       y    2    (  x  ). Hence show that       y    2    (  x  )  =      y    1    (  x  )      ∫                  x        0              x              W      (      t      )                      y        1            (      t              )        2              d  t  (  ∗  )The only first-order inhomogeneous differential equation I can get for       y    2    (  x  ) is simply       y    1    (  x  )      y    2    ′    (  x  )  −      y    1    ′    (  x  )      y    2    (  x  )  =  W  (  x  ). Which is trivial and simply the definition of   W  (  x  ), so I don&#039;t think this is what it wants. I can easily get to the second result by differentiating                     y        2            (      x      )                      y        1            (      x      )       and then integrating, but the wording suggests I&#039;m supposed to get to this result from the differential equation that I can&#039;t determine.So I would ask please that someone simply complete the above question, showing the differential equation we are to determine and how to get to the second result from this.Edit: The question later goes on to give the differential equation   x      y    ″    −  (  1  −      x    2    )      y    ′    −  (  1  +  x  )  y  =  0 call this equation   (  †  ), we have already confirmed that       y    1    (  x  )  =  1  −  x is a solutionHence, using   (  ∗  ) with       x    0    =  0 and expanding the integrand in powers of   t to order       t    3  , find the first three non-zero terms in the power series expansion for a solution,       y    2  , of   (  †  ) that is independent of       y    1   and satisfies       y    2    (  0  )  =  0  ,  y      ′              ′        2    (  0  )  =  1.I&#039;m stuck here too as I don&#039;t see how we could expand   W  (  t  ) as it is surely a function of the unknown       y    2  . Apologies for my ignorance, this is independant study, and aside from a brief look at some online notes that simply define   W  (  x  ), I&#039;ve never worked on these problems before.

soosenhc · Accepted Answer

Your arguments sound reasonable to me, and the formulation in the exercise a bit cryptic. But you should perhaps also add that      W    ′    =      |                                        y            1                                                y            2                                                            y            1            ″                                                y            2            ″                                |    =  −  p  W  ,from which   W  (  x  )  =  exp  ⁡  (  −  P  (  x  )  ) with   P  =  ∫  p. This ensures that in   (      y    2        /        y    1        )    ′    =  W      /        y    1    2   the RHS is just an explicit function of   x. Might help you on the last bit as well.

The function y ( x ) satisfies the linear equation y ″ </msu

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