Solve the following ODE by using the method of undetermined coefficients in which Euler's formula needs to be utilized: y ″ − 2 y ′ + y = sin ⁡ ( t )The way that I solved this doesn't involve Euler's formula, and I was wondering how I might use the formula here.My approach:The formula can be written as y ( t ) = y h ( t ) + y p ( t ) where y h ( t ) is the "homogeneous version" of the ODE and y p ( t ) is the particular solution that we'll obtain via the basic rule of the method of undetermined coefficients. y h ( t ):Putting r ( t ) = sin ⁡ ( t ) = 0 in the original equation, the ODE we need to solve is: y ″ − 2 y ′ + y = 0where we can set the general solution as y = e λ t and obtain the characteristic equation: λ 2 − 2 λ + 1 = 0which has a real double root, hence giving us the solution: y h ( t ) = ( c 1 + c 2 t ) e t y p ( t ):Judging by the fact that r ( t ) is shape k sin ⁡ ( ω t ) and we know that ω = 1 we can set the general solution to be of form: y p ( t ) = − K cos ⁡ ( t ) + M sin ⁡ ( t ) y p ′ ( t ) = − K sin ⁡ ( t ) + M cos ⁡ ( t ) y p ″ ( t ) = − K cos ⁡ ( t ) − M sin ⁡ ( t ) substituting these equations into the original equation and then simplifying gives us: y p ( t ) = 1 2 cos ⁡ ( t )And in conclusion, we can write that the solution to the given ODE is: y ( t ) = y h ( t ) + y p ( t ) = ( c 1 + c 2 t ) e t + 1 2 cos ⁡ ( t ) How would we be able to derive this conclusion via Euler's formula? Thanks in advance.

Question

Solve the following ODE by using the method of undetermined coefficients in which Euler&#039;s formula needs to be utilized:      y    ″    −  2      y    ′    +  y  =  sin  ⁡  (  t  )The way that I solved this doesn&#039;t involve Euler&#039;s formula, and I was wondering how I might use the formula here.My approach:The formula can be written as   y  (  t  )  =      y    h    (  t  )  +      y    p    (  t  ) where       y    h    (  t  ) is the &quot;homogeneous version&quot; of the ODE and       y    p    (  t  ) is the particular solution that we&#039;ll obtain via the basic rule of the method of undetermined coefficients.      y    h    (  t  ):Putting   r  (  t  )  =  sin  ⁡  (  t  )  =  0 in the original equation, the ODE we need to solve is:      y    ″    −  2      y    ′    +  y  =  0where we can set the general solution as   y  =      e          λ      t       and obtain the characteristic equation:      λ    2    −  2  λ  +  1  =  0which has a real double root, hence giving us the solution:      y    h    (  t  )  =  (      c    1    +      c    2    t  )      e    t        y    p    (  t  ):Judging by the fact that   r  (  t  ) is shape   k  sin  ⁡  (  ω  t  ) and we know that   ω  =  1 we can set the general solution to be of form:                              y          p                (        t        )                            =                              −                          K        cos        ⁡        (        t        )        +        M        sin        ⁡        (        t        )                                      y          p          ′                (        t        )                            =        −        K        sin        ⁡        (        t        )        +        M        cos        ⁡        (        t        )                                      y          p          ″                (        t        )                            =        −        K        cos        ⁡        (        t        )        −        M        sin        ⁡        (        t        )            substituting these equations into the original equation and then simplifying gives us:      y    p    (  t  )  =      1    2    cos  ⁡  (  t  )And in conclusion, we can write that the solution to the given ODE is:                    y        (        t        )                            =                  y          h                (        t        )        +                  y          p                (        t        )                                          =        (                  c          1                +                  c          2                t        )                  e          t                +                  1          2                cos        ⁡        (        t        )            How would we be able to derive this conclusion via Euler&#039;s formula? Thanks in advance.

repotasonwf · Accepted Answer

So, here Euler&#039;s formula means using   sin  ⁡  t  ==                    e                  i          t                    −              e                  −          i          t                            2      i       The particular integral will be found as  f  (  D  )  y  =      e          a      t          ⟹        y    p    (  t  )  =            e              a        t                    f      (      a      )      Where we have  (  D  −  1      )    2    y  =                    e                  i          t                    −              e                  −          i          t                            2      i          ⟹        y    p    (  t  )  =  (  2  i      )          −      1            (                  e                  i          t                            (        i        −        1                  )          2                      −                  e                  −          i          t                            (        −        i        −        1                  )          2                      )    =      1    4    [      e          i      t        +      e          −      i      t        ]  =      1    2    cos  ⁡  t  .      y    h    (  t  ) remains the same as you found.

Luz Stokes · Accepted Answer

Consider a new differential equation                    (1)                                                        d                              2                                      y                                d                          t                              2                                                    −        2                              d            y                                d            t                          +        1        =                  (                                    d                              2                                                    d              t                                −          2                      d                          d              t                                +          1          )                y        =                  e                      i            t                          .            Let   y  =      e          i      t        g  (  t  ). By the product rule                                          d            y                                d            t                          =                  d                      d            t                                    (                      e                          i              t                                g          (          t          )          )                =                  e                      i            t                                    d                      d            t                          g        (        t        )        +                  (                      d                          d              t                                            e                          i              t                                )                g        (        t        )        =                  e                      i            t                                    (                      d                          d              t                                +          i          )                g        (        t        )        .            Then the differential equation (1) becomes                              e                      i            t                                    [                                    (                              d                                  d                  t                                            +              i              )                                      2                                −          2                      (                          d                              d                t                                      +            i            )                    +          1          ]                g        (        t        )        =                  e                      i            t                              so that after cancelling       e          i      t       from both sides and expanding the expression in parenthesis on the LHS                              [                                    (                              d                                  d                  t                                            +              i              )                                      2                                −          2                      (                          d                              d                t                                      +            i            )                    +          1          ]                g        (        t        )        =                  (                                    d                              2                                                    d                              t                                  2                                                              +          (          2          i          −          2          )                      d                          d              t                                −          2          i          )                g        (        t        )        =        1            so that   g  (  t  )  =  c for some constant   c  ∈      C  . Therefore,   g  (  t  )  =  i      /    2. A particular solution of   y would then be                    y        =                  i          2                          e                      i            t                          =                  i          2                (        cos        ⁡        t        +        i        sin        ⁡        t        )        .            Since   sin  ⁡  t  =      I    m    (      e          i      t        ), then a particular solution to                    d                  2                    y              d              t                  2                      −  2            d      y              d      t        +  1  =      (                  d                  2                            d        t              −    2          d              d        t              +    1    )        y          p        =  sin  ⁡  tis       y          p        =      I    m    (  y  )  =      I    m    (      i    2    (  cos  ⁡  t  +  i  sin  ⁡  t  )  )  =      1    2    cos  ⁡  t

Solve the following ODE by using the method of undetermined coefficients in which Euler's formula needs to be utilized: y′′−2y′+y=sin(t)

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