Find the general solution of the equation d y d x = 1 + x yMy attempt: Arranging it in standard linear equation form d y d x + P y = Q, we get d y d x + ( − x ) y = 1Hence, the integrating factor(I.F.) = e ∫ − x d x = e − x 2 / 2 . Hence, the solution is y ( e − x 2 / 2 ) = ∫ e − x 2 / 2 d x + cHow do I solve this integral now?

Question

Find the general solution of the equation            d      y              d      x        =  1  +  x  yMy attempt: Arranging it in standard linear equation form             d      y              d      x        +  P  y  =  Q, we get            d      y              d      x        +  (  −  x  )  y  =  1Hence, the integrating factor(I.F.) =       e          ∫      −      x      d      x        =      e          −              x        2                    /            2      . Hence, the solution is  y  (      e          −              x        2                    /            2        )  =  ∫            e              −                  x          2                          /                2              d    x    +  cHow do I solve this integral now?

Beckham Leach · Accepted Answer

As a commenter mentions, the solution is:  y  (  x  )  =      c    1        e                            x          2                2              +            π      2            e                            x          2                2              erf      (          x              2              )  One thing to note here is that the erf(x) function is sometimes not all that well known. The erf() function is defined as:  erf(x)  =      2          π            ∫          0              x                  e              −                  t          2                      d    t  Now when you have:  ∫            e              −                  x          2                          /                2              d    x  It can be rewritten as:  ∫            e              −        (        x                  /                          2                          )          2                      d    x  Let&#039;s replace   x      /        2    =  t. This tells us that   d  x  =      2    d  t.Making the substitution:      2    ∫            e              −                  t          2                      d    t  Since limits are need to be inserted, it can be shown that:      ∫    0    t              e              −                  z          2                      d    z    =  ∫            e              −                  t          2                      d    t    +  CCombining all of that:      2    (            π        2    erf(t)  +  C  )  =      2    ∫            e              −                  t          2                      d    t  This means that:  y  (      e          −              x        2                    /            2        )  =      2    (            π        2    erf      (          x              2              )    +  C  )I trust you can simplify from there.

tapiat0wa4c · Accepted Answer

The incomplete equation is            y      ′        y    =  xthus  ln  ⁡  (      y    λ    )  =            x      2        2  and      y    h    =  λ      e                            x          2                2            the constante variation method gives      λ    ′    (  x  )  =      e                            −                      x            2                          2            the general solution is      y    g    =            (        λ  +  ∫      e                            −                      x            2                          2              d  x            )            e                            x          2                2

Find the general solution of the equation d y </mrow> d

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