Instead of using arrows to represent a planar vector field, one sometimes uses families of curves called field lines. A curve y = y ( x ) is a field line of the vector field F(x,y) if at each point ( x 0 , y 0 ) on the curve, F ( x 0 , y 0 ) is tangent to the curve.Show that the field lines y=y(x) of a vector field F ( x , y ) = P ( x , y ) i + Q ( x , y ) j are solutions to the differential equation d y / d x = Q / PFind the field lines of F ( x , y ) = y i + x j

Question

Instead of using arrows to represent a planar vector field, one sometimes uses families of curves called field lines. A curve   y  =  y  (  x  ) is a field line of the vector field F(x,y) if at each point   (      x    0    ,      y    0    ) on the curve,   F  (      x    0    ,      y    0    ) is tangent to the curve.Show that the field lines y=y(x) of a vector field   F  (  x  ,  y  )  =  P  (  x  ,  y  )  i  +  Q  (  x  ,  y  )  j are solutions to the differential equation   d  y      /    d  x  =  Q      /    PFind the field lines of   F  (  x  ,  y  )  =  y  i  +  x  j

Branson Perkins · Accepted Answer

If y(x) is a field line, this means that at every x you have that F(x,y) is colinear with the derivative of (x,y(x)), which is (1,y′(x)). So for each x there is a number   α  (  x  ) such that  (  P  (  x  ,  y  (  x  )  )  ,  Q  (  x  ,  y  (  x  )  )  )  =  α  (  x  )    (  1  ,      y    ′    (  x  )  )  .Thus  P  (  x  ,  y  (  x  )  )  =  α  (  x  )  ,           Q  (  x  ,  y  (  x  )  )  =  α  (  x  )        y    ′    (  x  )  =  P  (  x  ,  y  (  x  )  )        y    ′    (  x  )  .Thus      y    ′    (  x  )  =            Q      (      x      ,      y      (      x      )      )              P      (      x      ,      y      (      x      )      )        .When  F  (  x  ,  y  )  =  (  y  ,  x  )  ,the differential equation becomes      y    ′    =      x    y    ,with solution   y  =            x      2        +    y    (          x      0              )      2        −          x      0      2       when y&amp;gt;0 and   y  =  −            x      2        +    y    (          x      0              )      2        −          x      0      2       when y&amp;lt;0.

charlygyloavao9 · Accepted Answer

We consider the curve                    (1)                    α        (        x        )        =        (        x        ,        y        (        x        )        )        ,            with tangent vector                    (2)                              α          ′                (        x        )        =        (        1        ,                  y          ′                (        x        )        )        =                  i                +                  y          ′                (        x        )                  j                ;            if                    (3)                    F        (        x        ,        y        )        =        P        (        x        ,        y        )                  i                +        Q        (        x        ,        y        )                  j                    is tangent to   α  (  x  ) at (x,y), then       α    ′    (  x  ) is collinear with F(x,y); that is, there is some                    (4)                    0        ≠        β        ∈                  R                    with                    (5)                              α          ′                (        x        )        =        β        F        (        x        ,        y        )        ;            that is, by virtue of (2) and (3),                    (6)                              i                +                  y          ′                (        x        )                  j                =        β        P        (        x        ,        y        )                  i                +        β        Q        (        x        ,        y        )                  j                ;            comparing coefficients yields                    (7)                    β        P        (        x        ,        y        )        =        1        ,            and                    (8)                    β        Q        (        x        ,        y        )        =                  y          ′                (        x        )        ;            we observe that (7) implies   P  (  x  ,  y  )  ≠  0, hence we have                    (9)                    β        =                              1                          P              (              x              ,              y              )                                      ,            and combining this with (8) we find                    (10)                              y          ′                (        x        )        =        β        Q        (        x        ,        y        )        =                              1                          P              (              x              ,              y              )                                      Q        (        x        ,        y        )        =                                            Q              (              x              ,              y              )                                      P              (              x              ,              y              )                                      .            Now with                    (11)                    F        (        x        ,        y        )        =        y                  i                +        x                  j                ,            we obtain                    (12)                              y          ′                (        x        )        =                              x            y                          ,            or                    (13)                    y                  y          ′                (        x        )        =        x        ;            we observe that                    (14)                                          1            2                          (                  y          2                (        x        )                  )          ′                =        y                  y          ′                (        x        )        ;            (13) may thus be written as                    (15)                                          1            2                          (                  y          2                (        x        )                  )          ′                =                              1            2                          (                  x          2                          )          ′                ,            or                    (16)                                          1            2                          (                  y          2                (        x        )        −                  x          2                          )          ′                =        0        ,            whence                    (17)                    (                  y          2                (        x        )        −                  x          2                          )          ′                =        0        ,            which implies that                    (18)                              y          2                (        x        )        −                  x          2                =        C        ,                a constant        ;             a constant;(18)the field lines of (11) are thus the curves                    (19)                              y          2                −                  x          2                =        C        ,            which is a family of hyperbolas in             R        2

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