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 dy/dx=Q/P. Find the field lines of F(x,y)=yi+xj.

timberwuf8r 2022-09-28 Answered
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 / P
Find the field lines of F ( x , y ) = y i + x j
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Answers (2)

Branson Perkins
Answered 2022-09-29 Author has 7 answers
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>0 and y = x 2 + y ( x 0 ) 2 x 0 2 when y<0.
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charlygyloavao9
Answered 2022-09-30 Author has 2 answers
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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