Question on drawing a diff.equation

[transgalactic]

i got this equation

y'=1+x-y

i got y=x+(1-c)
for the isocline c=-1
i got tangent lines which are perpendicular

for c=0 the are not
why is that??
how do i find the slope of each isocline?

 

[HallsofIvy]

What do you mean you "got y= x+ (1-c)"?

 

[transgalactic]

this is a solution of this y'=1+x-y differential is y=x+(1-c)


for the isocline c=-1
i got tangent lines which are perpendicular

for c=0 the are not
why is that??
how do i find the slope of each isocline?

 

[HallsofIvy]

this is a solution of this y'=1+x-y differential is y=x+(1-c)

No, it isn't. If y= x+ (1-c), then y'= 1, obviously. The general solution to y'= 1+ x- y is
y= ce^-x+ x.

for the isocline c=-1
i got tangent lines which are perpendicular

for c=0 the are not
why is that??
how do i find the slope of each isocline?

 

[transgalactic]

ok
so generally how do i find the slope of each isocline
(those lines that we draw on the isoc lines)

??/

 

[HallsofIvy]

You don't find the slope of an isocline- you assign it. For example, with dy/dx= 1+ x- y, the isocline ("iso-cline": equal slope) corresponding to slope 0 must include points (x,y) such that 1+ x- y= 0 or y= x+ 1. Draw that line. The isocline corresponding to slope 1 must include all points (x,y) such that 1+ x- y= 1 or y= x. Draw that line. choose whatever values, s, you want for the slope. The corresponding isocline is the line dy/dx= 1+ x- y= s or y= x+ (1-s).