# Question on drawing a diff.equation

## Main Question or Discussion Point

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???

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

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
Homework Helper
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???

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

???????/

HallsofIvy