- #1

- 1,395

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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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- Thread starter transgalactic
- Start date

- #1

- 1,395

- 0

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???

- #2

HallsofIvy

Science Advisor

Homework Helper

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

- #3

- 1,395

- 0

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???

- #4

HallsofIvy

Science Advisor

Homework Helper

- 41,847

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No, it isn't. If y= x+ (1-c), then y'= 1, obviously. The general solution to y'= 1+ x- y isthis is a solution of this y'=1+x-y differential is y=x+(1-c)

y= ce

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???

- #5

- 1,395

- 0

so generally how do i find the slope of each isocline

(those lines that we draw on the isoc lines)

???????/

- #6

HallsofIvy

Science Advisor

Homework Helper

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