Question

Construct a tangent field for y'=1+x+y.
I just need someone to check if I'm on the right track. see attachment for picture

Answer 1

@UnkleRhaukus

Answer 2

what method are you using there

Answer 3

see attachment for pic. I just want to see if I'm on the right track

Answer 4

ummm manual like pick any point and plug in

Answer 5

wouldn't it be more fun to solve the DE?

Answer 6

I had to get the tangent field first

Answer 7

which are those lines

Answer 8

is it right???

Answer 9

oh and then it was like get the antiderivative which is like the linear solution of the equation

Answer 10

I almost got the answer which is y = -x-2 but something went off a bit. I'll draw it on paint and get it here.

Answer 11

what are those the blue horizontals on your graph?

Answer 12

thats very close, but you must remmber the constant of integration

Answer 13

when the slope is 0, it produces a horizontal line which is why I drew it blue

Answer 14

ah yes w, the blue horizontals are for when the constant of integration is zero

Answer 15

the answer is y = -x-2 but I'm wondering how they even got rid of the squared part or is that just individual ?!?!?!

Answer 16

thats not right

Answer 17

oh so how does it work?

Answer 18

wait is my graph in the right direction

Answer 19

alright the whole question is Construct a tangent field for y' =1+x+y. Find a linear solution by inspection of the3 tangent field and check it in the equation.

Answer 20

I've read the book and i saw anti-derivative

Answer 21

:O crap you mean I did all those lines for jacks?

Answer 22

if all the blue line were as ive drawn them, then they would fit the pattern of the other red line on tht upper right of the graph

Answer 23

ok but doesn't the lines create a curve?

Answer 24

i think ive worked the problem out

Answer 25

where do I go from here? I know dx x and dy y have to be in their own groups.