Question

Anyone knows how to draw phase portraits of second order differential equations? I just need the steps.. thank you!

Answer 1

for example: y"+2y'+2y=0

Answer 2

it is broken down to z=y' and z'=-2z-2y

Answer 3

Sure, so you are essentially turning this into two differential equations. We do this by picking y and y' to be two new functions basically.

u=y
v=y'

So now by taking the derivative of both of these we have:

u'=y'
v'=y''

We then rewrite these in terms of u and v.

u'=v
v'=-2v-2u

From here we simply make a vector with value (u',v') at the point (u,v) by plugging in points anywhere on a graph with u and v axes. So at the point (1,1) we'll have a vector pointing (1,-4). After you place enough vectors in your vector field, you'll start to see where to draw lines in your phase field of solutions.

Answer 4

okay, i get the substitution part, but where does the vector be pointing?

Answer 5

i would appreciate if you could draw the example you gave, sry for the trouble! >.<

Answer 6

No problem! So here I'll draw out a handful of points, give me a sec.

Answer 7

thank you! xD

Answer 8

u'=v
v'=-2v-2u

Several points I'm plugging in with their vector at that location

at (1,1) has vector (1,-4)
at (-1,1) has vector (1, 0)
at (1,-1) has vector (-1,0)
at (-1,-1) has vector (-1,4)
at (0,0) has vector (0,0) not very interesting here



See if you can put in another point like (3,1) and its vector

Answer 9

(3,1)-->(3,-4) is this correct?