Question

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Answer 1

i don't understand what you are saying..

Answer 2

so you mean that you want to classfy that your answer is correct or not

Answer 3

i have a differential equation, y'(t). i am trying to classify the roots i have found (stable, unstable etc) algebraically, without drawing a slope field diagram.

Answer 4

i know my roots are correct. i was wondering if when i substitute the roots back into y'(t), i can determine their nature

Answer 5

can i see your eq or some question...???

Answer 6

y'(t) = y(y+3)(y^2-1) ... which is equal to (y+1)(y-1)(y+3)

Answer 7

i am not really good at english so your question is you want to know y right?

Answer 8

no, i want to know whether or not the function will converge or diverge from each of the roots, and hence whether theyre stable or unstable..if you dont get it its fine :)

Answer 9

you mean you found the roots from the y' , but when you plug that roots, y is not correct . is it?

Answer 10

no, im trying to classify each critical point

Answer 11

critical point... y'=0 is critical point...

Answer 12

god.. i want to give you a help.. but.. my english ability is really pellet... really sorry...

Answer 13

i think drawing is best way of classifing the critical point but if you want to find it in algebrically, set the interval like \[x _{1} < x <x_{2}\] then, plug x1,x2 and figure out it is increasing or not

Answer 14

the only thing is, im not sure whether that will tell me the same information as a slope field of the function. i guess ill just do it drawing, thanks.

Answer 15

if you know the formula for y'(t) then you just need to find when y'(t)=0, so solve the simple algebra and find, in your case, y={-1,1,3} are equilibrium solutions
http://tutorial.math.lamar.edu/Classes/DE/EquilibriumSolutions.aspx

Answer 16

yes, but how can i classify them without graphing? whether the points are stable or unstable equilibria

Answer 17

check the intervals
in your example
y'=(y+1)(y-1)(y+3)
so for instance for y=-1, if for y>-1 we have y'<0 and for y<-1 we have y'>0, then it is stable

Answer 18

if it were y'>0 for y>-1 and y'<0 for y<-1 it would be unstable

Answer 19

the derivative would point away

Answer 20

thank you!

Answer 21

that's same as my explaination