Question

Consider the autonomous first-order differential equation y' = y(2 - y)(4 - y) Find the DISTINCT critical points and classify each as (1) AS for Asymptotically Stable, (2) US for Unstable or (3) SS for Semi-Stable. Enter your answer as a comma separated list of pairs consisting on a critical point and its stability type (e.g. your answer might look like (2,AS), (-3,SS), (7,US) ) Critical Point and Stability: For the initial value problem y' = y(2 - y)(4 - y), y(0) = 3 we have lim y(x) = x??

          Consider the autonomous first-order
differential equation

y' = y(2 - y)(4 - y)

Find the DISTINCT critical points and classify
each as (1) AS for Asymptotically Stable, (2)
US for Unstable or (3) SS for Semi-Stable.

Enter your answer as a comma separated list
of pairs consisting on a critical point and its
stability type (e.g. your answer might look like
(2,AS), (-3,SS), (7,US) )

Critical Point and Stability:

For the initial value problem
y' = y(2 - y)(4 - y), y(0) = 3 we have
lim y(x) =
x??
        
Show more…
Consider the autonomous first-order
differential equation

y' = y(2 - y)(4 - y)

Find the DISTINCT critical points and classify
each as (1) AS for Asymptotically Stable, (2)
US for Unstable or (3) SS for Semi-Stable.

Enter your answer as a comma separated list
of pairs consisting on a critical point and its
stability type (e.g. your answer might look like
(2,AS), (-3,SS), (7,US) )

Critical Point and Stability:

For the initial value problem
y' = y(2 - y)(4 - y), y(0) = 3 we have
lim y(x) =
x??

Added by Matthew N.

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Advanced Engineering Mathematics
Advanced Engineering Mathematics
Dennis G. Zill, Warren S. Wright 5th Edition
Chapter 11
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Transcript

-
00:02 In this question we are given here first three levels of ramburg integration, integration.
00:27 So as for the given table, we have to find our value for the log x in between when a is equal to 2 and b is given as 5.
00:38 So we have to find here first absolute error based.
00:43 So now let's see the solution of first part.
00:46 So our exact value, exact value, it has an integration of 1 to 5 for log x, which will be equal to 1 upon x and our limits are 1 to 6 and then we solve this is equal to 5 .7055681.
01:10 So therefore our absolute absolute error will be between two stymeters will be as e equals to 5 .744389364 and then 5 .7466.
01:52 So now we will get a value for e as equal to 0 .0046652.
02:00 Then after we will find here.
02:05 So now the value for absolute error between analytical integral will be as p is equal to as we have find it here 0 .0068975...
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