Y'=(y"+1)/(y-1) is stable at: y= infinity y= 0 y= 1 y= -1 Sorry, I meant: y'=(y"+y)/(y-1)

Question

Question

y'=(y"+1)/(y-1) is stable at:

y*= infinity

y*= 0

y*= 1

y*= -1

Sorry, I meant: y'=(y"+y)/(y-1)

math Advanced-Math

Answer 1

Answer #1

닝t dk dlJ dt dt d匕 -O. 933 d t dt d Y d七 d- 3 2 I dt 3 Po dt da Unstable : unstable cao

Answer 2

Similar Homework Help Questions

nsider the differential equation y'=1+cosly). y/o)=0 a) what is the value of y as x tends...

nsider the differential equation y'=1+cosly). y/o)=0 a) what is the value of y as x tends to infinity ? What is the value of t as x tends to - D? b) What are possible y-values such a solution can have? c) W hat are the equilibrum solutions of this differential equation Is it stable , unstable or, seni-stable? |
use the leading coefficient test to determine whether y-->infinity) or y--> - infinity or x--->infinity 3.T

use the leading coefficient test to determine whether y-->infinity) or y--> - infinity or x--->infinity 3.T
Im sorry I really had to repost these 2 questions, because I really couldn't understand any...

Im sorry I really had to repost these 2 questions, because I really couldn't understand any thing from the handwriting as it was not clear. thank you very much for your time 3. Find the inverse Laplace transforms: (a) e-{_5)} (1) c-'*:*}(0) "{G+236 +16}. 4. Use the Laplace transform to solve the initial value problem y' + 4y = e34 subject to y(0) = 1, for y(x).

sorry i meant to post number 14 but if you could do 13 that would be...

sorry i meant to post number 14 but if you could do 13 that would be nice too :D thank you Saramond BI U 12 AA Aa A EVEEV albe x x Ave A v 3 21 vv Ayu Styles Styles Pane 13a) Draw a figure showing the typical shape of the average fixed cost (AFC) curve (2 points) and explain why it takes this shape (2 points) 13b) Draw a figure showing a typical average total cost (ATC) curve...
Find the stable equilibrium solution of the following differential equation: + y - 1 = e2(...

Find the stable equilibrium solution of the following differential equation: + y - 1 = e2( y − 1). The stable equilibrium solution is y = Check Find the general solution to the differential equation: x + y - x115 = 0. Answer: y(x) = Check Solve the initial-value problem: dy = e ** - y, yO= dx Answer: y(x) = Check
Fx(x) = A{1-e^-x} for 1<x<infinity = 0 for -infinity<x<or equal to 1 a.) find the probability...

Fx(x) = A{1-e^-x} for 1<x<infinity = 0 for -infinity<x<or equal to 1 a.) find the probability density function of the random variable of the above function and sketch it b.) using the probability density function, find the probability that the random variable is in the range 2<X< or equal to 3 c.) using the probability density function, find the probability that the random variable is less than 2

– calculate Integral 0 to infinity xλe^−λx dx ? – calculate Integral 0 to infinity x^2...

– calculate Integral 0 to infinity xλe^−λx dx ? – calculate Integral 0 to infinity x^2 λe^−λx dx ?
I have tried this 10 times, and keep getting infinity; but it keeps saying infinity is...

I have tried this 10 times, and keep getting infinity; but it keeps saying infinity is wrong... (1 point) Evaluate the limit lim V x2 + 4x + 5 – x x
Doing integrals with Residues at Infinity specifically with Complex Analysis . Apparently if I split the...

Doing integrals with Residues at Infinity specifically with Complex Analysis . Apparently if I split the analytic function f(z)= into and . I am able to see where on the Complex Plane it is defined. But then somehow this problem uses information with solving it via the Facts of Residue's at Infinity. Yes, it is a Real Integral but I am to solve it using Complex Analysis and Branch Cuts. And lastly the fact with Residues at Infinity since it...

class: numerical analysis I wish if it was written in block letter Sorry I can't read...

class: numerical analysis I wish if it was written in block letter Sorry I can't read cursive = COS Problem 1: Recall that the Chebyshev nodes x4, x1,...,xy are determined on the interval (-1,1] as the zeros of Tn+1(x) = cos((n + 1) arccos(x)) and are given by 2j +10 Xj j = 0,1, ... 1 n+1 2 Consider now interpolating the function f(x) = 1/(1 + x2) on the interval (-5,5). We have seen in lecture that if equispaced...

Y'=(y"+1)/(y-1) is stable at: y*= infinity y*= 0 y*= 1 y*= -1 Sorry, I meant: y'=(y"+y)/(y-1)