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(Sece fohmula ffoh 'ivle9 Pokling fumchion by the Lagsrimge methodle Simfhfy cmd dsuw gJ? h 3 the Sesultfng Polymomicl ,X-[-1,-6 , 2 ,-s ] Y= [3 2 , 0 , 4 ]...

Question

(Sece fohmula ffoh 'ivle9 Pokling fumchion by the Lagsrimge methodle Simfhfy cmd dsuw gJ? h 3 the Sesultfng Polymomicl ,X-[-1,-6 , 2 ,-s ] Y= [3 2 , 0 , 4 ]

(Sece fohmula ffoh 'ivle9 Pokling fumchion by the Lagsrimge methodle Simfhfy cmd dsuw gJ? h 3 the Sesultfng Polymomicl , X-[-1,-6 , 2 ,-s ] Y= [3 2 , 0 , 4 ]



Answers

$$y^{\prime}=\frac{3}{4} y-3, y(0)=2, y(0)=4, y(0)=6$$

So for this question, were asked to determine if y is a function of X. The first thing that we want to do is go ahead and isolate why? So we're going to get three Y equals groups four minus two X. That we're going to divide both sides by three. So we get that Y equals 4/3 minus 2/3 X. So why is the function of X anytime? An input value which is X is matched with exactly one output value? Right. So any time that value of X, let's say ecstatic over 23 and why is equivalent to positive and negative six? Obviously in this case but assume we get to output values in that case we know that why is not a function of X because we're given, you know, two or more output values for that single input value. So in this case if we have X is equivalent to three, we get that Y is equivalent to 4/3 and then two thirds X times three is just going to be equivalent to two X because the threes are going to go ahead and cancel out and oh my bad. So we included this is going to be limited to because threes will cancel out. So we have four third minus two. So this is going to give us some, you know, just one single value. So because for any input of X right, we're going to only get one output value. We're not gonna get two or more output values. Because of that fact we can assume that this function two X plus three Y equals four is an example of why being a function of X. So this is going to be a function.

We have a number 66 in which we haven't provided the graph. Oh XX which is shown by red color. And now we need to find the graph skips the grapple. Why would go at off three x since this is a three and greater than one. So graph will be shrink by defector one by three along X axis, which means exporters will become one by three off what it is and wife corner will amend The same so exponents will be will be here minus one one. And the white water would amend the same So stretching will look like not a stating this is a compression compression will look like you have graphs should be between minus 1 to 1 only, just like like this They said the graph off at Joe Three years no, second graf is beeper. Why I could do that job one by three. It's so I would X coordinate should become three times what it is. Remaining y corners with him in the same. So this isn't minus name zero and the minus 1.5 minus 4.5 Somebody, This should be nine here and for 15 somewhere here Basic for somebody. Our crap should look like this. So they see the graph off. Why I called to that's one, but

Here we have why prime equals y squared minus three, y minus four. So we have a route at -1 and a route at four. So those will be our two equilibrium solutions. We have 1 -1 and 14. And we're looking for solutions to this differential equation when why, When the initial value for Y is zero and 3. So what we can see here is if we're out here why is negative? But the derivative why is negative? But the derivative is positive. So we're gonna be moving this way and the driver was going to get smaller and smaller and smaller until we approach and hit this point here out here, we're going to be going this way because the derivative is negative. So we're going the smaller values of Y and smaller values of why? Until we get the derivative gets very, very small and we approach this point um out here, we would actually be the Y is positive derivatives, positive system, going to higher values of why, where the derivative is greater. And so we're basically just running away. But they didn't give us a salute. Give us initial condition for why is greater than four, so we won't see them. All right. So when y is zero starts at zero, we can see that we just kind of assumed chaotically approach here, curvature is always positive When Y starts at three, which is here. So we have the slope is um is increasing down For a while until we get to about about two. And then the cover trip becomes positive. And then we asked until 2 -1 again. So those are two solutions what the two solutions look like.


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