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Question

For the following problems, determine how parameter $a$ affects the solution.
Solve the generic equation $y^{\prime}=a x+x y .$ How does varying $a$ change the behavior?

Robert Daugherty · Accepted Answer

Section four, not five problem to 59. Here in this section, we&#x27;re dealing with first order, linear, ordinary differential equations and solving these by finding and integrating factors. In this case, the key is to get this thing in standard form, which is why prime plus a function of X time is why equals another function off axe. So to get this one in standard form, that&#x27;s why prime minus X y equal a X The integrating factor. It&#x27;s just going to be e to the you take the anti derivative minus X dx that function p of X. So this gives us e to the minus. X squared over two is the integrating factor. So when I multiply the entire equation by that, I&#x27;ll have you to the miners. Why e to the minus X squared over two Prime is equal to a X E to the minus x squared over to, and now the key will be to integrate both sides of this equation On the left side, we just get why e to the minus X squared over two on the right side of this equation. So the most straightforward way to do this Let&#x27;s just scroll down. I need to integrate a x you to the miners X squared over two the X. So the most straightforward way of doing this is let you well, minus X squared over two. Then do you is going to be, um, Highness X DX so I can rewrite this one As this is equal to bring the A outside Put a minus there, minus x e to the minus, X squared over two t x. This is now just in the form of the integral of e of you, Do you, which is just e to the U plus a constant of integration. So what we end up with here is why e to the minus X squared over two is equal to minus a E to the minus X squared over two plus a constant of integration. Therefore, why is equal to minus a plus C e to the X squared over two. So that is my, um, answer to the, um uh, differential equation. And now we&#x27;ll go take a look at the graph. And so here you see the graph of the solution for this different. Your equation y is equal to So why did you go to minus a C E to the X great over two. So when C is equal to zero, um, you&#x27;re gonna have a horizontal line, So sorry. Let me change my mouse back. You&#x27;re gonna have a horizontal line, A to minus a. So as a varies, you just see horizontal lines when CIA zero and then if c were allowed to vary based on a my initial condition, then that&#x27;s where you see make this little bit clear. So if I scroll out, um, you see the exponential term, so you see the exponential term words increase decrease or a decrease increase based on the various values of A and C.

Robert Daugherty · Answer

Section four, not five. Problem to 60. We&#x27;re dealing with, um, first order, linear or new. Different joke wages here has to solve this one and then make some comments about what? Changing the value of a would do for this particular solution for the different equation. So we need to get this in standard forms of why prime plus a function of X times Why is equal to a function of X? So that&#x27;s going to be why prime minus a x Y is he going to axe? The integrating factor is going to be to the minus a x t X. So this is e to the minus a X squared over two. So that&#x27;s our integrating factor that will multiply by the differential equation and what we will see here. Sorry. Starting from this one, we will have. Why times e to the minus a x squared over two. Crime is equal to x e to the minus a x squared over to now. The solution would then be to integrate both sides of this equation. So on the left side, I will get why e to the minus a X squared over two equal to the integral x E to the minus a x squared over to t X. So the key to doing this is a use substitution. Let&#x27;s let you equal minus a X squared over two. That means that d you is going to be minus a X DX so you can see here that X DX is going to be equal to, um, xcx is minus one over a Do you So you&#x27;ve got X d. X. So what this turns into is you&#x27;ve got why e to the minus. A X squared over two is equal to minus one over a and then the integral of either that you d you So this is minus one over a E to that, you plus C, which is minus one over a e to the so you is what minus a X squared over two plus a constant. So that&#x27;s my answer. To solve this for why you&#x27;re I have why is equal to, um, you have minus one over a plus. See e to the A X squared over two. So that is the solution to my differential equation. So, analysis. Go take a look at it. Um, when C is equal to zero. Um, this just turns into the feel Look at it. IFC is zero then this is just Y equal minus one over a. So as a varies, you&#x27;re just gonna have a horizontal line or a constant. So, depending on what the value of of a is, you&#x27;re going to see the, uh, uh, the changes there in the constant value, and it&#x27;s reciprocal of them. Okay. And then you can see as I let&#x27;s see, very, you have various behaviors there in terms of what is happening to the exponential part of the of the function.

Robert Daugherty · Answer

So for about five, number 2 57 we have an equation here with constant. So why prime is a X plus y? Let&#x27;s solve this different equation and then see what the behavior of a could do for us of a changes. So to get this in standard form, why prime plus a function of X terms? Why is equal to another function of X So to get this guy in standard form that&#x27;s going to be why prime minus why equals a X. The integrating factor here is going to be e to the minus one, the X So this is e to the minus X is the integrating factor. So multiply this equation by the integrating factor to get either the minor sex prime. It&#x27;s equal to this a X e to the minus X. Then the key will be to integrate both sides of this equation On the left side, you have why e to the minus X and then on the right side. When you integrate this, you have minus okay and e to the minus x X plus one plus a constant of integration. Now, then, to solve this for why you have, why is equal to minus a X plus one plus see e to the X. So this is my different your equation. So what will happen? So obviously, if a zero, this is just an exponential function, but otherwise is the combination of a linear function and in exponential. So let&#x27;s just take a look at what this looks like. So if I let a very Okay, so I&#x27;m a values air going between negative 10 and 10. You see how it changes the behavior of this graph? Okay, Now, you also see that, um, secret Beria&#x27;s well, based on the boundary condition. So we&#x27;ve got several things going on here with his particular graph. Um, if you look at different values and then they could vary independently of each other different times, and so you get various behaviors based on how they are, how they&#x27;re changing

Robert Daugherty · Answer

section Ford up five. Problem to 58 were dealing with linear for soldier. Ordinary difference equations. The key is to get this veneer equation in standard forms a Y prime plus a function of X Times Y is equal to another function of X. So this just becomes why prime minus y equal a X. So the integrating factor here is just going to be e to the integral of minus one d x. So this is just e to the minus X. So this equation transforms into why prime e to the minus X. Excuse me. Um, that&#x27;s why E to the minus X prime is equal to a X E to the minus specs. Then the key becomes two different Jake shoes. We integrate both sides of this equation, so that gives those why e to the minus X is equal to. And then when you integrate the right hand side, you get minus a X plus one eat of minus X plus a constant of integration. Then to solve for why, why is equal to minus a X plus one? Um, plus C e to the x. Okay, So how does changing of a very here So if sea were equal to zero, you would see that this should be a line in ages, controls the slope. Otherwise, we&#x27;ve got the combination of a line and an exponential. So let&#x27;s just go look at this. So let&#x27;s verify that IFC were equal to zero. So if c is equal to zero, and there we go is equal to zero. There we go when C is equal to zero. Here, um, you see a line and then a is just controlling the slope. So when a is since I&#x27;ve got negative, a negative value is positive slope positive value A is a negative slope. And then when I very see that you see what happens is you got the combination of a line and then exponential. So if I let both of these guys just sort of very independently, um, you&#x27;ll see the behavior that you&#x27;re going to get for this particular, uh, solution

Robert Daugherty · Answer

Section 45 problem to 61 s. So what you see here is that we have, ah, first order linear, ordinary differential equation and were asked to solve this. The key to having this is in standard form, which this one actually is in standard form. So the integrating factor has a function of t is going to be e and then so minus one DT. So this is you to the minus T is my integrating factor. So this equation transforms into why e to the minus t prime is equal to e to the k t you to the minus t. So this is you to the t times K minus one. Now, in order to solve this equation, I will integrate both sides of this equation on the left side, I just get y e to the minus. T is equal to and then on the right side. When you integrate the issue, get E to the K minus one T and then times one over K minus one, plus a constant of integration. Therefore, why is equal to one over K minus one you to the K minus one e e to the T plus C to the T. And so this just becomes, um hi is equal to one over K minus one e to the k t plus c e to the t. Now. We had the boundary condition that why of zero, um, is equal to zero. That would allow me to solve for the constant of integration. So why to the zero is zero. So that means that zero is equal to one over, OK, minus one. And then e to the zero is just one plus ce to the zero on. It&#x27;s just simply one. And so what you see here is that c You&#x27;ve got a minus one over K minus one. So this solution becomes why the could one over K minus one you to the K T minus one, over K minus one. You to the t factor out that one over K minus one. So one overcame IRS one e to the K T minus e to the T. Now we&#x27;re asked to take a look and see what would happen as I let kay um, approach zero. Okay, so what would be the limit as K approaches zero in this case, Um, excuse me as can purchase one. So when K approaches one, you know that the denominator Here&#x27;s if I look at this, the limit is K approaches one of e to the K T minus e to the t over K minus one. So as K approaches one, I&#x27;m approaching zero and the denominator and I&#x27;ve had e to the team and is either the t I&#x27;m approaching zero in the numerator. So it&#x27;s the indeterminant form. So using low petals rule, I can differentiate and then take the derivative. So this is also going to be equal to the limit has que approaches one and we have to differentiate with respect to K. So differentiating with respect to K E to the K T. That&#x27;s going to be easy to the K t. And if K is my variable, that&#x27;s going to give me a t e to the K T minus. When you differentiate eat of the tea with respect to K that zero and then in the denominator when you different ship with respect to K, that is one. So this is the limit as que approaches zero of t e to the k t. Excuse me que approaches one I&#x27;m just so accustomed to zero as K approaches one. So when K approaches one, you can simply substitute that in. Now, this is just t e to the t. So that&#x27;s what the function turns into as K approaches one in this case.

Smita Praharaj · Answer

in discussion there, Astra. Find the general institution off the differential equation. Why Prime equals to y O operates. No. What? Why is considered as a function of next? So white crime would be d y over dx. And that&#x27;s why you&#x27;re Rex If we bring y to the bottom here and take the d x to the tough So we get d y o ver y equals two d X over Rex. Now integrate both sides of this previous equation. So we get Ellen absolute value off. Why calls to Ellen Absolute value of X plus C where C is any constant. Now I can write. See as Ellen off a for some constant A So we get Ellen. Absolute value of y equals two Ellen Absolute value backs plus Ellen, eh? Now, by using laws of logarithms, this boils down to Ellen Absolute value of a X. So Ellen, absolute value off by equals to Ellen, absolute value of the X or Y equals to a X. So this is the general solution to this differential equation. Here. We can also check if the solution is correct or not. If why is they extend the Y over DX equals to a which is same as eggs over eggs. Now a X we solved for is why. So that&#x27;s why Rex and that&#x27;s precisely what under potential equations.

Problem

For the following problems, determine how paramet…

Problem 259 Hard Difficulty

For the following problems, determine how parameter $a$ affects the solution.
Solve the generic equation $y^{\prime}=a x+x y .$ How does varying $a$ change the behavior?

Answer

$y=-a+C e^{x^{2} / 2}$
Hence a will have an effect if $C=0$ and $y$ will be a line parallel to $x$ axis which is $y=a$

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Gilbert Strang,

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Kristen K.

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Integration Review - Intro

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$y=-a+C e^{x^{2} / 2}$Hence a will have an effect if $C=0$ and $y$ will be a line parallel to $x$ axis which is $y=a$

Calculus

Grace H.

Kristen K.

Samuel H.

Michael J.

Integration Review - Intro

Definite vs Indefinite

Gilbert Strang,Calculus

Grace H.

Kristen K.

Samuel H.

Michael J.

$y=-a+C e^{x^{2} / 2}$
Hence a will have an effect if $C=0$ and $y$ will be a line parallel to $x$ axis which is $y=a$

Gilbert Strang,

Calculus