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Example 4.2.4. Solve the following initial value problem_3y2dy x = 0, 9(1) = 1. dx 2y4...

Question

Example 4.2.4. Solve the following initial value problem_3y2dy x = 0, 9(1) = 1. dx 2y4

Example 4.2.4. Solve the following initial value problem_ 3y2 dy x = 0, 9(1) = 1. dx 2y4



Answers

Solve the initial-value problem.
$ (x^2 + 1) \frac {dy}{dx} + 3x(y -1) = 0, y(0) = 2 $

Problem twenty eight we have d y o ver t acts equals faxed with one half much flat Why square? And we have wine off four It was nine and still separate Variables We have one over. Why square to you? Why you owe us axe to the while half t ax This's why To the negative too. So we'LL grow both side we have Why? To the ninety one over next you want Because axe to the three over to over three over two plus e Now we can simplify little bit We have negative want want over Why equals two x three over too over three plus e And if we plugging for nine let you want overnight it wass to multiply for to the three over too to all of us to replace See now for told this story over to his life for Cube squared off that sixty four service is it? This heart is eight but flat too sixteen over three was next year while we're alive So see you call us elective won over night blindness Waken transferred his toe a forty eight overnight So CEO Connective forty and I overnight his final answer is going to be ninety one over. Why equals two extra over, too over three. Subtract forty and I over just this far, but replace it by disquieting way. Help calculated, So does twenty eight.

So when this one I would like to separate the variables and then find the general solution and fill in the exact point to find my particular solution. So moving the Y squared over, I'm gonna call it wide to the negative, too, because I divided it over. And then I would multiply the d X to get it off of that denominator. Once the variables air separated, I need to integrate. So on this left side, I'm thinking of the power rule. If I add one to the X one and I get negative one why, to the negative First is on the denominator. Really? So I'm gonna write negative one over why and then on the right side. If I add one to this exponents, I get three halves dividing by that new exponents makes it 2/3 in front of it. And then I put plus C because I don't know what constant is added in yet Now I use the initial condition that 49 as the y value and the X value a Wiffle in nine is why and or is X and then solve for that plus c. So on the right side, it's where I have some stuff to simplify. Four to the three hats I think about first. The square root of four is two and then two cubed is eight. So I have 16/3. If I move it to the other side, I would want to subtract. But I also would wanna have a common denominator. So if I got a common denominator there, I would actually want to put it over nine. So 16/3 becomes 48/9, multiplying top bottom by three. So I get negative 49/9 as c using this value, I can fill it in to find my general solution. So the general solution one way to write it would be too right. Negative one over why equals 2/3 X to the three halves minus 49 overnight. That's one way I could write it, or I could solve it for y equals. So if I want to sell it for y equals, then I would first want to flip it, which means on this right side, I need a common denominator of nine. Actually have 6/9 instead of to over three getting that common denominator So this is my mental work here until math. So I flip it. I have nine over six X to the three hubs minus 49 and then to get rid of the negative, I've put it on the nine great divided over to the other side. So negative 9/6 extra three halves minus 49. So that would be my y equals form of the same answer.

We want to saw the initial value problem that we're given here, which is essentially, that's a way of saying we want to find some function. Why? Where is derivative is nine X squared minus four X plus five. And when we input negative one, we should output zero. So this is really just saying we want the point. Negative 10 All right, so first thing we could do to get why to appear is just integrate each side of this. So that's gonna give us that Why is equal to so we can use the linearity of intro down to distribute this across the addition subtraction as well as factor out any Constance. So this is going to be nine times central of X squared, minus four times integral of X plus integral of five. Now, each of these, they're gonna be powerful to integrate, so the first ones will be nine times, so we add one to the power so it becomes three, and then we divide by the new power, and then the next one says maybe four x. So this is really extra. The first power Add one to the power. You get scared. Divide by two and then Constance, we could think about his ex to zero. So that would just be a plus one for the zero. So it be by X, and we divide by one. But divided by one doesn't do anything. And then we add our constant of integration. C Let's go ahead and clean this up a little. So we should end up with three x cute minus two X squared plus five x plus c now. So this is gonna be our general solution for why, But we want to know in particular what function would have this point. So what we could do, go ahead and plug it in. So why is zero X's negative one? So we have negative one cube, which would give us negative three negative one squared. So they'll just be negative. Too thin, minus five plus c. So negative three minus two minus five. That should be negative. 10. Add 10 over. We get tennis. See? So to get our solution, we're gonna take this plug it in for why? And so that's gonna give us that our solution. Why is three X cubed minus two X squared plus five X plus 10. So this solves the initial value problem.

Okay, So to solve this, uh, initial value problem first we're gonna need to find the general solution of this. So writing that in differential operator form, that's gonna be like this. So nine x e to the two X now, first or next, we need to find B. Um, complementary solution. That's gonna be D squared minus one. Why? C is equal to zero. So we set that right hand side equal to zero first, then our corresponding auxiliary equation P of our has won t be able to r squared or sorry, R squared minus one is zero. So we get our is going to be able to plus or minus one. So are complementary solution. Why Sea of X is going to take the form c one e to the X plus C two e to the negative x si two e to the negative X like so next we need to annihilate this part. Here's we have f of X. Is it legal to nine x g to the two x So here we noticed that Ah, we have a root of two here and then we also have a multiplicity of two here. This tells us that are a NY leader is going to take the form of, uh, D minus two and then squared. So applying this to both sides here Well, actually, just get that our trial solution why P of X is going to be equal to and then Ah, a We have a one e to the two X and then we're also gonna have a sorry a not each of the two X and then plus an a one x heats of the two X like so, which we can also rewrite as, um so a knot plus a one x e to the two X, like so. So our general solution, it's gonna look like why of X, uh is equal to C one e to the X plus C two e to the negative x plus a knot plus a one x e to the two X so to solve for a not in a one. We're gonna plug it into the original equation here, So we need to find why p prime of X. This is going to be equal to now. We do first times derivative of a second. So it's going to be too a not plus a one x e to the two X and then plus the derivative of the first, which is going to be just a one and then times e to the two X so this can simplify to become ah, if we combine the, uh this with this here we get to a not plus a one and then plus to a one X and then e to the two X. Then we also need to find why double prime my p double prime Thanks. So again, the derivative will take this time to drive over this. Listen to be two times that, so we'll have four a knot plus to a one plus for a one x times each of the two x than, plus the derivative of this times This So it's going to be plus to a one e to the two X. Okay, so we can further combine this, uh, well to a one each of the two X to be here. So we get that's gonna be able to four a not plus for a one plus for a one X is e to the two X, like so, so OK, plug bees in now here and we get So we had four a not less for a one plus four a one x e to the two X and then minus the original, the original being, uh what was it? A Not A not minus or plus a one x he to the two X And then this is going to be able to our right hand side, which was nine x e to the two X, like so. So let's combine these two. So we haven't minus a. Not here, so that's gonna become three. A. Not, and then we have a minus a one X, so it's gonna come from here. So we have that plus four a one and then plus three a one x, this is times E to the two X is equal to nine x E to the two X. We can cancel out the each of the two X from both sides. So we need to match this, uh, this with this here, and it's gonna be an imaginary zero. So we'll have 381 It's gonna be able to nine, which means that a one is equal to three, and then we also have three a knot plus four. A one is equal to zero. So a one, um, is equal to three. So this is 12. So we have a negative 12 and a three A. Not here. So then a lot isn't a vehicle to negative four. So write that down A not is able to negative four. So our general solution if we go back, so there's gonna be negative four and then plus three here. So three x, or we can write it as three X minus four, like so minus four each of the two X. Now, this solution we're going to need to solve for C one and C two by using the initial conditions. So let me, uh, start doing that on this side here. Okay? So the initial conditions, where why of zero is equal to zero and then why a white prime of zero is equal to seven. So we're going to need to also find why, um, first, we can try plugging in. Why a zero? So why of zero? It's gonna be able to so a C one plus C two and then negative four is equal to zero. So this gives us the equation C one plus C two is equal before next, we need to find why prime of X. So if we do y prime CNBC one e to the X minus C two e to the negative x and then now the driven of this, we need to use product rule. So that's gonna give us plus. So first times were of a second is gonna be six x minus eight e to the two X, then plus, then we're gonna have three e to the two X so this can simplify down to so or well, we don't need to really simplify. We can just now plug in. Why? Prime of zero, we get C one minus C two and then this just goes to zero negative eight times. Um, negative eight times one. So get minus eight minus eight and then plus three because each of the zero is gonna be one one. So negative eight plus three is gonna be negative. Five. And this is equal to seven. So we have C one plus C two is equal to four and then see one minus. C two is equal to 12. If we add the two equations we get to see one is going to be able to 16. So see, one is able to eight. Now, see, one is eight. So we have eight plus C. Teams will do four, therefore see two Must be negative for so get that C two is equal to negative for so we can plug that in here. So that's eight. And then we have a minus four here. So this is our final solution. Uh, here.


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