5

Dy/dx + (1/x)(y)=xy^4 , y(1)=1/2solve the differential equation over the interval [1,4]using Heun's method with step size h=1.5. Round off to 5decimal places....

Question

Dy/dx + (1/x)(y)=xy^4 , y(1)=1/2solve the differential equation over the interval [1,4]using Heun's method with step size h=1.5. Round off to 5decimal places.

dy/dx + (1/x)(y)=xy^4 , y(1)=1/2 solve the differential equation over the interval [1,4] using Heun's method with step size h=1.5. Round off to 5 decimal places.



Answers

Determine the general solution to the given differential equation. $$y^{\prime \prime}-14 y^{\prime}+58 y=0$$

All right to solve this differential equation. Let's first write it in linear operator form. So that's gonna be the square and minus 14 D. That's in privacies plus 58 of why is equal to zero. So then the corresponding differential or sorry, the auxiliary equation is on the Beagle two r squared minus 14 R plus 58 like so? So we said that he was zero. Now, if you want to try to factor this, we need to take a look at the factors of 58 that add up that add up to 14. Okay, so the factors of 58 include ah one and 58 which clearly does not add up to 14. Next is going to be two plus 29. This also cannot factor into anything over. This does not add up to 14 and cannot factor anymore. So we're going to need to use quadratic formula here to solve this. So we get our is equal to, and then we have negative B, which is going to be plus 14 plus or minus plus or minus and then square root of 14 squared minus four times 58 all over on then to so out front here. First we have seven. So 14 divide by two is gonna be seven, and then it's gonna be plus or minus this. Um, 14 square is 196. All right, so that's 196 and then four times 58 is 232 somewhere in a 232. So then now this becomes negative. 36 little things to square to that is going to be equal to six I. And then we have divided by two on the bottom. So that's gonna be plus or minus three I on the bottom. So this is a here. This is be here. So our general solution Why of X is going to be able to see one e to the A. X. So seven X, then sign of B X. So and then plus C two, e two, the A X again and then co sign of B X, which is cosigner three X here. So this is going to be our general solution here

Okay, So if you saw this problem, we're first going to solve the corresponding homogeneous equation. So the initial equation here is gonna be able to r squared and then plus a one minus one here, a one. It's minus or negative one. So then minus one is gonna be negative, too. So we're gonna have a minus two are. And then here as to our and then plus a two, which is five. This is equal to zero. So here we can factor this since the, um or we can't back to this. Since here, the coefficients or the factors of five are one and five, which doesn't add up to two. So we're going to need to use quite ahead creations to solve this. We get negative, too. Negative. Negative twos. That's when he posited to plus minus B squared is for and then minus. And then here, we're gonna have a B, uh, for a c. Sorry. A is one, and then be us five here. Okay. Square to that. That should be a five. Sorry. And then all over two A, which is just too okay. And then out front, here we have a one. Is this gonna be one plus or minus and then we have here four minus 20 which is negative. 16 square root of that is going to become for I And then we have a divided by two down here. So that's gonna be plus or minus to I so are complementary solution. Then it's going to be able to see why see of X is going to be equal to see one X to the A, which is just one times cosigned of to Ln x and then plus and then we'll have si two and then sign of two Ln x Not sorry. And then this should be an ex out front as well. For this problem to solve the right hand side here, notice that this is a solution to a specific differential equation as well where we can take the Annihilator. So we're gonna choose the Annihilator A of D. This is going to be such that it turns this into zero or the derivative. This is gonna be a solution to this differential equation. A D is equal to zero. So here we noticed that if we have a equal one right, because it's actually a and then this here well are actually just gonna people to one. And then this Ln x square tells us we have a multiplicity of three. So r a f d is going to be such that this is gonna be a root of some Annihilator. So applying the Annihilator a of d to our differential equation, Let's just call it p of the why is he with zero? This is going to give us an extra roots of our and multiple three. So are solution are full solution. Why of X? It's in a week or two C one x co sign of two Ln x year plus C two X sign two Ln x and then we're gonna have her coefficients here. So first we have a not X And then times or sorry, it's gonna be X times a not plus a one Ln X plus a two Ln X squared. Okay, like this l in Squared X. Okay, now we're going to use This is our particular solution here, and we're gonna plug it back into the original differential equation. So this here is why p of X. So let's take y p prime. So why p prime of X. This is evil too. So we'll do first times the derivative of this. So that's just gonna be ah, the derivative of or Sorry. So x times the derivative of that is going to become now. Here we have the derivative of this. Okay, we have a one. Ah, okay. That's an a plus. Sorry. It's gonna be a not plus our Sorry, not a not. It's just gonna be a one Times X to the negative one. And then we'll have the derivative of this is going to be equal to so we plus to a to and then l n of X and then times X to the negative one like that. So and then now, plus again, the first for the second time. Servitude first. So we just have then plus and a not plus a one Ln x plus A to L in squared X. This can actually simplify. Will have, uh, when we do that, we'll just have a one and then plus to a two Ln x so I'm going to group the terms. So here we get a to, um Ln x m n and a one. So I'm gonna factor in l. A next from those two terms, we're gonna move. This may not actually out front here. So a not and then we have a plus a one here and then the Ln X And when a factor out from these two terms. So we have this here and then just times Ln x and then this rest here. Okay, then next we're gonna take Y p uh, double prime. So that's going to be equal to Okay, so this will just go to zero. Then we're going to take ah. So the river this is going to be able to we'll have to A to A plus a one times X to the negative one. And then plus the derivative of this is going to be to a to times X of the negative one times l n of X. So Okay, now we can plug all of these into this equation here, so x squared times this. Okay, let's, um, plug that in. OK, so first, let's take X squared times This So that's gonna become to a to on let me do this in a different color. So I have hay to eight to plus a one times X and then plus to a two x l n of x and then minus x times this here we have minus a not plus a one X and then minus to a to plus a one x Ellen of X. And then plus we're sorry minus and then a two x Ln X squared. I'll move that over to the next line here. So Ellen squared X and then now plus five. Why so Plus five A knot plus five a one and then plus and then we have 10 a to okay here, 10 a to plus okay, a five a one times Ln X and then plus five A to l n squared x For all of these terms, actually here I forgot There's also an X Times, or multiply by X out in front of all of this as well. Let's move out to the next Patris simplify. Okay, here. Okay, Now let's write some of this out. So this is equal to to a two X. So let's find our eight or ex terms first are without only have X. So that's includes this this and this here. So we have Okay. Five A. Not okay. Five a. Not an A minus a Not so that's gonna be for a not X. So for a not first came then. Okay, Now we have a one and then minus a one and plus five a one. So that's going to be a, uh, oops. Sorry. Here. I don't know why I have this here. Let me just rewrite this. This should just be five times X. Uh, five x a not plus a one. Let me write it as five times. Um, plus five a not plus five. A one Sorry. Five A not X class five a one x l n of x last five A to Ln squared X. Okay, so that's here in 75 a not X plus five a one x Ln X plus five a two x Ln squared X. Okay. So actually, the only extremes we're gonna be these against the a one cancels out and the next we have a to a two term. So we'll have now a plus to a to that's Times X. Next, we're gonna look out at our excel and ex terms, so that's gonna be this term here this term and this term here. So we're gonna look at the coefficients here. So are a one turn first is going to be this and this. So we have a five minus two a one. So that's going to be so it's gonna leave a plus. So far, we have a four a one of X Ln x eso again. Let's take a look at this here. So then we have a minus two a to an A plus a two. So that's going to just be Harris County Council out. So we're just left with for a one x l n of X Next for our X Ln squared X terms, that's gonna be this and this here. So that's just going to give us a five A to minus a two is gonna be a plus for a two x island squared X. And remember, this is going to be able to eight x l in squared X. So first we're gonna have that. This term is going to match this term here. So we have four. A two is gonna be able to eight so a two is equal to two. Next here, we're gonna have that this term should match with a plus Imaginary zero x Ln x. So we get that for a one is equal to zero. Next, we're gonna have four a knot plus to a to should be able to zero because there's a plus zero x term here. Okay, so a two is equal to two. So this is going to be equal to four. So a not it's going to be equal to negative one then. Right? So we have This is four which when we moved to the other side, becomes a negative force. We have four a not equals negative four. Divide by four. We get negative one. All right. So our particular solution, it's going to be able to Why P is equal r of X is gonna be able to x times a one just one. So then plus two Ellen squared X. So that's gonna be negative. X plus two x l in squared X. So are general solution. Why of X? This one is equal to and C one and then, um X co sign to Ln X. Okay, See up. Sorry. C one x cosigned of two Ln X plus c two x sign of two Ln x and then minus x plus two x l in squared X

To start solving this problem. First, we're going to solve the homogeneous solution, which means finding in dish the initial equation, which is R squared. And then we're gonna have plus a one minus one. So a one is able to four and then we take minus one, just three. So we're gonna have plus three are and then plus two plus two here. Now, this can factor since the factors of to our one and two, which add up to three. So it's gonna be our plus two are plus one is equal to zero. And then now we have our is able to negative to negative one Now are complementary function. Why sea of X. It's gonna be able to see one x to the negative two plus C two x to the negative one. So I'm gonna choose why I want to be extra native to why two. It's gonna be extra the negative one. Now, for the particular solution, this is going to take the form Why P is equal to u one y one plus you to why to now to solve this we're going need to find ah well you want in you two are going to satisfy then the following creations. You one. Why one or you one prime? Why one and then plus why to you To Prime is equal to zero. And then why one prime? You one prime plus Why too prime you to prime is going to be able to and then this should be able to f of X where f of X here, we're going to need to put it into standard form, which means having this move over to the other side, we need to divide this whole equation by X squared. So here are f is going to be able to four Ellen of X, divided by X squared. So then now we can find you one prime and you threw prime. So you one prime there it's gonna equal to W one divided by W and the new two primes gonna be w two divided by W According to Cramer's rule W one or W. First fall is going to be called Ron skin, So it's gonna be why one just accidentally to why one prime, which is able to x a negative two x of the negative three. Why two is excellent native one. And then why? To prime it's negative x to the negative too. So this is gonna be equal to So we multiply this times this So we get negative X to the negative four and then minus negative. So plus two x to the negative four. This is able to just x to the negative four. Next R W one this is gonna be able to We're going to replace the first column here by this here. So we have four. And then, all right, it is X to the negative to Ellen of x zero. And then we have X to the negative one and native, excellent native to. So this is gonna be equal to this times, this zero and then minus this times this So we'll have negative X or negative four x to the negative three. Ln of X next w two. This is gonna be equal to we have X to the negative to negative two x of the negative three and then four x to the negative, too. Ln of X and zero. Okay, so this Now. Okay, we won't buy this. Buy this minus this Times this this is just zero. So then this times this is going to be able to four X to the negative four Ln of X So now are you one. So you want is going to be able to w or sorry you one prime is gonna be equal to then we have negative four x to the negative three l n of X, divided by X to the negative four so we can bring up this up to the top. So it's like multiplying by X to the fourth. So we get negative four X Ellen of X as you one prime. So then you one is gonna be called to the integral of that negative for X l n of X, the X. So we're going to need to use integration by parts. So I'm going to use Ah ah Acronym lie eight. To determine which of these functions I need to set as you Since l is near the top, I'm going to choose you to be equal to Ellen of X. So you is able to Ellen of X, do you is equal to one over x dx. Then I have TV is gonna be able to negative four x t x so obvious in a vehicle to native two x squared. So I end up having UV u V is equal to negative two x squared l n of X. Yeah, and then minus this times this so minus integral. And then we're gonna have a plus two x dx. So this becomes, um, X squared. So native two x squared Ln of X plus X square and I'm going to move the X Square now, just over to the other side. And then now we can also factor out on X squared from this and this so that we have one minus two Alan of X. Like that. Okay, Okay. Now we can find you too. So you two prime. First of all, this was equal to here. Um, we had four or four X to the negative four island of X over X to the negative four. So that's a negative. Force will counts out. So we just have four Ln of X. So YouTube's gonna be the integral of four Ln of x, the X. So here we're also going to use use of our integration by parts, and we're gonna let u equal. Ellen of X d U is one of equal to one over x t x and then Devi is gonna be able to just four the ex Soviet is gonna be able to four x So this is gonna be able to We have four x l n of x and then minus ah v d u So this times this so we just get the integral of four d X that's going before X allen of X minus four X. We can factor out a four x from both of their terms. So when we factor out the four X, this becomes a one and then times that okay, next, why P of X This is going to be equal to So you won was X squared times one minus two Ln x and then, um, you or why one was excellent. Negative too. So multiply by X Negative two and then plus, Now we have you two. So four x times l n of X minus one and then times y to why to is just X in the negative one. So these will cancel out the X to the negative one. Here, we'll cancel it. This and this. So what? We're left is we have a one and a minus four when we multiply this out. So we have one minus four. So that's gonna be minus three. And then here we have a minus two and a plus four. So that's gonna be plus two Ellen of X. So have negative. Three plus two. Ellen of exes are particular solution. So then now our total solution Why of X? It's going to be equal to our complimentary solution, which was C one excellent native to plus C two x of the negative one. And then we're gonna have plus our particular solution, which is minus three plus two Ln of X.

Okay, to start this problem were first gonna find the initial equation. That's gonna be our square. And then we'll have a plus. A one minus one are so everyone is negative three. So that minus one is gonna be equal to negative for we're gonna have a minus four r and then plus four, there's just a to Z with zero. So this can factor into our minus two squared. That's gonna be able to zero. So our is gonna be, too with a multiplicity of to So are complementary function. Our solution is gonna be able to see one have an excellent too, then plus C two x squared and then times L n of X next, Our particular solution. Why P is going to take the form you won. Why one less you to why, to where? Why one is going to be X square and why two is going to be X squared. Ln X next, you one YouTube are gonna satisfy the following The equations will have. Why one times you one prime plus y to you to prime some people zero. And then why one prime? You one prime Plus why? To prime you to prime is gonna be equal to f of X. We're here f of X. It's gonna be the right hand side of our differential equation, except we need to put it in standard form. So we need to get this to one. Do that we divide by X word on both sides. So we're gonna have now won over Ln of X. And I'll rewrite that as Ellen to the negative one or, um, Ellen of X to the negative one. So by Cramer's rule, you one prime is gonna be, um this w one divided by W you to prime is gonna be able to w three Divided by W where w is the Ron skin of why I wanted night too. So that's gonna be able to we'll have X squared and then two X and then we also have X squared Ln of X and then we're going to use product will to find this derivative. So we'll do first times derivative a second, which is one over X. So that's going to leave us with X and then now plus second times or second times the derivative the first so two x Ln x okay, then here When we multiply fear we get an X cute Plus it's x cubed plus two x cubed Ln X and then minus This times this so minus two x cubed Ln of X which this is going to be able to This too will just cancel out here so we'll just be left with X cubed next to find w one. We're going to replace this first Call him here by this here so we'll have zero and then Ln x to the negative one x squared Ln of X then x plus two x Ln x So this times this is zero and then we'll do this times this So we're just gonna be left with negative X squared, then w two Okay, we'll leave the first column and then here. Now we replace the second column and here will have x the native one zero. Okay, so there's gonna be equal to X squared times Ln x So the negative one. So are you want or yeah, you won Prime The son of equal to now we're gonna have W one, which is negative, X squared, divided by X cubed. So that's just gonna leave us with negative one over X. Next are you to prime? This is gonna be equal to So we'll take our X squared and then Ln x to the negative one. Help divided by X cubed. So it's gonna be able to one over x times l n of X to the negative one. So you want prime and you two prime Let's I'll copy that over to the next page so you won is going to be equal to the integral of negative one over x the X So that's just gonna be able to negative l innovex here and then now for you to here, you two is going to be the integral of that times l n of X to the negative one dx We're gonna make the substitution that you it's gonna be equal to l N of X, which means that be you is gonna be one over x dx. So this here is gonna be, do you and then we have you to the negative one. So we're taking the integral of you to the negative one, do you? And then now there's gonna be Ln of you so we'll have Ln of you. But you itself is Ellen. So we'll have Ellen of Ellen of X. So now. Ah, Where, um, particular solution. It's gonna be good to you one times. Why one. So why one is X squared? So we'll have negative X squared and then Ellen of X, and then we're gonna have Plus why two was X squared Ln X and then we're most played by Ellen Ellen of X. Then we can actually factor out on X squared Ln x from both the two terms. And then we can put the minus over here. So that becomes minus. That was a minus one. This here. Okay, so now our general solution is gonna be our complimentary solution, which was C one x squared and then plus C two x squared Atlanta of X. And then now we also have a plus X squared Ln x times Ln of Ellen of x island of Ellen of X minus one. So this is our full solution here.


Similar Solved Questions

5 answers
Cma cmn obyedt locates converging lens with local length picture and locate the located at ?0 cm from the lens a) draw the ad the imbe height, €) image b} calculate the irage distance inverted upreht, real or virtual; determine whether the image Jarger or smaller _
Cma cmn obyedt locates converging lens with local length picture and locate the located at ?0 cm from the lens a) draw the ad the imbe height, €) image b} calculate the irage distance inverted upreht, real or virtual; determine whether the image Jarger or smaller _...
5 answers
A)Costs are (increasing/decreasing) at the rate of $_ per week at this production level b) find rate of change of revenue; use implicit differentiation with respect to t c) find rate of change of the profits, use implicit differentiation with respect to t
a)Costs are (increasing/decreasing) at the rate of $_ per week at this production level b) find rate of change of revenue; use implicit differentiation with respect to t c) find rate of change of the profits, use implicit differentiation with respect to t...
5 answers
E d AL L 1 ZeuD
E d AL L 1 ZeuD...
5 answers
Question 7. (30 points) and that J € R gatisfies: Suppose that Ta, 6] = R is integrable L(P f) < J < U(Pf) For &ll partitions P of [a,0},Prove that f(r) dc 2 ],
Question 7. (30 points) and that J € R gatisfies: Suppose that Ta, 6] = R is integrable L(P f) < J < U(Pf) For &ll partitions P of [a,0}, Prove that f(r) dc 2 ],...
5 answers
1 Prove Ineorem 1.7 by Snowing tnat quadrilaleral Aecb iS a parauelogram ; f 40=CD ard AD=BC. Teorem ,,7: In quadrilatera| ABCD , Soppose that 48 CD and AD= PC Twn ABco (< Parallelogram:
1 Prove Ineorem 1.7 by Snowing tnat quadrilaleral Aecb iS a parauelogram ; f 40=CD ard AD=BC. Teorem ,,7: In quadrilatera| ABCD , Soppose that 48 CD and AD= PC Twn ABco (< Parallelogram:...
5 answers
True or False. logs' (2 23) 2logs* logs23Logy(sab) LosaS tlogqa t leg_bLogyS tlog n los_(xew)10. What Is a logarithm?
True or False. logs' (2 23) 2logs* logs23 Logy(sab) LosaS tlogqa t leg_b LogyS tlog n los_(xew) 10. What Is a logarithm?...
5 answers
Question 183 ptsUse implicit differentiation to solve for dy from the expression cy? = 3y = 4 drdy dx d 2y 3 d 2cy d 3 y
Question 18 3 pts Use implicit differentiation to solve for dy from the expression cy? = 3y = 4 dr dy dx d 2y 3 d 2cy d 3 y...
5 answers
Q7: Generally, we may use 70-76% alcohol to:0 D: A and BB: Sanitize bench workA: Clean bench workC: Sanitize surrounding air0 E: A and €
Q7: Generally, we may use 70-76% alcohol to: 0 D: A and B B: Sanitize bench work A: Clean bench work C: Sanitize surrounding air 0 E: A and €...
5 answers
Find the derivative dx by solving the given implicit equation for y explicitly in terms of xry = 7OA
Find the derivative dx by solving the given implicit equation for y explicitly in terms of x ry = 7 OA...
5 answers
(II) A Ferris wheel 22.0 m in diameter rotates once every 12.5 s (see Fig. 5-9).What is the ratio of a person's apparent weight to her real weight at ($a$) the top, and ($b$) the bottom?
(II) A Ferris wheel 22.0 m in diameter rotates once every 12.5 s (see Fig. 5-9).What is the ratio of a person's apparent weight to her real weight at ($a$) the top, and ($b$) the bottom?...
5 answers
Ict R? R * R have the usual topology and let M {(I,w) € R? " = 0} then M is closal and every pojnt of M is a boumlary poitt.A Trute B FalseA 0B 0
Ict R? R * R have the usual topology and let M {(I,w) € R? " = 0} then M is closal and every pojnt of M is a boumlary poitt. A Trute B False A 0 B 0...
5 answers
Obiect @ 0sce the wir wih An wpward telecity 0f & Jce RT rrondfmom Lhc er buiklng fect huiph Thc litc HEakes %oxthe object ta hit th? pround bclox E gn Urcneil c" Cel hic vnela cAneucraikFtic lnm Ra€ ,bus rcmcmb Ik ncrrnc may Tejccted }FHCMIUacylow
obiect @ 0sce the wir wih An wpward telecity 0f & Jce RT rrondfmom Lhc er buiklng fect huiph Thc litc HEakes %oxthe object ta hit th? pround bclox E gn Urcneil c" Cel hic vnela cAneucraikFtic lnm Ra€ ,bus rcmcmb Ik ncrrnc may Tejccted } FHCMIU acylow...
3 answers
1. Calculate and describe how to prepare write the procedure 50.0 mL of2.SOx102M solution of Morphine for preparing the solution: remember to note how (CITHI9NOB) and how much water to add (hint: it is not 50mL many grams of your of water)? dnug
1. Calculate and describe how to prepare write the procedure 50.0 mL of2.SOx102M solution of Morphine for preparing the solution: remember to note how (CITHI9NOB) and how much water to add (hint: it is not 50mL many grams of your of water)? dnug...

-- 0.024225--