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Homework: Third Assignment Score: 0 of pt 2.3.21Solve the initial value problem.39112 43 144cos X)dx y sinx= Txcos 2x, 9{5) The solution is y(x) =Enter your answer ...

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Homework: Third Assignment Score: 0 of pt 2.3.21Solve the initial value problem.39112 43 144cos X)dx y sinx= Txcos 2x, 9{5) The solution is y(x) =Enter your answer in the answer box and then click Check Answer:All parts showingType here t0 search

Homework: Third Assignment Score: 0 of pt 2.3.21 Solve the initial value problem. 39112 43 144 cos X)dx y sinx= Txcos 2x, 9{5) The solution is y(x) = Enter your answer in the answer box and then click Check Answer: All parts showing Type here t0 search



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Solve the initial value problems in Exercises $71-90$
$$\begin{array}{l}{y^{(4)}=-\cos x+8 \sin 2 x} \\ {y^{\prime \prime \prime}(0)=0, \quad y^{\prime \prime}(0)=y^{\prime}(0)=1, \quad y(0)=3}\end{array}$$

So saw this first order linear or new difference request in This is a separable equation. Um, so I can rewrite this as D Y is equal to three acts minus the co sign of X plus two times DX. So I've separated the variables and now integrate. And so what you get on the left side is just simply why? And then you get three x squared over two. Will you integrate the coastline function? You get the sine function so minus sign of X plus two x plus a constant of integration. Now we're given that Y zero is equal to four, so four is equal to zero minus zero plus zero, plus a constant. Therefore, my constant is equal to four. So this solution is why equal three x squared over two minus the sign of X plus two x plus four. And that is the final answer.

So we're solving this first order differential equation. This is a nonlinear differential equation, but because I have a wide squared term, but it is a separable equation. So I can write this as, um, we can go ahead and write it as, um why prime over three y squared, they go to x plus co sign of X. So another way of thinking of this is de y the X 1/3. Why squared is equal to X plus co sign of X. So you have one over. Why squared do you want is equal to X plus co sign of X, the X Now it's separated the variables and we can integrate on both sides of this equation. So when you integrate on the left side, you end up with minus one third and you have why, to the negative one. So minus 1/3 why is equal to And then you end up with X squared over two and you integrate the coastline function, you obtain the sine function. So plus a constant of integration. So now to solve this guy for why, um you're going to have Aiken if I multiplied by negative three have won over. Why is equipped in negative? Three X squared over two minus three. Sign of X plus, some constant of integration. Still gotta figure out what that constant is in a moment. So to solve. For why? Why is equal to the reciprocal of all of that? So one over. Negative three X squared over two minus three. Sign of X plus a constant. Now, the boundary condition. The initial condition was wives. Zero is negative. Two. So why zero equal? Negative to. So this means that negative too is equal to one. And, uh, negative. Three times zero over two, minus three times zero plus my constant. So this tells me that, um, So what we see here is negative two. You gotta 1/0 plus zero plus C. Negative two is one. Oversee, therefore C is equal to negative 1/2. So this solution is Why equals one over Negative three X squared over two minus three. Sign of X minus 1/2. Um, you could if you don't like the fractions, they're making it a complex fraction. You could write that a little bit differently. Um, we could do is to say Okay. Well, we could, Right? This is why this equal to let's factor out a negative three and no one ever negative three. And then you got what? X squared over two minus sign of X minus 1/2 and then what I could do it said, Well, I could also multiplied by 2/2. And so then you would get why is equal to one over negative three x squared. What? You would have a two here, um, minus to sign of X minus one and correction there once a factor that that mine is three. This was a positive sign. Um, that you had this, um And then now you see, you got negatives in front of all of those terms there, so you could just write. This is negative to over three x squared plus two sine X plus one. Um, like there. So that would be a final answer. So we had a good final answer when we were right here. It's just that if you want to play with it to not have mixed fractions in the denominator, that fraction, um, made a couple of errors on the way they got to the right answer here. Finally, by factoring out the three and then multiplying by a common denominator

In this problem. We want a solution to this differential equation with these initial values. So to do that, we're gonna have Teoh and a great over here a couple times. So we confined. Our white prime is going to be the indifferent indefinite integral of co sign of X, which is going D X, which is equal to sign of X plus some constant C one and then we have we can integrate again to see that why will be equal to the indefinite and a girl a sign of X plus c one D x, which is equal to negative co sign X plus C one X plus C two All right, so we have that. Why equal to negative co sign X plus C one x plus. C two is a solution to our differential equation, But we also need our initial values in here. So we want that Why of zero is too sweet and go ahead and plug that into here. So we'd say to equal to negative co sign of zero plus C one zero plus seed to So then we consult for C two here and we get to Z equal to no co sign of zero is one so negative one plus C two. So then we get that C to Z equal to three. Now we consol for C one by plugging in. Why prime of zero is equal toe one. So we take our equation for y prime, which is about here we'll rewrite rat So sign my prime equals sign of X plus C one and we want right Prime of zero equals one So one equals sign zero plus c one sign of zero is zero So one equals zero plus C one Yes, we get that C one is equal to one. Now we have it solutions for both our C two and R C one And we have our general solution right here so we can get our solution from our initial value. Why equals negative co sign of axe plus X plus three

To start solving this differential equation. I'm going to rewrite this differential equation in terms of its linear operator form. So it's gonna be D squared plus nine. Why isn't be able to five co sign two X Now we can get the corresponding complementary solution by setting the right hand side with zero so b squared. Why carpentry is he with zero? So now, to solve this, we find that the auxiliary equation p of our isn't B r squared plus nine is equal to zero. So we get that are is going to be able to plus or minus three I This gives us a complimentary solution. Why of X equal to C one e to the, um sorry. See one sign of three x plus C to co sign of three X Now, to get rid of this right hand side here, we're going to find its annihilator. So after Lex is equal to five co Zain two X So then, uh, the Annihilator. Okay, we have here that b is equal to two. Also, A is equal to zero. So remember, our Annihilator is gonna take the form The squared minus two a D plus a squared plus B squared. So we get that, um so 80 So this just becomes here. Then we have two squared, which is gonna be equal to four. So we get these squared plus four as our Annihilator. Now we apply our Annihilator toe our original equation here. So we get t squared plus four and then B squared plus nine of why this is equal to and then a pie are Annihilator to this side here and that will be equal to zero. K is equal to zero. Now, solving this equation here our auxiliary equation p of r zero to R squared Plus for our square plus nine is he with zero? So we get our sea with you plus or minus two I and then our is equal to plus or minus three I. So the plus or minus three I were to get from our complimentary solution So you have and why of X would be equal to see one sign of three x plus C to co sign of three X and then we're gonna have a plus now the plus or minus two I that's going to give us on a not sign of two X and then plus B not Oops. Be not cosigned of two X. So next we need to solve for a not And why not? This is our trial solution here. So we're gonna plug this into our original equation differential equation here and then find then use this to calculate a not and be not so. First, we need to find the second derivative. So Y p prime first. Why? P prime of X. This is gonna be able to to a not co sign of two X and then plus to be not Ah, sorry. Minus to be not sign of two X Now, why P double prime of X is going to be equal to all right is equal to, um, sir, Right by P double prime. So we get negative, right? For negative for a not sign two X, uh, sorry, minus two. Ah, sorry. Four b not cosigned two X like So now we can plug these into this formula. Here. It's We get negative for a not sign. Two x minus four B, not cosigned two x and then plus nine times a not or nine times this. So nine a not sign two x then plus nine b not cosigned two x and remember, this is equal to five co sign of two X So combining our sign terms Here we get, uh five A. Not to get five, a not sign of two X and then plus five b not cosigned of two X is equal to five co sign of two X notice here that there's no signed her plus zero sign. So that means five a not is gonna be zero. So we had five a Not is he of the zero And then we also get that five b not is going to be able to five So you're not zero and be not is equal to one. So since a not a zero Weiqing uh, cancel that out and be not distressing with the one. So this is going to be our general solution here. Now we need to use our initial conditions to get, um, this C one and C two. Okay, so that's right. Our initial conditions again. So why double prime plus why a nine y his people to five cosigning two x okay. And then our initial conditions Why zero is equal to two. And then why Prime a zero is equal to three. Okay, so our general solution this why of X is equal to see one. See one signed three x plus C two co sign three x and then plus cosigned two x. So let's plug in. Um, Well, first we need to find why Prime. Oh, are. Actually, we can go ahead and plug in zero here, we'll get why of zero is equal to so sign of zero is zero. So we'll end up getting C two and then, um co sign of zero is one and then co sign of zero is also one. So get C two plus one is equal to two. So that means C two is equal to one. Okay, so now that we know that C two is equal to one, we can, uh, plug it back in here. Okay, so that's just gonna be one here. Now we need to find why prime? So we can use this. So why prime of X is going to be equal to So now we're going to use, um Okay, that's gonna be three c one sign of three X. Well, just like that are sorry. Co sign Coastline three X and then we're gonna have minus three. Sign three x minus to sign two X So then plugging in. Why? Prime of zero or at zero? So co sign of zero is gonna be sorry. Coastline of Zero is gonna be one. This is zero and zero. So we just get three C one and that's equal to our initial condition here, which is three. So we get that C one is also equal to one. So our final solution, it's just going to be this. Ah, this equation here.


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