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Solve the differential equation y ~ 3y' + 2y = Cos (t) _ 1....

Question

Solve the differential equation y ~ 3y' + 2y = Cos (t) _ 1.

Solve the differential equation y ~ 3y' + 2y = Cos (t) _ 1.



Answers

Find the general solution to the differential equation.
$$\frac{d y}{d t}=y \cos (3 t+2)$$

In the problem we have been given do you square while Plus by 30 equals cassocks. So it is equal to the square Plus one. That it was zero And M is equal to plus 0 to. So complimentary solution will be seven sign X plus C two six. No further. The particular solution will be X. Cynics plus B. X. Lot six. Now we have The hope that equals two and do sign X plus X. For six. Plus being too Cossacks minus X. Findings. And The square wipe that equals two in two cassocks plus Cossacks minus six. Cynics plus B into minus cynics minus cynics minus X. Cossacks. Hence after putting this in the revocation we have a into two. Cossacks -X. Cynics plus B into minus two cynics minus X. Cossacks plus X. Cynics yes B. X. Kazakhstan equals two for six. No, here we have do we. Cossacks minus to B. Cynics that equals two. Cossacks. So to equal to one and minus to be equal to zero Should be equal zero and equal half. Therefore Y is equal to Y C plus Y P. And further they were equals see one Cynics Plus C two Cossacks plus half X. Cynics. And this is the equation or the answer.

So we are given the executive differential equation. Do I. T. T. Is he go to of course T. And E. To the puff sign T. And we have to find a general solution. So integrate double size with respect to T. On our left we get Y. Is equal to the integral of course. T. E. To the power off sign T. T. T. Now here we recognize that the right hand side is in the form the integral of stay here. Do you E. To the power of U. D. T. And our solution here will be E. To the power of you plus C. And so the more of these you work out, the more you will be able to recognize this form when you see it. So since if we differentiate sine we get course. So we have just proved that it is really in this form And our solution will be why is he co two E to the poor offside. T plus a constant of integration seek and this is our final solution. Mhm.

Today we're going to be solving the separable differential equation. Dy over dx equals the fraction X minus E. To the negative X divided by Y plus E. To the the white power. And the first thing we need is put all the white terms on one side and all the extremes on the other side. So we're going to have Y plus E to the Y D Y equals x minus E. To the negative X. D X. And now we just need to take the indefinite integral of both of these sites. So if we're going to have y squared over two plus E to the Y plus C one equals X squared over two. Uh and that will be minus times minus which is a plus E to the negative X plus C two. And the C one and C two are just arbitrary and don't really matter what you call them. So now we're going to move all of the variables one side and all the constants the other side. So we'll have y squared over two plus E. To the y minus x squared over two minus E. To the negative X equals c. Two minus c one. Except I'm going to replace C to minus C one with just plain C. Um because it's just a random constant and it doesn't really matter what you call it, but you can add subtract whatever and it will still just be a constant. As long as you don't put any variables in there, it'll just be a random constant and we can just call and see. So now we're just going to get rid of the fractions. So we'll have uh y squared plus two E. To the y minus x squared minus two E. To the negative X equals C. And if you'd like we can take out the two and a group these together. So I'll do that. So y squared minus X squared plus two times E. To the y minus E. To the negative X. E equals C. And both of these, have you considered correct? Um I think this one would just be a little bit more simplified, so we'll go with this one as our final answer.

This question asks us to solve the differential equation. Eat of the Y minus one times why one which we can write his D y over DX is equivalent to two plus co sign X. Now we know that we want the wise on one side and the exes on one side. So in other words, we need to bring DX over to this side. We bring it over to the right hand side by multiplying book sides by Deac Sweet cancels off from the left. As you can see I'm doing right now. Take the integral of both sides Integrate left hand side to be e the y minus y on the right hand side were adding a variable 2 to 2 acts, plus integral of Cosa nexus, sign axe and then plus C or constant of integration.


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