5

Y" _ 2y' + 2y = -5 cos t y(0) = 1, y(0) = 1....

Question

Y" _ 2y' + 2y = -5 cos t y(0) = 1, y(0) = 1.

y" _ 2y' + 2y = -5 cos t y(0) = 1, y(0) = 1.



Answers

Solve the initial-value problem.
$$
y^{\prime \prime}+y=0, \quad y(\pi)=1, \quad y^{\prime}(\pi)=-5
$$

Of instantly know that the general solution is given by the headroom empty Y D. C. So if we separate the variables we end up with D Y is equal to one plus two over tee times. DT. So if we take the anti group But about the left right, that's I'd be enough with y is equal to you t plus the anti derivative off to T is two times ln of the absolute value of tea plus c So using the initial value here more wise people to fire Tzu Once the wise people five t is equal to one and then we're solving for up T is equal to one. Then we're solving for C. This term goes to zero so we end up with four is equal to see. So our final answer is why is equal to t plus two l end of the absolute value of tea plus four

Start about discretion here to solve initial value problem or D. Wire or D. T. And plus two guys go to one. So let's bring this to our to the other side by subtracting to where both sides and this is a proportion. So let's do a cross multiplication so that the terms are separated. Uh this can better than us. Dy over minus two can be taken out. So this can be written as why minus 1/2 and this is equal to D. T. And this can be easily integrated. Now I have kept minus one or two anyway outside because that's a constant. So the integration becomes -1 or two with other tools, Integration of one over via -1 or two as natural log of I minus one or two. Integration of dtc plus the constant of integral A. C. C. But we need to find the constant ridiculous as well because that's an initial value problems were given some conditions, this condition means that when he is zero, y is five or two. So let's substitute that if y is 5/2. So this becomes five or two minus one or two and 50 0, this is zero plus C. So this becomes -1 or two weeks ago to natural about 5 -1 is four. Forward with two. As you can see we've got the value of C. So if we substitute does finally we get the final answer is minus one or two. Natural log of. Why minus one over to physical to t minus 1/2. Natural log of If you want to re decorate this, perhaps we can bring this one or to the left side. So that's one or two. Natural love of two minus one or two. Natural log of y minus discomfort. And as to why minus one or two because they're just taking a calcium and this is equal to T. And finally we can club both the logs so one or two is a common factor and log A minus log this log of A over B. So that's log off to over two minus one over to this is equal to T. And finally this can be denounced. Who is bringing two can be returned over to the other side, applying to both sides and this becomes four over Absolute value of the Y -1. So this is the final answer. Thank you.

Our first step here is going to be taking the LaPlace transform of both sides of our differential equation. And when we do that, we're gonna get to the class transform of the second derivative of y plus two times with the loss, transform the first derivative apply plus five times the loss. Transform of why being equals class transform Delta Party minus pi over two. And now we can use the rules that we have for the loss transform to simplify this equation. So waste have been the rule for their football assurance. From the second derivative, we're going to get s squared times of loss. Transform of why minus s times Wyatt zero minus the first derivative zero plus two times asked times that Well, flash transformer Why minus y zero plus five times of loss transform Why being equal to e to the negative are a spy opportune, so negative high over to terms s And now we can gather terms and insert our values for y of zero live primary zero. You can see that y 00 and why prime a zero is too. So this whole terms gonna be zero and this will turn was going to zero. So therefore, when we gather terms, we're going to get factoring out the capital by we're gonna get s squared plus two s plus five all times capital. Why s minus two is equal to e to the negative by over two times s. And now we're going to solve for capital. Why, it will give us e to the negative by over two times s plus two divided by a squared plus two s. That's fine. And now, in the in the denominator, we can see that this is almost s plus one. Oh, squared is that is s squared plus two s plus one where that came from. The fact that we have a two s here, we're gonna match the s, and we're gonna right this as each of the negative ir two times s plus two all divided by as Post born squared. But we only had a plus one. There's we need to add four to get to five. So plus four and we can separate this in the two separate factions. Namely, you need to the negative by over two times s times one over X plus one squared plus four, which is two squared plus to or s plus one elsewhere plus two squared. And now we're gonna use some of last transform rules. So we take the class transform of the sign of B times T This is just being over s squared plus b squared. So we look at this second term, take the bus transform or inversely slash transform. Using the inverse on both sides brings A If we have the universal boss transform of B being able to over a squared plus two squared, this will be equal to you. The sign of B is too so to cheat. And that is what this second term is right here. So this is the sign of two T or the boss transform from the siren to t. And we can modify this again since we have the inverse of class transform of one over s squared plus two squared. If we want to figure out what this term is, but only have an s instead of ask plus one, you could work our way to this whole thing. We want a two in the numerator such that we can just use this rule again. So if we multiplied by one multiplied by 1/2 times to this is just one. But if we pull the 1/2 outside should be 1/2 times the inverse LaPlace transform of this, which will be a 1/2 times the sign but to t. And now we can use this again, but added a plus one toe s. And here we're going to use the first shifting here, Um, which says that this will be the inverse of class transform of what we had for just regular s So 1/2 terms, the sign of two team but multiple i e to the negative A where a is one negative one time. See, that's the first shifting fear. Um, And now we're going to use the second shipping theory, um, to get our expression for the first term in our equation, which was e to the negative high over to times s times two overexpose born, all squared plus two squared. And if we have the second shifting, the're, um this says that this will be the heavy side function at what he's raised you, which is by over two of tea times. Little flash transform of this which was 1/2 times sign of to We have to subtract by over two. From everywhere we see t negative, T minus pi over two. And now these are all of our terms. Now weaken right out our solution. Why? Of tea? After taking the universal flash transform of this whole equation. So we're going to get 1/2 times the heavy side function at high over to of tea times the sign of to terms team minus by over two times each in the negative team, minus by over two. Oh, plus sign of to tee times each and the negative. T You can actually simplify this a little bit. Shall we carry in? The to we'll get the sign of to t minus pi. We can use the identity that this is just negative sign of Tucci to get that. Our final answer will be the sign of to thine times Each of negativity minus 1/2 every side of probably were too of tea times the sign of two t times e to the negative T minus pi over two

Why Prime Minister Wide five times she Why has life here are one. And is it for why you Mais s where 60? Yes, right. You just gave me kill as to s means squared course for going to kid que thank you.


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