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Question 2 (1 point) The transfer function associated with the differential equation ay" (t) + by (0) + cy(t) = g(t) has the formG(s) as2 +bs +cas2 + bs + cas2...

Question

Question 2 (1 point) The transfer function associated with the differential equation ay" (t) + by (0) + cy(t) = g(t) has the formG(s) as2 +bs +cas2 + bs + cas2 + bs + c G(s)

Question 2 (1 point) The transfer function associated with the differential equation ay" (t) + by (0) + cy(t) = g(t) has the form G(s) as2 +bs +c as2 + bs + c as2 + bs + c G(s)



Answers

$1-10$ Solve the differential equation.
$$\frac { d y } { d t } = \frac { t } { y e ^ { y + r ^ { 2 } } }$$

Um this was said we have the paper duty as it could go P square p on this people, cities were minus one. So that brought did be very deeply. Is it here to the square minus one misty square minus one. This is a little too square minus one key plus one. I felt therefore, 80 on the plus on his head T square With this one. Duty taking tickets on both sides with your over keep this fun is brought to you by three Honesty escape. So I have a B plus one. Is it quicker to the power to be killed? A tree Manistee this yeah period with bar key to this one. They're scared business, huh?

Let's solve this problem by trying to simplify out the right hand side. So we'll have that. Dy DT equals to one plus T squared divided by one plus Y squared. And the way I got this was by distributing this square, the whole square to the individual parts and the denominator and numerator. And so now what I can do is I can multiply both sides by D. T. And also by one plus Y squared. So in the end I'll have one plus Y squared. Do you? Why equals to one plus T. Square T. T. And here I can actually take the integral of both sides. So now we can try and integrate both sides. So I'm going to end up doing a use substitution for the left hand side. So I'm gonna do you equals to one plus Y. And D. You equals to one something to plug that in. So have U squared to you on the left hand side and the right hand side them into the same thing except them to use a different variable. I'm going to use the V. Here. So I'm gonna v swim plus T. And the D. V. Equals to one. So when we rewrite the integral we're gonna have v squared. Mhm. And so now we can take the integral of both sides and we'll get you cube divided by three. And feast cubed divided by three plus seats. And we can substitute back in U. And V. So in this case we said that you was one plus Y. One plus Y cubed divided by three. We said that V equals one plus T. So we have the one plus T cubed divided by three pussy. And so we can both play these both sides by three. And so it will end up with this one plus why cubed equals to one plus T cubed plus C. And we can take the cube root now so I have that one plus Y equals to the cube root one plus T cubed plus C. You can subtract one from both sides will have that Y equals to the cube root of one plus T cubed. Let's see minus one. And now we can go to the initial condition which is why I have zero equals two. So where we see a white term we're gonna plug into and where we see A. T. Term we're gonna plug in zero. So we'll have one plus zero which is one. So one cube for C. And minus one. So we're now we're going to add one on both sides will have that three equals 23 Their route third grade of um one pussy. We can cube both sides and we'll have three acute which is 27 equals to one plus C. And we subtract one from both sides will have that C equal to 26. And so with this we can actually write our final solution, which is why equals to the cube root of one plus t cubed plus 26 minus one.

Let's find the differential of dysfunction here. Let's re write this as a function of three variables. Keep these in order. And then, if I want, I could maybe just take that denominator and rewrite it as a term to the negative first power. Well, see why that's a good idea in a moment. So now the formula for the differential DT take those partial derivatives, multiply them by the corresponding differential and then add them together. There's our formula for the differential, But now, as we can see, we'LL have to do three calculations, one for each partial derivative. So here, let's go into t you first. So you on ly appears in the parentheses, so we'LL just go ahead and use the power rule here and then we'LL have to go inside and then use the chain rule as well. So if you must supply the thing underlined in red with respect to you, you could be w And let's go ahead and well, we can simplify this while we plug it into the formula over here so you could combine those bees. Go ahead and put this turn back in the denominator if you like. Don't forget to square it and then we'll still need our do you there So that's the first one Two more to go next One derivative with respect to be so same idea here Similar idea We'Ll use the power rule on this but also as we can see here because we're doing with respect to be let's actually differentiate the original Let's use the quotient rule over here. So the quotient rule Recall that Alright that over here for you Quotient rule So here are numerator is just be so the derivative of that is one and then we'LL have our gr denominator in the numerator is the first term and then we subtract f times g prime So fsb and then she prime is just the derivative of the denominator with respect to be you w and then square that denominator And you could also cancel outs in terms here in the numerator So we'LL go ahead and plug that in over here is our next term and then we have one more ago that's derivative with respect to w So this time it might be easiest to just use this formula for tea over here so you don't have to use the quotient rule. So use the power all to bring down the one multiply by. The derivative of inside the Prentice is with respect to w giving you UV And then when you simplify that you could right, that is a minus sign and then combined those fees again and then rewrite that exponents in the denominator. And don't forget DW. So there it is. That's our differential, and that's our answer.

Hello. Everyone today were to determine if X equals co signer to tea. It's a solution to the differential equation DX dt plus T X equal Signer Tootie. So first we're gonna start out by taking the derivative of X with respect ity from which we obtain two times negative Sign in to tea, which you can rewrite as they have to sign of to t. So now we plug in X and d x d t inter differential equation So when we have DX DT plus tee times X equal sign of to t we substitute the values and we get negative to sign of to t first tee Times co signer to t Equal Sign of Tootie Now we're gonna combine negative to sign of to t with signing to T by adding by to sign to tea on both sides and we obtained a T co signer to t is equal to three time signing to t now notice that tee Times Co signed to T and three times sign of Tootie are two completely different functions and there's no known relation between tee times, coastline to T and three sign or two tea. So this is not true. And so the answer to our question is no


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