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Given ODE with initial conditions:y" + 2y' + y = 0 AI 91 92 te y (0) L4 y' (0) = 2 find C2...

Question

Given ODE with initial conditions:y" + 2y' + y = 0 AI 91 92 te y (0) L4 y' (0) = 2 find C2

Given ODE with initial conditions: y" + 2y' + y = 0 AI 91 92 te y (0) L4 y' (0) = 2 find C2



Answers

Solve the given differential equation with initial condition. $$y^{\prime}=2 y, y(0)=2$$

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.

Hello and welcome to another differential equation problem. And this one we have Y double prime. When it's too wide prime plus two Y equals zero. To solve for this, we have to use the characteristic equation and from section 3.1 we learn that if we want solutions of the form E. To the R. T. We can actually simplify this using the characteristic equation of R squared minus two are plus two. Where each power of our is equivalent to a derivative of the differential patient of above. All right now let's solve this. Uh quadratic by completing the square and we'll get our minus one quantity squared Plus one equals 0. All right, let's subtract one and take the score. You get AR -1 equals plus or minus I. And lastly we will add one to both sides. So we are equals one plus or minus. I. Now due to this plus or minus term here, we'll have two solutions to this. Ah The differential equation and the optimal solution will just be a linear combination of those two solutions. Um So if we remember this characteristic equation will get us results of the form by equals C. E. To the RT. So let's substitute RN for our discovered value from this characteristic equation. So why someone look right with green Is going to be C1 E to the one plus I times t. Now, it doesn't matter which one we do first. I'm just gonna do plus uh for this case. Now, the other solution, all right in blue Will be white to let's see two E to the one minus I team. Great. Now let's uh let's use oilers pharma. This is what we learned earlier in this chapter. And that's if we have the uh if we have E to the I theater, we can actually rewrite this as kusenov theta plus I sine of theta. All right. So I'm going to uh do that here. Start off with the left side. So if we note that we can take out this E. To the T. Because there's a there's a real component. So why one equals c. one eat to the T. But times is uh E to the I. T. So that means in this case uh theater will just be T. Because we'll have I times T. Because I hear the data in this case must be T. This will be times co sign of T Plus I Sine of two. Great. Now for the second case we're gonna we're gonna have a very similar situation because the the real part will factor out. But the only difference is instead of tea it will be negative T. Because we have this negative sign right here. So I'm gonna write that out now. We'll have y sub two the C. Two E. Two T. Times. Cosine of T. Plus I signed up co sign of negative T. Plus I sign of negative T. Great. Uh Now what's the next step? Well the next step is actually to relate these two values. Um We we know that uh they're the real solution. The most broad solution will just be a linear combination of these two solutions. So what we can do is we can uh take all of those linear combinations and just put it into uh these variables here, these constants there. Uh those will absorb all the factors. So the next real step is to resolve this intersection here. So we can really add these two together and get why one plus Y. Two. So let's first think of what is co sign of negative T. Well if we look at the graph of co sign of negativity, we see I'm going to write it up here. Co sign of negativity will look something like this where this is the Y axis and this is the exactness. And if we take the negative, if we substitute any value of X. Access for the negative X. X. For its negation it won't change value. So co sign of negative T. Is just co sign of T. So I can erase this negative there. On a similar note, we note that uh the negation of sign will actually be it's negative because the sine function looks like this for its initial values X. And that's why. Um So when we have a negative inside the sign will be negative sign. And on the outside so we can do negative on the outside and raise this intersection. Great. Now let's take the some of these two uh components. Why one plus Y. two. This is the uh we'll get the solution to our differential equation. Uh And this one will be E. To the T. Because that's coming to both um Plus two co Santee. Because we have co sign of tea on both both of them. So call it a C. One co sot of T. Plus. Well this uh sign I sign of two will have a C. One value attached to it. Um So we can't uh we don't know what this value is compared to this negative uh term over here. So we are still going to have to include it uh in our final solution. But we'll have to see to sign to T. They're a sign of two and that concludes our problem.

To serve this problem by rewriting why primacy? Y. T. T. That equals two T. Squared E. To negative three Y. And now let's divide both sides by each of the negative three Y. So I'll have that one over each of the negative three. Y. Dy equals the T. Squared T. T. Because we're gonna multiply both sides by T. T. As well. And now we can rewrite the left hand side E. To the three Y. Dy equals the T square DT. And let's integrate both sides. So the left hand side will now have that one third each. The three Y equals two T. Cube divided by three plus C. And let's multiply both sides by three. So have at least three Y equals two T. Cubed plus C. Now we can take the natural log of both sides to eradicate this E. Term will have three Y equals to the natural log of T cubed plus E. And we'll divide by three. Still have the Y equals one third natural log of T cubed plus E. And now we can go to our initial condition which is why of zero equals to two. So wherever we see a wide term we're gonna plug into and where we see a. T. Term we're gonna plug in zero. So if we plug in zero here we're all we're gonna left with is C. And so we can multiply by three on both sides will have that six equals to the natural log, see rather E. To the sixth equals to see. And so with that we can actually build our final solution here. It's our final solution is going to be Y equals to one third the natural log of T cubed plus E to the sixth power. So that's

Problem. Five of section 7.1 asked for the solution of the following differential equation. So the first thing that we're going to do to solve this as we're going to write, why Prime s d Y t X and isolate variables. So what we're going to do from here is we're going to divide both sides by why, and multiply both sides by D Y t x. Sorry, just by the X. So doing. So we eliminate that DX is on that side and the wise on the other side, leaving us with d y over. Why is equal to two d X now? We're going to integrate both sides, leaving us with 11 of the absolute value. Why is equal to two X plus e now we're going to do is we're going to solve for just why so To do so, we need to have a Basie and this will simplify out to be wise, equal to eat to the two x times e to the C. So what we can do from here is we can replace this with a variable. We'll just call it a. So that leaves us with the equation. Y is equal to a times E to the two X. So now we need to go back and satisfy this initial condition. And when we do this, we will be solving for a So we're going to set y equal to two and X equal toe one. So during So we get a times E to the to cause two times want us to So a is equal to two over e squared. So our initial sorry, Our final equation will look like this. Why is equal toe to over e squared times e to the two x?


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