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For each of the differential equations in exercise set up the correct linear combination of functions with undetermined literal coefficients to use in finding a par...

Question

For each of the differential equations in exercise set up the correct linear combination of functions with undetermined literal coefficients to use in finding a particular integral by the method of undetermined coefficients. (Do not actually find the particular integrals.)$$frac{d^{2} y}{d x^{2}}+6 frac{d y}{d x}+13 y=x e^{-3 x} sin 2 x+x^{2} e^{-2 x} sin 3 x$$.

For each of the differential equations in exercise set up the correct linear combination of functions with undetermined literal coefficients to use in finding a particular integral by the method of undetermined coefficients. (Do not actually find the particular integrals.) $$ frac{d^{2} y}{d x^{2}}+6 frac{d y}{d x}+13 y=x e^{-3 x} sin 2 x+x^{2} e^{-2 x} sin 3 x $$.



Answers

Find the general solution for each differential equation. Verify that each solution satisfies the original differential equation.
$$3 x^{2}-3 \frac{d y}{d x}=2$$

Yeah. So for a question area have divide over the X dressy calls to the integrate off 12 X plus four. So d y over the X just equals 26 Squire plus four x with a constant. So integrate both sides. We'll have y equals two to act to the power of three plus two x choir. Let's see one x You forgot the X here, Reese constantly to and for B We'll have integrate off the term Amanda life that that's just equals to the y over the X So we can just write you I over the x here integrate offi to paragraph X sign next the X so D y over the X just equals two. With our constant and now less integrate, both sides will have white just equals two c one x with C two. Yeah, and foresee We'll have the y over the X equals to integrate off extra power of three actual power off minus three. So do you buy over The X just equals two x two powerful over full minus 1/2 extra power to plus the constant. And we integrate both sides. Okay, we can see that it's the integrate off the y. And this is the integrate. Yeah, See, Wan T X. So why Jesse kills Thio acts the power off 5/20 Last one hour two x and constancy.

Hello 100. Today we will discuss second order. Essentially English. So the question tight John want to discuss is a by the cash. Must be by rash. Massie. Why? He was two products. It is a second orderly near ordinary differential equation off the non homogeneous type that constant coefficients a VCR. Constance, we're here on our X. Is any function off next? We will try. Who stole this type of question today? Okay, just all this. We have Vegas methods. I didn't using one off them through an example. Okay, let's constant examples. Let's take some. Why Nash Nash was okay. So you solve this. I'm going to use the metadata. So you sold by any matter? First of all, we need to find out the complementary function. Just the solution off the corresponding homogeneous equation. So the corresponding homogeneous equation is where our X is where Who's you? So I substitute our expert zero. That is why number nationalist too wide. Ash was you now so I can write down my axillary question. Auxiliary equation is in some real valuable. I have our squad find out. It's so I can see that it's B R minus. So this gives us are complementary function or the solution off are promoting this equation as see one buzz to depart minus two weeks. This is our solution. Questions just function. Find out the particular solution. We will be using the metal ovation off para meters where people most awful You will use the new independent functions which we have computers in our complementary functions. Complementary function I hard white one as one on white US e to the minus two x right. These were the functions I have already opened in my complementary function. This I'm going First of all, I will sign drawn steno five and come away. That is close to that. Don't minute off. Why? Like to five days later on find minus do u minus two weeks. This is gone skin. It's now to compute the but Nicholas solution by invitation off para meters I'm going to use a formula. Stays minus one Integration all of light into politics divided Find that on Stan The X bless Why? To integration buying This is the formula to find out the particulars division by the nation off para meters I was substitute the values What was our why one by one is minus one integration off part minus two weeks into my Rx our exes need to the bar three x divided by Ron Skin that is minus minus two x The X plus y suis depart minus two X Integration one Into our access to depart three x divided by minus minus two X sticks now simply find the stones and invigorating. I will get minus one by two Last one by two Integration off E to the par three x DX minus one by two e minus two X integration off e five x t x This is what our immigration like now computing these individuals three x and fire. Next I get like to e g X minus one like to place did the par minus two weeks into fighting's. Okay, so this is what I get after indicating because indignation off to depart all my exes heated battle sites divided by. So that is what I get after intonation. So that will be won by six e three X minus one by 10 3 x Just again my light Take the issue so and see him off. 10 and six will be totally get so fais minus 33 x, which wants to bite hors d because it was toe one by existing a three egg. So this is the final particular solution that I have nothing. So the final solution will be. Why were y c classes white? That is what Arjuna solution will look like. Why equals y c plus y b where y c is c one plus c to me to the part minus two x plus one by 15 Even deeper three x That will be our final for the given on.

So let's say we're looking at this different show equation three X squared minus three D Y. Everyday X equals two. And if I want to take it back to the general solution, then I want to get rid of that. D y reacts. So actually, my first move would be to try and isolate do whatever the X so I would subtract that three x squared to the other sign, and then I would divide by the negative three. So d y over the X is by itself dividing by the negative three gives me a one X squared minus 2/3 dividing both terms by negative three. So at this point, I am ready to separate the variables. To do that, I'm going to multiply the D X to the other side, right? That allows me to continue to reverse across us of the derivative here by taking the anti derivative. So on the left, I have the integral D y, which is like one times do I so that integral goes to y equals and then, on the right side, the anti derivative here, I would add one to the exponents and then put that exponents that new exploding on the denominator. I'm doing that reverse of the power rule for the anti derivative. So 1/3 x cubed and then minus 2/3 x number of you have just a constant in the integral. It becomes a coefficient for your anti derivative, and we have to put a plus C. I put plus C to represent any possible constant number that could have been there. So there could have been a plus five plus four that would have gone away when we took the derivative. But I put a placeholder for it right now because I know there could have been something there in the original. This is a pretty easy process to check. If you're not sure if he did it correctly to check it, did I do my anti derivative? Right. Well, take the derivative and see if you go back to where you started. So remember the derivative of why is do I Buddy X with respect to X right? Think of it like implicit differentiation. As I do the power rule. I get X squared for that first term. I get minus 2/3 for the second term and I get a plus zero for this constant term because any constant number there right plus two would have a derivative of zero. So any plus c there has a derivative zero, and then I can just undo the steps that we did before I could multiply by the negative three instead of dividing. And then I could add over the three x squared into this, attracting so literally just doing the opposite steps that I did originally. And I verified that what I wrote can go back to the original if I take its derivative.

Our goal in this video is to make sure that all of the equations given match with this differential equation. So first off, taking the derivative of the first function of B why prime would be, um e negative e to the negative X substituting that back in to the function on top. We'd have to times that so we could just go ahead and say negative to B to the negative X and then we still have plus three Why, which is just this equation. So three e to the negative X. Now we want to make sure that this does, in fact, equal even the negative X well based off of the constants in front here. Then, of course, this just gives us one either the negative X or just e to the negative X. So this first one did work and did satisfy the top equation. The next one that will look at is taking the derivative of this function, and we'll get negative e to the negative X and then we're going tohave minus three have in to the negative three heads X. Now this one is going to be a bit longer, but I think we could fit it in the next step here, so it's gonna be two times all of this. Let's go ahead and write it out. E negative E to the negative X minus three pounds. The so the negative three hands x all of that now plus three multiplied by the original function, which is in blue there we just got and put it back in blue. So it's easier to see e to the negative x plus. So I'm negative three pounds specs and all of this. We want to be able to simplify and to just get you to the negative X here. So let's see what happens. What we've got to e negative to eat of the negative x positive. So these two will combine to e to the negative acts. Let's see about these next term's here. Well, the twos canceled here from the front and so that would be negative. Three. This second one here would become positive three. So this one positive three that negative three this and so these ones would cancel. So that would just leave us with each of the negative X And then from here we just have a more generalized version of it below. And so let's just jump right into it. So we know it's going to be two times y prime, which is negative e to the negative X and then minus three on D. C mm to the negative three hands X and all of this plus three. We'll go back and say, plus three times the original function that you could just put back in blue here. E the negative ax once. See he the negative three. It's, uh, okay. The first two terms negative to either the negative X positive three. Either the negative XP's would combine e so the negative X the next term's the two would cancel with this, too. So that's negative. Three c, etcetera. This one's positive. Three. After distributing, see etcetera, positive, negative negative cancels with this positive. So we are in fact left with this expression as well. So all three of these do fit this differential equation here, and all of them satisfy that solution


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