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Verify that the given function is a solution to the given differential equation. In these problems, $c_{1}$ and $c_{2}$ are arbitrary constants.$$\begin{aligned}&\diamond y(x)=c_{1} e^{x}+c_{2} e^{-x}\left(1+2 x+2 x^{2}\right)\\&x y^{\prime \prime}-2 y^{\prime}+(2-x) y=0, x>0\end{aligned}$$.

is a solution

Calculus 2 / BC

Chapter 1

First-Order Differential Equations

Section 2

Basic Ideas and Terminology

Differential Equations

Oregon State University

Harvey Mudd College

University of Nottingham

Boston College

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this problem will, given a differential equation, a differential equation and a possible solution. And we want to confirm that this way of X is a solution. First, we need to check that it's at least twice differential, which is so now we can go ahead and differentiate it. Um, we will start with taking the first derivative. So why? Prime of X is equal to see one eat the X plus. This is going to be a product roll C two e to the minus X derivative of the first part times. This part stays the same. Plus first part stays the same. We take the derivative of the second part two plus four X. All right, then we confined. Why? Double prime of X is equal to see one eat x. And again, we're gonna use the product role here. Ah, So this will be plus C two e to the minus. X times A second person is the same one plus two x plus two X squared minus C two e to the minus X take the drive of of the inside of these parentheses is two plus four X and then finally, we're gonna take the derivative of this last term again using the product role we have minus C two e to the minus X times two plus four x and we will add that to see to eat the minus x times the derivative of the inside of the prince sees just for all Right now we have expressions for why my prime Why double prime? And we can plug them into our differential A differential equation up here and make sure we get zero. So starting with X times y prime we have I am going to go ahead. And as we do this, we're multiplying X times y double prime Gonna go ahead and distribute this as we go So x times why double prime is equal To see one need to the x Times X plus C to eat the minus x times X plus two x x squared plus two x cute plus. See, um, sorry. It's gonna be gonna be minus c two e to the negative x two X plus four X squared minus C two e to the negative x two X plus four x squared then um plus C to eat the negative x times four X. All right, so that was X times y double prime. Now we want negative two y prime. So similarly, student negative two y crime is equal to negative. To see one you the x plus two c two e to the minus x one plus two X plus two X squared plus C two times e to the negative X Um, sorry to in there you should actually be in negative too. So actually own. But the syriza here lips No, I want to write minus two, See to eat of the Native X Times. Oh, yeah. Okay, we're right here. So minus two C two e to the negative x Times two plus four x And then finally, the last part of our differential equation is two minus x times. Why? Where are y of x is right here. So tu minus x times Why is equal to I'll just go ahead and write this one out like this two minus X See one he to be X plus C two e to the minus x one two X plus two x where All right, So now we're adding all of these things together and hoping to get zero, adding this X Y double prime, the negative two y prime and the two minus x times. Why? So here. Let's see if we can combine some like terms here. Well, we have. Let's look at the C two e to the minus. X is so here we have four x of them, and here we have another acts of them here we have minus two X of them and minus two more x of them. So let's rewrite this as, um, see to eat the negative X Times four five, 67 Oh, sorry. So 45 on minus two is three minus two more because these negative signs is one. Yes, we have X those. Then let's look at the ones that have a ex weird. So plus C two, eat the negative x times How many expert we have with two x squared minus four X squared minus four X squared. So that's negative. Six X squared and that takes care of that term that term. And finally, we just have a plus C one each The x times X and a plus c two e to the minus. X times two x cute. Similarly, with this one we have. This is equal to zero. Um, okay, we have C to either the negative X and negative C to eat of the negative X. Um, let's put this to on the inside. Make this a to 44 in here. Let's make this positive and make this to go away. Negative for negative eight. All right. So what this We see we have well, first will rewrite native TOC one each x and then we have plus C two e to the minus X. We have tu minus four of them. So it's negative too. Plus see to eat of the minus X, we have four x of them in minus eight. Acts of them is negative. Four, that's and then, finally we have see to eat the minus x times. This last term we have the four x squared four x squared and that's all of our terms. Lastly, let's just multiply out this last one just how it ISS So we yet two c one e to the x plus to see Teoh eat the minus x one plus two X plus two X squared minus x C one Eat of the X minus accede to eat of the negative X times one plus two X plus two X squared. All right. Now, when we combine these three blue equations here what we get, Let's see first, if we can have anything cancel because I see here we can rewrites. Well, okay. How should we take this? Yeah, let's go ahead and rewrite these two equations as to see to eat in the negative. X plus four, see to eat of the negative x times X plus four C to eat of the negative X times X squared And this one we can rewrite as well as negative X C two times e to the negative X minus two X squared C to eat of the negative axe minus two x cute. Si two e to the negative X that takes care of that. All right, so now we can look for cancellations in these three terms. Um, so here we have a minus two C one you the x and a positive to see one. Either the ex. Here we have a C to e to the minus. X Times X. Where else do we have those? See Teoh e to the minus X Times X Well, here we have four of them, and here we have negative one. My goodness. All right, let's just go ahead and rewrite everything. We have one more time. All right? Si two e to the minus X. I'm starting up here plus well, actually. Okay, See To eat of the minus x times X minus six X squared Time C two e to the minus x plus C one e to the x Times X plus C to eat in the minus x two of them Times x three We already noticed these canceled. So we'll go ahead and leave Rat off and shoe plus were minus see to either the minus X right here, right now minus four x c Teoh eat of the minus X Now we're here plus four c two x squared Eat of the minus X Now we're down here Plus to see to eat in the minus X plus four C to even the minus x times X plus four C to eat in the minus. X times X squared minus x C one You the x and finally, these last three terms minus x times C two e to the minus X minus two X squared C to either the minus X minus two X cubed C two e to the minus X All right, now we're looking for are cancellations B to e to the minus X times X Okay, so here's a positive one and a negative one. So those cancel Ah, here we have six x squared times. See to do than times even the negative X um that cancels with that cancels partially with this four c two times X squared times, even the minus X. So now we have negative. Two of them left. But there's positive. Two of them. Um, there's negative. Two more. Hang on. Okay, so So there's negative two of them here when we cancel the minus six of them in the positive four of them. And here's four more of them. So we're at positive two of these terms, and then here were at negative two of these terms, So all of the X squared terms canceled. Now let's check the X terms. So, um, the X terms with C choose. So we have four x times C two e to the negative X slurs. That's a minus. We've minus four of them and then we have plus four of them right here. And that's all I see in terms of those here we have to see to either minus x times X to the third and negative two extra third C two times e to the minus X those cancel And now we have our constant one. So minus two c two e to the minus x in positive C two, uh, times e to the minus axe and finally we're left with C one times of the x Times X, which cancels with X minus X Dempsey one times the of the X. So finally we see that this is all equal to zero and therefore we do have wrapped this differential equation is satisfied by this solution here, but

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