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Find three linearly independent solutions olne given third-orde differential equation and write genera soluton 35 an arbitrary linear combinaton 0fthe252" 1252...

Question

Find three linearly independent solutions olne given third-orde differential equation and write genera soluton 35 an arbitrary linear combinaton 0fthe252" 1252 =A general solution z) =

Find three linearly independent solutions olne given third-orde differential equation and write genera soluton 35 an arbitrary linear combinaton 0fthe 252" 1252 = A general solution z) =



Answers

Determine three linearly independent solutions to the given differential equation of the form $y(x)=e^{r x},$ and thereby determine the general solution to the differential equation. $$y^{\prime \prime \prime}-3 y^{\prime \prime}-y^{\prime}+3 y=0$$

For this differential equation. We need to find solutions of the form. Why is able to eat the R X? Therefore? Ah, why Prime is able to are either the Rx. Why a double prime is equal to r squared either the Rx And then why Triple prime is going to be our cubed e to the R. X. So we substitute these into this differential creation here, and we get our cute either The Rx minus three R squared either the Rx minus minus R e to the R X plus three needed. The R X is equal to zero. Now we can factor out and eat of the Rx from every term. Get either. The Rx times are a cubed minus tree R squared minus are plus three is equal to zero. Since even the Rx cannot equal zero, we just need to set this part of the equation equal to zero. So we get our cubed minus three r squared minus R plus three is equal to zero. Now, this is 1/3 order equation, So we're going to need to factory, or we're gonna try to factor this by the rational root beer. Um, we take a look at the, um, factors of this, uh, to try to find the roots. So the factors of this, um, are just either going to be plus or minus one or plus or minus three. So we're going to use synthetic division to try to find out which of these is actually a factor. So I'm going to try, um one first, see if one works. So 111 alright. Then becomes negative too. One times negative to then is going to be, uh, negative, too. Adding these two numbers together. Now we get negative three negative. Three times out of one is they get a 30 since we got zero here. This is a factor of this equation, or this is a root of this equation here, so we can factor it into R minus one. And then what's left over is going to be. These coefficients are squared minus two ar minus tree are is equal or sorry. Minus three is equal to zero. Then this weekend factor. So we just need to look at the roots of three here, um, so this can actually factor into are plus one and R minus three we look at, the bigger this negative tells us the bigger one is gonna be negative. Okay, so that's equal to zero. So therefore are three routes are one negative one and three. Therefore, we confined three solutions e to the X E to the negative x and E to the three X. So our general solution why cvx is gonna be see one either the X plus c two e to the negative x plus c three e to the three x.

Okay, so we have this differential equation here and we need to find solutions of the form A Y of X is equal to eat of the Rx. So first we need to find why prime. So why Prime X is equal to R E to the R X, Then why double prime of X is equal to our square either the Rx and finally, why Triple prime of X is equal to r cubed e to the R. X. So plugging all of these into this equation here we have, ah y triple prime is our Q E to the R X plus three r squared E to the Rx minus four R e to the r X minus 12 e to the R X is equal to zero. We can factor out and eat of the Rx from all of the terms. So we have either The Rx seven R cubed plus three R squared minus four are minus 12 is equal to zero. So now we, um, since we know that either the Rx can never be equal to zero, then we just need to set this part equal to zero. So that's our cue. Plus three R squared minus four R minus 12 is equal to zero. Um, now, to factor this, we can use the rational luthier. Um um, since, um, this condition here is one we only need to look at the factors of 12 here. So the factors or 12 or one, You know, plus or minus one close to minus to plus or minus three plus or minus +46 And then finally just, uh, a six or plus or minus 12. Right. So let's use synthetic division to try to, um, see, which of these factors are going to be a factor or Yeah, read. So let's put a co efficiency or 13 negative four Negative 12. And let's just start with, um, I'm gonna guess one for now. Okay. So one always just carries down here, then we'll take one and one. So that becomes one. You had it. 44 here become 00 And the negatives. Well, this last number is not zero. So one is not going to be a route. So let's start over. Okay, let's try a negative one. Try negative one. Right. Let's go through this. So native three plus negative one is gonna be negative, too. A native one times negative two is positive to negative two and this becomes to hear negative 10 again, not zero. So let's let's move onto the next one. Let's try positive too. Okay, so two here. So to want two times one here. So then 52 times five is 10. Then this becomes I'm sorry. It's become 62 times sixties deposited 12 0 Great. So now we found one of our roots is too, so we can factor that into our minus two. And then r squared plus five are plus six is equal to zero again. We can factor this one here. Six is going to be equal to two times three huts and then two plus three is five. So we can turned this into R plus two and then our plus three here groups R plus three is equal to zero. So our roots we're going to be to native to a negative three. So our three equations are going to be e to the group's e to the two x needs of the negative two x and eats the negative three x

For this equation, we need to find solutions of the form wise equal to eat the Rx. So next we need to find why Prime, which is gonna be our either the Rx by a double time, which is equal to R squared either the Rx. And then why Cube, which is equal to our cube either the Rx. So now we substitute these into our differential equation here. So we get our cube. You know, there are X plus three R squared. Either the r X minus four are either the Rx minus 12 e to the R. X, and that's equal to zero next week. In fact, out in either the Rx from all four terms here. So we get either the Rx and then what's left overs are a cube plus three R squared minus four. R minus 12 is equal to zero. Since either the Rx can never be equal to zero, we need to set this part of the equation included. Zero. That's our cue. Plus tree R squared minus four. R minus 12 is equal to zero. Since this, um, constant coefficient here. Is it with one by the rational root beer? Um, we only need to look at the factors of 12. So includes plus or minus one plus or minus 12 plus or minus to plus or minus six plus or minus three and poster minus four. Right. So we're going to go and try to find the roots of this by using synthetic division. So 13 negative four and negative 12. So we contest which of these are going to be roots. Okay, so I want to start with one to see if one is a root of this equation. So one. So I bring down the 1st 1 then one times one is one. Then I add those to you. I get four. Um, then one times four is four. That becomes 00 and negative. 12. Since I don't have a zero here, one is not a route. So I'm gonna start over, so erase those. So now I'm gonna try a negative one. Okay, so let's try. Negative one. One times. Negative One is negative. One that becomes three plus negative. One is to hear one times negative. Two is negative. Two is becomes negative. Six, which comes positive. 12. Um, sorry. Um, this becomes positive. Six and again it's not. Ah, so this is not a route as well. Okay, um, so now let's try to okay, Race nous. So let's try to here so to I'm gonna try to times one is two. They gives five here. 10 um, so 10 minus four is six than two times six is 12 0 So two is a root of this equation here. So if we factor that we get our minus two and then are left over coefficients and we are square plus five r plus six is equal to zero. So now we can factor this further. The factors of six at up to five are two and three, so that's gonna factor into our plus two. And r plus three is equal to zero. Therefore, our roots are gonna be r equals too negative too. And negative three. So we're gonna hav e to the two x e to the negative two x and e to the negative three x as are linearly independent solutions Now our general solution. Why Sea of X is going to be equal to C one e to the two x plus c two e to the negative two x plus c three e to the negative three x

Okay, so we have this differential equation here, and we need to find the equation or solutions of the form Y of X is equal to eat the Rx. So we need to find why Prime Vex, which is able to are either the Rx. Then why double prime of X, which is r squared e to the R X and then why Triple prime of X, which is our cubed e to the R X. Now, um, we need to plug these all into our different situation here. So we're gonna get our cube, he to the Rx minus r squared e to the R X minus two R e to the R X is equal to zero now weaken background and eat of the Rx from all of it. You have eaten the Rx, then our are huge minus R squared minus two r is equal to zero. Since either the Rx cannot people zero. All that is left is this two equals zero. So are acute. Minus R squared minus two. R is equal to zero. We can factor out and are from all the terms we get. Our times R squared minus or minus two is he gonna zero this can then further be factored into. All right, So we need to look at, um this to hear the factors of two or just one into, and then we're taking the difference. The difference is equal to one, and then the negative one is going to be the larger of the two. So we have Our times are plus one times ar minus two. This is equal to zero. So then that means our is equal to zero negative one or two. So our solutions are gonna be the form eat of the zero e to the negative one X and eat of the two X. Remember, Either the zero is just going to be equal to one. So our solutions are gonna be one e to the negative X and then need to the two X like so


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