Question
For the following differential equation; find the integrating factor; the general solution; and the particular solution satisfying the given initial condition. Y' +y = 12x8 y(0) = 16The integrating factor is I(x) =The general solution is yThe particular solution is
For the following differential equation; find the integrating factor; the general solution; and the particular solution satisfying the given initial condition. Y' +y = 12x8 y(0) = 16 The integrating factor is I(x) = The general solution is y The particular solution is
Answers
Determine the general solution to the given differential equation. $$y^{(i v)}-16 y=0$$
Okay to start solving this offensive question, let's first write it in differential operator form. So that's gonna be D to the fourth minus eight d squared. Plus 16 of why is he with a zero? So then our corresponding on So the recreation p of our is going to be able to our to the fourth minus eight r squared plus 16 and we're gonna set that equal to zero. So we're going to have, um, a difference, Or this is kind of a quadratic equation here. If we instead we had r squared and are, um so let's see if we can factor this. So 16 the factors of 16 uh, that we need to find factors of 16 that add up to eat. So that includes one in 16 which does not add up to eight. Then it's going to be to an eight again. Also does not add up to eight. Let's try for and four. Okay, so four and four does add up to eight. So we can do is factor this and we have our squared inside. Instead of our R squared minus four squared, it's going to be equal to zero. So there now, since we have R squared minus four squared in this square causes that anything we get here is gonna have multiplicity too. So we have our squared is equal to four. And then now we're gonna have our is able to plus or minus two. And then again, multiplicity too, though. So our general solution, Why have X is gonna be C one e to the two X first and then R multiplicity to her that we just want to buy an Exxon front X E to the two X and then plus C three, then e to the negative two x and then plus are multiplicity. Two of that is gonna be plus C four x e to the negative two x So this is going to be the general solution of our problem here.
Alright so to start off this problem let's go ahead and substitute out these white terms. I am a replacement with R squared plus A. R. That's 16 equals zero. We can do a little bit of factoring here and we should get our plus four times are plus four were r plus four whole squared And it equals to zero. And from this we can derive the root is gonna be negative for -4. So in other words we have real and repeating roots. So with this information we can actually build our solution. Our solution is going to be Y equals two K. One Each, a negative four x plus K. Two E. To the negative four X. And since we have repeating roots we're gonna have an X. Term here. So let's rewrite to make it a little bit better looking. So I have Y equals two K. One X. Eat the negative four X plus K two E. To the negative four X. And so that's our answer.
Okay. We'll start this problem by rewriting why prime is Dy Dx and this equals to why and then we can move terms around. So let's keep the wives to one side. So I. Y. Dy sorry one over. Why do I? It was one of the access to the other side. So have one over X. Dx. Okay integrate both sides. So on the left hand side we should integrate to natural log the absolute value of Y. On the right hand side we'll integrate to the natural log of the absolute value of X plus C. Now we raise both sides to eat power. Well have Y equals to see X. And that's our answer.
To find the general solution. Let's first right this in differential operator for So that's gonna be be to the fourth minus 16 d Square plus 40 d minus 25. And then why is he got zero? So there. Now we confined the corresponding auxiliary equation. That's p of our is equal to our to the fourth minus 16 r squared plus 40 ar minus 25. So and we set that equal to zero. So to find the roots of this equation, I'm going to attempt to factor this first. So according to the rational root beer, um, all I need to do is look at the roots of 25 which are only 15 and 25 and we're gonna do the plus or minus of them. So I'm gonna write out, be, um, the coefficients. So I can use synthetic division to test the, um, to test the factors. Remember, Here, this is zero for the ark you term. Um, so there we go. Now, that's going to, um, Now I'm going to test a few factors, so I'm gonna try plus remind it's one first, so let's go with one. So I put in one, get 111 Then I add these two, which is one that I multiply. It's one that negative 15. So I get negative 15 that becomes 25. And then, uh, times 25 get zero. Since I have zero here, um, then that means zero is also a factor of or since I have zero here, that means one is a factor of this here. Now, here I have a cubic term. This is our cubed plus R squared minus 15 R plus 25. So I'm going to attempt to factor it some more. Okay, so I'm gonna factor this polynomial here now again by rational root beer. We only need to look at the factors to 25. So I'm gonna try plus or minus, um, one. So let's try one again. Let's see if one is a factor of this. A polynomial here again. So one bring down the one multiply one by one, we get one, then two and then so next we get one. Times two is to discuss negative 13 when they get a 13. Okay, so this doesn't This is not even zero. This is equal to 12 eso one is not going to be a factor, so let's try. Ah, different number. Let's try, uh, negative one. So negative one. Now. Time perspective one. That 00 negative. 15. 15. That's gonna be 40. So, again, not negative one. So now let's try five. Okay, So if we try five here we get five. This becomes six. Six times five is 30 and then negative. 15 times or plus 30. Gonna be negative. Are sorry. It would be positive. 15 and then 15 times five is 75. This is Ah, hundreds or not, Not five. Let's try. Negative five. So one times one is our Sorry one time saying, if I was thinking to five, this becomes negative for negative four. Uh, negative. Five times negative. Four is going to be positive. 20. This gives a positive five here. Positive five. Now, five times negative. Five is going to be negative. 25. Great. So now we found another factor of, um of this polynomial here. So then who we get are factored form? Um, it's gonna be r squared. Or so I are, uh, minus one for this first, this first factor here, then R plus five for this factor here. And then the remaining terms are gonna be the coefficients of R squared. Minus four are plus five is equal to zero. So let's try toe factor. This here. If we get the factors of five, um, they need to add up to five. But the only factors of five or one in five which clearly do not add up to four. So this cannot be factored. So we need to use a quadratic equation to solve this one here. So first, let's through the 1st 2 routes. The 1st 2 routes are 1st 2 routes are R equals one and negative five, and then here we're going to need to use court articulation. So negative b is for plus or minus and then square root of B squared, which is 16 minus four times A, which is one time, see is 20. So of 16 minus 20 here all over too. Okay, so this first part here gives us just to so is equal to two and then plus or minus Ah, 16 minus 20 is negative for square root of negative four is going to be equal to two I and then divided by two. Is this going to be plus or minus? I Okay, so then now these are our four factors. 15 and two, plus or minus. I. So our general solution, it's going to be why of X is equal to actually let me move this one and set up here. Okay? And then our general solution is going to be why of X is equal to and then we'll do the r equals once that c one e to the X, then plus C two e to the negative five x and then plus C three. And since we have a imaginary in solution, okay, so first we do, uh, this is a and then B is going to be one. We have C three e to the two X and then sign, uh, one x then plus C four be to the two x Times Co sign one X Hopes B X, but be is just one. So this is gonna be our general solution here.