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Q4. Write the types of given differential equations_0) 2dx + (y2 + 2y)dy = 0b) y' + 3y =yty' = Sy +y-2e-2xy' = 1 + 8+42...

Question

Q4. Write the types of given differential equations_0) 2dx + (y2 + 2y)dy = 0b) y' + 3y =yty' = Sy +y-2e-2xy' = 1 + 8+42

Q4. Write the types of given differential equations_ 0) 2dx + (y2 + 2y)dy = 0 b) y' + 3y =yt y' = Sy +y-2e-2x y' = 1 + 8+42



Answers

Solve each differential equation. $$y^{(4)}-y=0$$

All right. So this time we are working with the differential equation of the second, distributive of why -4 times the first derivative plus four times Y. B zero. And so this one is a little bit unique because when we factor it for working with the auxiliary equation, which is going to be r squared minus four R plus four. Factoring this gives us ar minus two square, right? Because um negative two times negative two of those A positive four. And then anyway, so we have is that our our one is equal to R. R two which is equal to two. So when we go to put it in our general solution, we're gonna do something a little bit different uh in the past few problems. So we'll take Y and we're going to take the first step is the same and we have a constant times E. To the power of their first are two times X. And we add that to a constant. But this time due to the multiplicity, we're going to take that constant times X. And multiply that by E to the power of two X. Okay, so the general solution has this extra X in here. So it's good to be aware of this property when solving these differential equations. Otherwise, this is going to be our journal solution, since we have no initial values to solve it any further.

Okay. Okay. So here we are again um solving some of these differential equations and this time what makes this sort of questions? The next few different is that we get a higher derivative. So here we have the fourth derivative of y plus three times the third derivative of y minus four times a second derivative of Y. And we set that equal to zero. And again we're going to use the um auxiliary equations to solve it. So we replace why with our and then the derivative with a power right? So it's going to be our to the power of four plus three. R cubed minus four. R squared equals zero. And we want to solve this equation for the roots so we can pull out in R squared. He would get our square times three just by getting ahead of myself here. So R squared times R squared plus three ar minus four. Okay, so that's going to be equal to R squared times the factor this second expression here, which is going to be our plus four and ar minus one. Because this means our roots are going to be the first end. A second route. So R one and R two are going to be equal to zero from the R squared. Alright, so you drew is going to be equal to negative for and the fourth fruit is going to equal to one. So we can plug this into our general solution which tells us that why is going to be equal to some constant times E. To the first route times X. Person. And since zero has multiplicity to you, we're going to take a second constant times X times E. At the power of the route times X plus some third constant times E. To the power of the route which is negative four times eggs plus some fourth constant times E. To the power of that which is one times X. Okay then we can simplify it. I'm going to eat of the power of zero is equal to one. So this gives us C. One plus C two X. Same thing happens there Either the power of zero is equal to one plus C. Three or third constant times E. To the power of negative four X. And then plus our fourth constant times E. Then the one times X is just X. So that these are general solution. We're gonna stop here. Since we don't have any initial conditions, we can't solve it any further.

Right. So here again, we are using the auxiliary method to find solutions to some of these homogeneous equations. And here the equation is the second derivative of Y plus the first derivative of Y plus Y. And so are auxiliary equation is R squared plus are plus one Equals zero. Again, no roots are jumping out at me as being obvious. So I'm going to use the quadratic formula. Quadratic formula states that for an equation X squared plus bx plus C equals zero. We can find the roots umm by solving for negative B plus or minus the square of B squared minus four times a times C. All over two times a. Okay, so applying that to our problem now we get negative one plus or minus the square roots, B squared is going to be 1 -4 times eight times one. She's also one. So it's me four times one times one minus four, All over two times once. That's just too, we simplify, they get negative one plus or minus the square of negative 3/2. And so I'm going to rewrite this so that it's in the form alpha plus or minus better. I Um so just give us negative 1/2 plus or minus U squared of 3/2. I All right. So we can basically what happens, I rewrote the square of negative three as the squared of three times the square root of negative one um squared of negative one being I. Okay. And so now we can use the general solution for when we have complex roots. Just applying that. We'll get that the general solution is going to be equal to some constant times E. And it's going to be the real part. So for us this negative one half times X times co sign of the imaginary part. So that's gonna co sign of Square to 3/2ves times X. Okay, We add that to a second, constant times E. Again To the power of our real part negative 1/2 times x. And then times sine. And then sign um will be the complex parts. So that's gonna be negative. Uh square 23 has no negative um times X. Okay. And then with this problem we were not given any initial conditions so we can't solve it any further. So this is our solution here.

In this problem, we have to solve the differential equation. Why? Double prank minus four. White crane plus 13. Why is it called to zero? Now we can associate the characteristic equation, which is going to be D square minus fourth Dynasty plus 13 is equal to zero. Now let's try to solve this quadratic equation. So T is going to be negative. Be so negative. Off negative four plus minus Squire rode off B square, so negative force Squire minus for a seat four times one times 30 divided. Fight two times a day. So two times one. So this is going to be four plus minus square root off 16 minus 52 divided by two which is going to be four plus minus. Squire rode off Negative 36 divided by two. So finally I will end up getting too plus minus three heart iota What I owed a negative one. So now we see that the roots Salter characteristic equivalent are complex numbers in this case. So let's recall that you're in. So if our one is it called too Alfa. Plus I'm Peter and are too. Is he called to Alfa minus Iveta Art, though routes off the characteristic excavation characteristic equation. So then we know the solutions are going to be you to the poor Al flicks Time See one times call sign Beata X plus C two Time Sign off B takes seeing our kids. We see that our Alfa is a call to to and our beater Is it going to treat? So then in our kiss, the general solution is going to be, he told the power. Two weeks time. See one time school sign off three X plus C two, then sign off three x So that's going to be oh, general solution for the differential equation.


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