Repeat eSign Order with airSlate SignNow

Repeat e sign order

section three point four is a page 124 and this section deals with repeated roots and this reduction of order process we'll go over that first let's get so repeated root status when solving a quadratic equation ax squared plus BX plus C equals zero there are three possible solutions the three outcomes that is you could have two distinct solutions and they are real you could have two imaginary solutions or you could have one solution so we dealt with three one three three three four I look at this problem and I noticed that this is a linear second order differential equation with constant coefficients I let y equal e to the RT hopefully we're all comfortable by that by now take a derivative take another derivative and this would be nine R squared e to the RT plus 6ar e to the RT plus e to the RT equaling to zero if i factor e to the RT out I'll get nine R squared plus six R plus 1 equals 0 e to the RT cannot equal zero so it must be that and this is three R plus 1 into 3 R plus 1 equals 0 and R will equal negative 1/2 multiplicity 2 so if I want to talk about a fundamental set I would say e to the negative T over 3 and e to the negative T over three and I don't need to take linear algebra to figure out that obviously this is a multiple of that meaning those are not linearly independent solutions so those are not living independent the column from insurer still one is a multiple of other so it turns out if this does happen and you do have multiplicity to really make those solutions livering linearly independent guess what you just multiply by T and in chapter 4 when we have a third fourth fifth if this happened three times Yuki you multiply the next one by T squared then by T you keep on multiplying by T every time and that guarantees that those this would be a solution to this equation this would be a solution to that equation and those are linearly independent and the general solution would be y equal c1 e to the negative T over 3 plus c2 t e to the negative T and it doesn't matter really where you put c1 or c2 and let me have a formula there there is a definition if the characteristic equation yields one solution are equal a we're talking about of the character characteristic equation of a second order we have to mention that if dr. character characteristic equation of a second you know what let me type it sit 30 here we are looks better evacuate the characteristic equation of a second order differential equation yield one solution are equal a with a multiplicity of 2 then fundamental set is generated by e to the 80 because that's a and T times e to the 80 and the general solution will be as follows now that we have all three cases well they cannot mix the order here they mix the pot so it's not like you know every section has its own here are they they look at number four and there we give you a page 130 all right okay look at number four well this is again second order linear differential equation with constant coefficients I let y equal e to the RT I need a derivative and I need a second derivative R squared e to the RT minus TR e to the RT plus 10 e to the RT equals zero factor e to the RT out get the characteristic equation in order so e to the RT can never equal zero it must be that and if I look at if I look at this doesn't factor so instead of using the quadratic I like and being the square half of the middle number squared is a 1 this is R minus 1 squared because mine so square root both sides and move the 1 over plus or minus 3i my fundamental set would be e to the T the cosine of 3 T and e to the T sine and my general equation our general solution I should say my general solution would be y equal c1 e to the t cosine 3t plus c2 infinity sign alright next problem up problem number 10 the difference here they give us initial conditions same deal we look at this this is a second order linear differential equation so it's linear that's what comes with constant coefficients I let y equal e to the RT be a solution Y prime would be our e to the RT and y double prime would be R squared in totality and I'll get R squared e to the RT minus 6 r e to the RT plus 9 e to the RT equal 0 factor out e to the RT I'll get R squared minus 6 R plus 9 equal to 0 so e to the RT can never equal 0 and R minus 3 squared equals 0 I'll get R equal 3 multiple city two so the fundamental set is actually e to the 3t e to the 3t but to make those linearly independent I'm not by one Almighty and I say the general solution is y equals c1 e to the 3t plus c2 t e to the 3t now they gave me initial conditions if I apply those Y of 0 Y of 0 would be c1 plus 0 and that equals 0 that means C sub 1 equals 0 that means that's gone and Y prime if I take Y prime that is the product C 2 is a constant derivative of the first times a second plus the first times derivative the second and if I take Y prime of 0 that is C 1 plus 0 equaling 2 so that means C sub 2 equal 2 so my solution is 2t eat that would be the solution to this differential equations that satisfy both of these cases combined I want to do one more before we move on to the last version of this section consider the following modify modification of the initial value problem no problem as an example do whatever that means it doesn't make any difference modified not modified we could solve anything this is a linear differential equation with constant coefficients guess what can you guess I'm gonna let y equal Authority they're modified not modify whatever they say find a solution as a function of B okay then determine the critical the critical value of B that separates the solution that remains positive to negative from those that are eventually become okay we're running way too much I'm not reading any of this I see find a solution I'm gonna find that first then I'll worry about all of that stuff let y equal e to the RT y prime is our e to the RT and y double prime is R squared e to the RT and this becomes R squared e to the RT minus re to the RT plus one-fourth e to the RT equaling zero that would become factor that's the characteristic equation of course R squared minus R plus 1/4 equal to 0 and e to the RT we already know that well can never equal 0 so it must be that R minus 1/2 squared equal to 0 right and that would mean R equal 1/2 the fundamental set would be what eat of 1/2 T eat of 1/2 T but with an extra team and the general solution is really y equals c1 e to the T over 2 plus C 2 T e to the T over 2 alright so again I'm gonna find a solution like I did on every single problem find c1 c2 and you're done Y of 0 is C 1 and that equals a 2 Olmsted it's going to be 0 that equals 2 so this is a 2 and if I take Y prime if I take a derivative Y prime since that's two that's gonna be 2 times e to the T over 2 times 1/2 plus and here it's gonna be the product rule it's gonna be C 2 times 1 times e to the T over 2 plus T times e to the T over 2 times 1/2 and if I take Y prime of 0 that would be a 1 plus C 2 into 1 if I put a 0 there that's gone and that equals a B so C 2 will equal B minus 1 well there it is now what I would look at y equal C 1 to e to the T over 2 plus B minus 1 t e to the T over 2 well let's look at this and figure out how this works since since te tivity over 2 dominates over e to the T over 2 as T approaches infinity it turns out this coefficient is key B minus 1 equals 0 would indicate that be equal 1 is equilibrium if the B minus 1 is bigger than 0 if that's positive then this is gimmick rope and if B minus 1 is less than 0 as Y approaches infinity this is dominating this is TE 30 multiplied by a negative so you're gonna reflect it down and that's gonna approach negative infinity so sort of a graph that looks like this actually and sort of a graph that look like this and the last portion of this section deals with reduction of order something we're gonna play with in Chapter I believe 10 or 7 it's really toward the last week reduction of order is extremely important we know how to solve the ydt because that's calculus and if it's not we could use that trick if it's linear now if we have y double prime we could reduce it to a Y Prime and use calculus for right there we learned in the beginning of chapter 2 on it if we have a third degree we could reduce it to a second degree and reduce it to a first degree and use the same idea what does that look like that looks like linear algebra this is how you get a system of equations well we're not gonna get that much in depth we're gonna talk about specifically second-order and later on we touch on third order suppose that we know one solution well this is the catcher you must know one solution and that solution happened to be y sub one if you know one solution y sub one right and that's the solution to the first order which is calculus not everywhere 0 of this second order differential equation to find the second equation we let y equal V of T this is y FD V of T right and then plug him in and soul and what that does that reduces the second order to a first order and I'm going to show you how that works we're going to work out a couple of examples I'm going to look at number 18 first and notice what they have now what's the difference this is a second order linear differential equation but it doesn't have constant coefficients y equal e to the RT doesn't work on this y equal e to the RT works on a linear differential equation of any order with constant coefficients those are not constants anymore so if I play this right I'll say well let's see they're given me y1 equal T squared I'm gonna let Y 2 and it's right there equals V times y1 which is V times T squared basically and all what I'm gonna do now I'm gonna take this take two derivatives a derivative e to the T and plug it in if I take one derivative so be patient with this that would be product rule derivative of the first times the second plus the first times derivative of the second and if I take a second derivative power to V double prime times T squared plus the V prime times 2t and the product will take the two out derivative of the first times the second plus the first times derivative the second so y double prime is actually the V double prime T squared plus two the V prime T plus another two that makes it four plus two now if I substitute this let me go by color-coding dos y double prime right there I'll be looking at t squared times V double prime T squared plus four V prime T plus two B and if I use if I use why prime right there that would be minus 40 times and there that's gonna be V prime T squared plus 2 T V and if I replace Y by my traditional right I will be getting 6 times V T squared and that equals a 0 and if I clean this up a bit I'm looking at V double prime T to the fourth plus 4 the V prime T cubed plus plus 2 T squared V minus 4 T cubed V prime minus 8 T V plus 6 V T squared equals 0 I'm looking at V double prime T to the fourth now if I look at V Prime four of those minus four of those those cancel out and if I look at those that's six seven eight of those if I'm not mistaken that would be right when I multiply those I forgot to make that a squared and those cancel out I'm left with V double prime plus T to the fourth equalling simply zero that's it that's all I have so what I do I come in here and I say and this is the reduction part you guys pay attention I came in and I say let any time I get to this level I say let W equal V Prime and W prime equal V double prime so this would become the w prime T to the fourth equals 0 which means W prime equals zero which means the wdt power reduction if I integrate I'll get the W to be a constant plug that right there and so that would mean the V DT equals SC that means the DV equals C DT and V will equal C T plus D and guess what that's pretty much it that's a tricky if you take this right here you will say why to equal it was V times T squared and we just found the V to be C T plus D so the second solution to this problem is actually Y to equal C T cubed plus DT so that would be the second solution to this differential equations and I would like to work out one more problem I'm giving it you on the homework so I want to do two for the notes again we have the same issue if I glanced at this this is not this is a second order linear differential equation but this is another constant coefficients which means y equal e to the RT doesn't work they're giving us a solution and the directions on those use the method of reduction of order what's the reduction of order reduction of order is really not going to come up until you really get every Prime in there V double Prime minute use it so how's this gonna work you're gonna let y 2 equal V times T to the negative 1 whatever this is and what you're gonna do you're gonna take a derivative now the whole goal is for you to find V plug it in and you're done you're gonna take a derivative you're gonna say well that's a product will V prime times T to the negative 1 minus V T to the negative 2 and if you take a second derivative the product rule V double prime times T to the negative one minus the V prime T to the negative two here the product rule dorita the first times the second plus the first times dirt of the second so Y double prime turns out to be the V double prime T to the negative one minus two of those plus two of those and now if I again color code those I assume there's Y prime so if I plug that in I'm getting T squared times the V double prime T to the negative 1 minus 2 V prime T to the negative 2 minus 2 V T to the negative 3 that's y double prime plus 3t times and if I go by what would I get that would be F V prime T to the negative 1 minus V T to the negative 2 and of course plus y and assume Y is right there and by the way that's a prime and that would be VT to the negative 1 and that equals zero and now just distribute that will be V double prime t minus 2 V prime minus 2 V T to the negative one plus the reavie prime minus 3v T to the negative 1 plus V T to the negative 1 equal 0 V double prime T yeah that was only 1 and let's see what we have left if I really glanced at those minus 2 of those plus 3 of those is one of those and minus 2 of those right just give me one second yeah negative times a negative is a positive my mistake so positive two of those plus one of those is three of those minus three of those cancel out this is what I have and you guys this is power reduction right at this stage at this stage you say let W equal V prime this is power reduction that's what they're talking about reduction of order of sorry of power reduction what am I thinking this is the reduction of order so this is a second order this is reduction of order you're letting w e 12 V Prime and W prime equal V double prime that means this becomes W prime T plus W equals 0 and now this is chapter two two point one or two thank you so the W DT times T equal negative W the W times T [Applause] equals W DT the W / W equal DT / T I'm gonna integrate natural log of absolute value W equal natural log of absolute value of 2 + C so I could say W equal e to the natural log of T plus C which is absolute value of T actually let me write that I told you that we did that already but let's make sure we're seeing that that's a constant so I could say W equal this is absolute value of T times C but I noticed that they specified that T is positive and now I would go to my you know what I'm getting is yellow and I'm gonna say this right here equals V prime so dv/dt is gonna equal CT VV is gonna equal ctvt and I'm gonna integrate and I'm gonna get V 2 equal C T squared divided by 2 plus C okay I made a small mistake I'm gonna catch up with that I could redo the whole thing that that's gonna take a while it's much easier just to follow it you see that negative right there I forgot that's a negative that's a negative that's a negative so I'm just gonna fix this part that negative becomes a power that's e to the natural log of absolute value of e to the negative 1 times e to the C the same steps and that would mean what that would mean that C equal this is a constant so that's C and this is under 1 over so this is OK W is gonna equal C over absolute value of T but since T is positive C over T and now I really follow the same steps nothing changes except the function is slightly different here I would say well look this is this but the W is so that is C over T and that's DV DT equals C over T or DV equals C over T DT integrate and I'll get VT VC and natural log of T I don't need the absolute value is always positive plus D and that would be V and now we noted you go back to the very original assumption and you say the second solution is V which is C natural log of T plus D and that is multiplied by T to the negative one or you could say that is C natural log of T over T plus T over T it really doesn't matter it's the same thing and there is the homework