undetermined coefficients, diff eq, sect4.5#19

okay I'm going to solve this differential equation for you guys as you can see we are talking about second order linear differential equation with constant coefficients and you see that the right-hand side is not equal to zero so this is a non-homogeneous situation and we will not be using work on alpha in this video we will just go ahead and do it all the way and you know the deal we had to have the three steps right so let me put it down right here first we are going to find the YH that means at look at this y double prime minus three y prime plus two Y could be zero like this and solve that so to solve this we change this into R square and this is minus 3r and this is plus two North this is equal to zero back to our time right this is going to be R minus one times R minus two because the zero and you know R will be one and two from here you know why H is going to be C one e 21p so let me just put on key and then plus C two e to the two T so here we have the by H and now here we are going to be using the method of undetermined coefficient to find YP for the particular solution so that's the step two so we are going to find y P right here well YP depends on what can the form that we have on the right hand side and let me just put this down right here for you guys first of all remember whenever you have sign in this case your sign key we must have cosine key to help us out as well so based on this time heat I will just go ahead and put down sine T and cosine heat right and this is being multiplied with E to a t so right here I will have to multiply this by into a key like right here e to the T okay so this is pretty much a building block for this particular solution but I don't know the coefficient so so I'm going to put down a right here and B right here and then I'm going to add map so you know the deal we have take this and differentiate twice and I'm plugging the second the repartee onion first three of the answer to the original into the differential equation and then match the coefficient and then find out what am B's alright another remark before we do that you see yes here we have e to the T and yet here we have e to the T as well but this is just a constant multiple with e to add heat right here we have a times e to add heat and this is also be multiplied by sine T in other word this and that they are not accounting multiple of each other this and that are linearly independent so we don't have to multiply by any extra T right here which is a great news isn't it anyways derivative in action YP prime and let me just put this down like this I will look at this as my first function and then this is my second function and likewise for the second term this is my first and that's my second you know we added to the product rule right anyways right here I'm going to first keep my first function which is a e to the T and the derivative of sine is cosine T and then we add with we will keep the second function which is sine heat so let me put it down right here okay and we multiply PI DT of a e to the T which is just a e to the T okay and the next we are going to be using a platter right here as well so I will put down well let me just put down the first function right here which is B e to the T and I will multiply by the derivative second the relative cosine T is negative sine T okay so we have the subtraction right here and we add the second function which is cosine T a multiplied by Delta to the first the real could be e to ax T is just exactly B to a T and as such a beauty anyways this is what we have and if you allow you can kind of combine the constants for the a and B maybe you should do that because right here we have four times already and you know this is multiplying right there are multiplying for two functions if you want to just look at the expression is s how D is and then do the derivative you know you have to use the product rule four times and you end up with a really slow is pressure for the second derivative so thirsty which is just a collect sound terms people will do the next derivative this is still my first derivative okay so what I would like to do is let me collect all the terms that has e to the T times sine T so I'm talking about this one right this is a term that has e to the T sine T and likewise we can also combine it with this well I'm just going to put down the function part right here e to the T sine T any coefficient is a minus P rather than to put it down like this a minus B like that cool and then you see this and that can be combined as well I'll be adding them together the coefficients a and the coefficient here is B so it's a plus B and then I will put down e to the P right here times cosine T like this okay to do the next derivative once again we are going to use the product rule and now you know a - piece just a number do not put a down-cycle another consonant we must have a and B together like this okay we have to work with a and B I'm going to just pair up a minus B times equal T as the first cultural and scientific my second likewise out do the same right here and this is just a fad for multiplication here we go I'm going to keep the first function and then D I'll put down a minus B times e to the T this is clinically my first function and then I would differentiate the second function which is going to be cosine key and I will keep the second function which is sine T and then I will differentiate the first the question right here is that if the work the first is just that because it's the constant multiple each would key so it will still be a minus B e to the keep like this okay and we'll do the same thing right here I'm going to first keep the first function so I will write down a plus B e to the key okay and we'll differentiate cosine which is going to give me negative sine key like that and I will keep the second function so put it down right here cosine T and what differentiate this could you steal that right so wha a plus B e to the T just like that that's the second derivative and can we combine terms yes we can write Y P double prime this is songul this is going to be that's what with the et sine keepers which is one to get in the middle like this and that right so allow me to just put out right here e to the T sine T first and then I will figure out the coefficient right in the front okay so do we care for these this is positive a times positive times negative P right so this is right here it's going to be a minus B okay this is a minus B like that but for you to strip the negative here this is going to give you negative a and then negative P right so in other words you see this a and that it will cancel and then negative B negative B will be negative 2b and that's the coefficient in front of this part the second derivative and we do the same thing for e to the T cosine T so let me just open a parenthesis for the coefficients in a minute and we will have e to the T cosine T like that so right here this is just a minus B nothing special and right here this is also a plus B nothing special value C minus B plus B cancel and we just have a plus a which right here this is 2a okay so in fact the second derivative it's actually better than the first derivative isn't it anyways here we go we're plugging everything into the original right so now I'm just going to draw an arrow like this so plugging YP for the second derivative I will just do that right I will put this down in blue so this is going to be negative 2b times a so let me just write down now you can see to be e to the T sine T and then we add it with to a e to the T cosine T this right here is for the second derivative right and we will let me just write down the next line so minus 3 times the first derivative which is this right here okay this right here and I will just have to write it down for you guys right here this is going to be a minus B and we have the et sine T and then we have the plus a plus B and we have e T cosine T like this like that right and then this is for the second term like that and we have to add two times y which is so we have to plug in Y P so it's 2 times that which is a e T sine T plus beta3 like this ok all this shall be equal to e to the T sine T ok now on the left hand side you know you can collect nut terms so that you have something times e to a key Sankey and i quess something e to the T cosine T and that's what we have to do right here so this right here is what we have to do and they may just set our first let's look for all the terms that have e to the T and let's put down some peepers right I'm not sure what's in front get and then we'll add it with something and we have e to the T cosine T and this should give us this e to the T sine T okay and now we'll figure out what you're going to the parentheses in red so let's look for all the terms that have eg sine key first of all I'm looking at this right this right here has et sine T and this right here also has et sine T but then be sure you see we have a times negative title so a minus B like this but we have the negative in front so be sure you take negative three distribute negative three times a this is negative three a negative three times negative B is plus 3b okay so I'm looking at this turnout and lastly we have this right here as well and we had to multiply this by two so I'm just looking at the coefficient so that's just two a alright so now let's see what do we get let's collect a first a - a right here - three a so that's going to be negative a right to a minus 3 a is negative a and then 3 B minus 2 B this is going to be plus B so that's what we have okay and now we'll be looking for the eg cosine key term which is this first I have the two a right here let me just make this $10 little bit more and the next you have to be careful this is the turn that were looking at we have a plus B that we still have this negative 3 in the front previous negatives we put it here and then distribute so you are talking about negative 3a and then distribute it you get negative to be like this once again you have this negatives we put it here because you're multiplying and then this is another parenthesis so you have to distribute right here and you get out okay at the end you have 2 times this which is to be like that so we're looking at this turns out okay that's to the a first we have 2a minus 3a which is negative a again and then this and that together negative 3b plus two B is negative B like that isn't it so this is very exciting because now you see this right here this is the coefficient for e to the T sine T that must match with 1 because the coefficient right here is 1 okay so what you're saying is that you must have negative a plus B equals to 1 and this is the coefficient for et cosine T but we don't have any regular that means is 0 so I'll put down negative a minus B equals to 0 all right and you can just go ahead and do this real quick you see you can do that you know combine them you get negative 2a that's equal to 1 right and let me just write out my two better this is a negative 2a is equal to 1 so in other word is equal to negative 1/2 plug in negative 1/2 into maybe this one so we are talking about negative negative 1/2 for the a minus B I don't know yet is equal to 0 I'm going to move the B to the other side so this is going to be B right here negative negative becomes positive so you're talking about B is equal to positive 1/2 oh my goodness oh my goodness we are done with this right and see I didn't even use a calculator so you guys can do this too anyways this is going to tell you why P or I should still put a sign group for the YP right okay well let me just put it down here YP okay YP is going to be a we found it which is negative 1/2 so let me just put it down right here and then this is with E to the key sign key okay and then P right here is partly 1/2 so we'll just put down positive 1 and then e to the T cosine key and once again here is my Y P and we got our Y H right here earlier okay so at the end Y this is the step 3 so let me just put it down right here this is my step Street remember Y which is the final answer for the very original this is equal to by the superposition principle y h plus y p right so in other work I'm just going to write it down right here Y is equal to C e to the T plus C 2 e to the 2t and then I will have that which is minus 1/2 e to the T sine T and M plus 1/2 e to the T cosine T and ladies and gentlemen here is the final answer don't you feel still accomplished that's it

Channel: blackpenredpen

Views: 189,190

Keywords: solve a non-homogeneous second order differential equation with constant coefficients, blackpenredpen, method of undetermined coefficients

Id: nkXdvPUki-A

Length: 16min 41sec (1001 seconds)

Published: Tue Mar 28 2017