 equation first we will understand the definition of linear differential equation so what kind of od what kind of ordinary differential equation is actually categorized as linear differential equation so let's look into the definition first so linear differential equations are those in which the dependent variable in which the dependent variable dependent variable you can call it as y let's say if you are if you are having dy by dx or d2y by dx square such kind of differential coefficients present in the od then y can be called as a dependent variable and its derivatives occurs only occurs only in first degree occurs only in the first degree now what are the meaning of occurs only in the first degree that means no differential coefficient should have a power more than one on it okay exactly power one should be there whether it is dy by dx whether it is d2y by dx square in fact the dependent variable also should occur in the first degree that means there should be no y square y cube kind of a term okay and one more thing and they should not be multiplied together and they should not be multiplied together not multiplied together so a general linear differential equation would actually look like this let's say I go to a nth order linear differential equation so it will look like this plus some p1 dn minus 1y by dxn minus 1 plus p2 okay dot dot dot till let's say pny equal to some x okay now here please note the coefficients that you see here p1 p2 etc they are all either functions of x or constants including your x1 so p1 p2 p3 etc till pn including your x are either constants or functions of x remember constants can be also termed as functions of x in fact constants can be functions of any variable you want but in this lecture my dear students I am going to only talk about linear differential equations with constant coefficients here so I'll be talking about only those situations in this video I'll be only talking about where your p1 p2 etc are constants so I'm going to talk about linear differential equation with constant coefficients with constant coefficients okay I'll be not going into any other higher version of linear differential equation why I'm mainly interested in LDE with constant coefficient is because you will find that these equations are very much used in the study of electromechanical vibrations and other engineering concepts okay I'm sure you would have all done concepts related to simple harmonic motion right motion of simple pendulum or you would have done oscillation of springs or you would have done oscillatory electrical circuits like lc circuit lcr circuit right lc circuit with emf lcr circuit with emf in engineering we study the deflection of beams whirling of shafts right lot of application is there for linear differential equation my dear in the field of physics and engineering so I'll be only talking about those cases where you have these p1 p2 till pn etc as constants and of course x could be some function of x see a linear differential equation solution is called a complete solution okay which we write it as cs okay complete solution is made up of two things one is called the complementary function okay and the other is called the particular integral other is called as the particular integral so I will tell you how to find both of these components whether complementary function or particular integral for different cases okay and those will be the only cases that will be coming for your problem solving in physics or any engineering exam so cs what we call it is made up of cf and pi okay so complete solution is made up of cf plus pi i so first I will talk about how to find cf how to find cf complementary function before you understand complementary function you have to understand a very small thing which is called the operator d okay so operator d what is operator d normally what do we do is we use some operator to define such operations in calculus for example d by dx is it in like operator d d2 by dx square this is called d square okay d3 dx cube this is called d cube okay and so on I hope you would have got the pattern so if you have the nth derivative happening with respect to x or respect to the independent variable we call it as dn for example d2 y by dx square plus 2 dy by dx minus 3 y this you would write as d2 plus 2d minus 3 y okay now I will tell you why we use this operator d in order to find complementary function we frame something which we call as the auxiliary equation also written as ae okay let me give you an example for example the previous question let's say I want to find out the complementary function for this differential equation okay so what do we do first we write this like d2 plus 2d minus 3 y okay equal to 0 now this equation that you have over here this equation when you create like this this is called the auxiliary equation ae complementary function depends upon the roots of the auxiliary equation how it depends let us look into this example so if there is an auxiliary equation like this we factorize it so let's say the factorization of this will be d plus 3d minus 1 okay from here we get two roots for d okay what are the roots here that we get for d the roots that we get for d are minus 3 and 1 correct now I will not go into the details of how do we get this but the complementary function that we will get will be nothing but c1 e to the power minus 3x plus c2 e to the power 1x where c1 and c2 where c1 and c2 are arbitrary constants are arbitrary constants okay now how to find these arbitrary constants you would be given some initial conditions or boundary conditions into the question through the use of which you can find this c1 and c2 remember don't forget your basics any nth order differential equation will always have n arbitrary constants into it okay now this is a very simple example but let me give you a generic view of this let us say let us say you have got a differential equation which looks like this okay I'm just taking a nth order linear differential equation for you okay let's say this goes till k and y equal to 0 okay so to find complementary solution what would you do you will find the auxiliary equation auxiliary equation would be dn k1 d to the power n minus 1 k2 d to the power n minus 2 up till kn okay put it to 0 put it to 0 now it is not because there was a 0 over here let's say even if there was any function of x you would do the same thing to get the auxiliary equation okay so don't get me wrong here you have not put it 0 because there was a 0 here I have changed it to x x could be some function of x okay so now let us say the following cases will arise let us say this n degree equation in the operator d is factorizable okay and I get these n factors from there okay so when n distinct factors are coming here so let's say it's a case where the roots where m1 m2 etc etc all these values are real and distinct real and distinct okay then in such case your complementary function is given by c1 e to the power m1x c2 e to the power m2x and so on till cn e to the power mnx where again c1 c2 till cn etc are arbitrary constants okay so nth order differential equation will have n arbitrary constants okay don't forget that okay case number two case number two in case number two what if any two of the roots become equal so let us say out of those m1 m2 m3 till mn roots two of them are equal okay let's say m1 and m2 are equal and the other are distinct okay so let's say the others are distinct okay in such case again I will not go into the derivation part because I know you are here to get the idea how to solve a linear differential equation with constant coefficient you don't want to get into too much of theory else this video will become too long in such case your solution will become your complementary function will become c1x plus c2 and the roots which were repeated let's say I write m1 and m2 are same so you can write c to the power m1x and the others will be the same as it was before that is c3 e e to the power m3x c4 e to the power m4x and so on till cn e to the power mnx now a natural question will arise in your mind what if three of them were repeated right so here I took an example of two being repeated let's say if three of them were equal if m1 is equal to m2 is equal to m3 and others were all distinct let's say m4 m5 etc till mn then your complementary function would become c1x square c2x c3 e to the power any one of the name you can take m1x let's say and others would remain the same let's say c4 e to the power m4x c5 e to the power m5x and so on till cn e to the power mnx getting the point so now you would have got an idea that if four of them are equal what to do then there would be a cubic equation coming over here c1x cube c2x square c3x for c4 e to the power m1x then c5 e to the power m5x and so on and so forth okay so you can continue from here right now case number three case number three okay so complementary functions will be required when you are dealing with those kind of circuits where there is no external emf applied or those kind of simple harmonic motions or those kind of spring oscillations where there is no periodic there is no external force applied to the system okay yeah third case is if one of the pair of roots is imaginary now remember we are dealing with k1 k2 as real numbers so remember if imaginary roots will exist they will exist in conjugate pairs so let us say while you are evaluating the root there were two roots which were conjugates of each other alpha plus i beta and alpha minus i beta and the rest roots were like m3 m4 normal your real roots okay so these are all your real roots in such cases what happens your complementary function will become your complementary function will become e to the power alpha x c1 cos beta x remember alpha was the real part of m1 and beta was the imaginary part of m1 okay plus c2 sin beta x and the normal process will work for m3 m4 that is c3 m3 e to the power m3x c4 e to the power m4x and so on till your cn e to the power mnx okay this is the one which is basically used in the simple harmonic motion let's say unforced simple harmonic motion remember you still per harmonic motion used to get your x as a function of t which was having cos and sin term okay so this falls into that category now a natural question will arise in the mind of the students what if what if two of the imaginary roots are equal okay so what if there is a repeated or imaginary root so let us say we have a case where m1 and m2 is your alpha plus i beta and let's say m3 and m4 are alpha minus i beta and rest roots are your normal ones so let's say m5 m6 etc till mn are your real roots coming out from the roots of the auxiliary equation so in such cases again there will be a minor change in the complementary function it will become e to the power alpha x now see what will happen a small change instead of this c1 you will write c1x plus c2 cos beta x and instead of this c2 you would write c3x c3x plus c4 sin beta x and other roots will continue in general like they used to continue in the previous cases so you will have c5 e to the power m5x c6 e to the power m6x da da da da da all the way till c and e to the power mn okay so you will not get more than these cases my dear to find out the complementary function now let us look into few examples i think examples are the best way to understand these concepts let us say i want to find out i want to find out the complementary function for this differential equation okay now thankfully i would take a very simple example here okay so i will take a very simple example here let's find out the complementary function for this okay and let us say i give you some initial conditions that x0 is 0 and let's say the derivative of the function let me write it like this derivative of the function at x equal to 0 is 15 okay now these are your initial conditions since this is a second order differential equation with the dependent variable as x and independent variable st i need two initial conditions or two boundary conditions to get my arbitrary constant okay so let us find out the cf in this case okay now for cf we first need to make we first need to make the auxiliary equation okay for auxiliary equation we get d square plus 5d plus 6 equal to 0 okay this is factorizable okay i get two roots from here which are minus two and minus three okay so your complementary function will become your complementary function will become in this case c1 e to the power minus 2t remember now your independent independent variable is t and c2 e to the power minus 3t okay now luckily in this case your complementary function will be your complete solution because there is a zero on the right side see try to understand my dear students here that particular integral will come into play when there is a function over here okay when there is a function of x over here or some other constants over here right now since there is a zero over here there will be no particular integral so your complete solution would be same as your complementary function okay role of particular integral will come when there is a function of x on the right side of this equal to sign now how do i find c1 and c2 because if i get that i will get my complementary function and hence i will get my complete solution so let us use these boundary conditions x zero is zero means when t is zero the value of x is zero so now remember x will become your c1 e to the power minus 2t c2 e to the power minus 3t so put t equal to zero and put x equal to zero you will get c1 plus c2 that's number one case number one equations and the derivative of it dx by dt if you do you will end up getting minus 2 c1 e to the power minus 2t minus 3 c2 e to the power minus 3t put value of dx by dt as 15 and x and t value as zero you'll get minus 2 c1 minus 3 c2 solve these two simultaneously to get c1 and c2 your c1 value without much waste of time will come out to be 15 and c2 will come out to be minus 15 so in this case your complete solution will become 15 e to the power minus 2t minus 15 e to the power minus 3t so this would be your complete solution for this differential equation again there is no particular integral here because there was a presence of zero i will come to particular integrals when i take such a scenario let us take one more example let us say we want to we want to solve this differential equation again i'm keeping zero on the right side because i don't want to involve particular integral right now we'll talk about it little later on in today's session so if you see the auxiliary equation here auxiliary equation will be d square plus 6t plus 9 equal to 0 here you realize that your different your auxiliary equation will give you repeated roots okay so the roots here are minus 3 minus 3 isn't it it's a repeated root so in such cases your complementary function and hence your complete solution will become c1 plus c2 x now remember there is no x there is a t here so c1 plus c2 t e to the power of minus 3t it will become okay so this will also become your complete solution okay since i have not given you any boundary conditions to find c1 c2 we will leave the answer in this form only okay let me take one more quick example for you let's say i want to solve d3 y by dx cube plus d2 y by dx square plus 4 dy by dx plus 4 y equal to 0 okay again i have purposely taken a zero on this case so that we don't get any particular solution so if you write down the auxiliary equation here what will it be d cube plus d square plus 4d plus 4 okay put it equal to 0 it is not see you're not putting it 0 because there is 0 over here auxiliary equation always is formed by writing the operator d equation equated to 0 okay so this will give you if you factorize this this will give you d square plus 4 times d plus 1 so what is happening here you get one root one root as minus one and you'll get two imaginary roots from here okay so the other roots are 2i and minus 2i correct yeah so in this case your complementary function also the complete solution in this case will become c1 e to the power minus x okay and remember the real part of these two conjugates are zero so you'll have e to the power zero x which you don't need to write actually c1 cos 2x c2 sin 2x i have written it here so that your understanding is clear so ultimately your complete solution will be y equal to c1 e to the power minus x okay sorry i should not use the same name of the arbitrary constants here there will be three arbitrary constants because it's a third order differential equation so the complementary function would be this okay i hope this example was a good enough example for you to understand things all right students let's talk about particular integrals so far we were talking only about finding complementary function now we'll talk about particular integral so particular integral will make sense when you have instead of zero and x over here right so let us say this is your original linear differential equation given to you with a function of x over here okay once you have a function of x over here then only particular integral makes sense remember if there is a zero over here then your complete solution can only be found out by using complementary function so what i want to say is if your x is zero then your complete solution will be just your cf but if your x is some function of x let's say i call it as g of x then of course your complete solution will have a complementary function plus particular integral okay so how do we find out this particular integral let's talk about it so first of all the process is very similar i mean the beginning part of the process is similar to what we use in complementary function first we need to frame up the left hand side of the auxiliary equation which is actually dn k1 d to the power n minus 1 k2 d to the power n minus 2 and so on till let's say k okay so what i'll do is i'll call this guy as f of d okay now in order to find out the particular integral please note you have to do the inverse of this operator on the function x given to you okay so this function x will be given to you in the question you have to do this operation now how do i do this operation and what is 1 by f of d okay so don't worry about it i'll explain you in simple terms let me write down this as let's say f of d is a polynomial in d which is factorizable let's say these are these are the roots of f of d equal to zero okay so let's say i factorize it like this so f of d is factorizable like this so first what do we do is we treat this as if we have a case of a partial fraction scenario like this okay we all have done partial fractions right in class 11th you would have done partial fractions while doing indefinite integrals you would have done partial fractions so let's say you just you break this up into partial fractions like a1 d minus m1 a2 d minus m2 and so on till let's say an d minus mn okay find out your a1 a2 till an by you know your smart substitution method or comparison of coefficient of various powers of d there are so many ways to find out the a1 a2 a3 etc once you have found out let me give you a small formula by which you can compute this okay the formula is here 1 by d minus a of any function is given by e to the power ax integral of x e to the power minus ax dx okay so this is all you need to know okay and you will be able to manage all questions related to particular integrals so please make a note of this now how is this going to help us in solving particular integral see now that you have broken 1 by fd in partial fractions so what are you trying to calculate you're trying to calculate 1 by or inverse of fd on x function okay whatever is your x function that means you're trying to do this a by d minus m1 plus a2 by d minus m2 and so on till an by d minus mn on the x function okay so you can break this up as a1 d minus m1 on x plus a2 d minus m2 on x and so on till an d minus mn on x okay so this is nothing but a1 okay now treat as if you're doing 1 by d minus m1 on x okay so how will you do this very simple we already discussed the formula over here 1 by d minus a on the function x or you can say inverse of d minus a operator on function x will lead into this integral e to the power ax integral x e to the power minus ax dx so in a similar way this fellow it will result into e to the power m1x integral of x times e to the power minus m1x dx okay and same will be true for the others also correct so you will have next one is a2 e to the power m2x integral x e to the power minus m2x okay okay and so on okay so the next term would be a3 e to the power m3x integral capital x whatever that function given to you that depends upon the function so I'm just writing as capital x as of now but you will act as per the question given to you okay so finally the last term also let me conclude here since I've started writing this okay this would be e to the power mnx mnx integral of capital x e to the power minus mnx dx okay right now many times this computation will take a lot of your time so there are certain standard cases that we have already formulated so there are three four cases and those three four cases are always asked in the exam so what we have done we have simplified this act for you through cases okay so I will run you through certain cases by which you can find out particular integral very quickly okay so I'll take case number one first I will start with that case where your function given to you is a exponential function like this okay in such cases your particular integral your particular integral that you're trying to evaluate that means 1 by fd remember fd was the left-hand side of the auxiliary equation okay this term will just become e to the power ax by fa but this is provided this is provided your f of a is non-zero okay so what you have to do is you just have to put this a okay in the left-hand side of the auxiliary equation that you had got so remember you made an auxiliary equation so left-hand side of the auxiliary equation that is what we call as fd the polynomial in d right there you need to put instead of d this value of a but this is to be used only when fa is not zero okay for example let us say let us say I have given you this particular differential equation d2 y by dx square plus 5 dy by dx plus 6 y is equal to e to the power x okay now let us do a complete question on this let's say I asked you to find out the complete solution for this so complete solution will have complementary function plus a particular integral correct now particular integral will feature in over here because there is no more zero on the right side as you can see they have given you a x for that capital X that is the function of x okay so complementary function you already know that you have to form a auxiliary equation look at this and tell me what is the auxiliary equation that you will get for this you'll get d square plus 5d plus 6 equal to 0 correct and its roots are we can all know its roots is minus 2 and minus 3 so complementary function would be c1 e to the power minus 2x c2 e to the power minus 3x so no problem with complementary function now particular integral particular integral will be 1 by fd which is this term remember I'm calling this as fd this whole term here is fd term okay this is your fd term okay so you're trying to evaluate something like this now as this particular method says you just need to put this a in place of d over here so I'm going to put one in place of this will become e to the power x by 1 square plus 5 into 1 plus 6 which is actually e to the power x by 12 so what will be my complete solution my complete solution would be c1 e to the power minus 2x c2 e to the power minus 3x plus ex by 12 now many people ask me sir there is no arbitrary constant in particular integral no there is no arbitrary constant in particular integral please make a note of this a particular integral will not have or does not contain any arbitrary constant does not contain any arbitrary constant please make a note of this very important okay so I hope with this example it is clear that how do we deal with particular integral when your capital X function is an exponential function like this okay so now people say what if sir f a was 0 then how would your formula change so if f a is 0 if f a become 0 please note there will be a small change in the particular integral it would become x into e to the power ax by f dash a okay but this is again provided provided f dash a is not equal to 0 okay again what if f dash a is 0 then what to do then if f dash a is 0 then again not to worry your particular integral let me write it in different color your particular integral will become x square e to the power ax by f double dash a okay and it can go on and on like this so on okay so don't worry even if your f a was becoming 0 or f dash a was becoming 0 or f double dash a was becoming 0 this is the trend which we need to follow now I would not go into the proofs for this because this is a short video I would like all of you to just see in order to deal with your physics problems now case number two case number two I will talk about where your right hand side function is either a sine function you can say sine ax plus b type of a function or it's a cos function cos ax plus b type of a function now here please pay attention it is slightly tricky we need to follow this formula f of d square that means inverse of f d square sine ax plus b is basically sine ax plus b upon f of negative a square okay but this is when your negative f of negative a square is not zero provided f of negative a square is not zero okay similarly 1 by f d square cos ax plus b is given by cos ax plus b upon f of negative a square again the same condition provided f of negative a square is not zero now many people fail to understand this what is the meaning of f d square up till now it was f d how all of a sudden you started talking about f d square and all those things okay let me explain this with a simple example let us say let us say I want to solve I want to solve this differential equation d2 y by dx square let me take slightly more complicated version d3 y by dx cube so that you people will understand you'll you'll think that sir took dx d square because he had a d square in the question okay so let me take d3 y by dx cube plus y is equal to let's say cos of 2x minus 1 I want to solve this differential equation okay so as you know complete solution will have two parts complementary function and particular integral okay for complementary function what do we do let's quickly revise what we had learned we form an auxiliary equation like this okay so what would be the roots of this auxiliary equation minus 1 would be one of the roots okay so let me write down the roots for this your roots would be minus 1 minus omega and minus omega square I hope you all know the complex number minus omega and minus omega square so minus omega is negative of minus half plus i root 3 by 2 and minus omega square is negative of minus half minus i root 3 by 2 okay in short your roots will be half minus i root 3 by 2 and half plus i root 3 by 2 okay now remember for this your complementary function will be c1 e to the power minus x but what happens in case of complex roots you'll have e to the power ax which is going to be to the power half x c1 cos c1 cos beta x plus c2 sin beta x okay remember we had seen this so this is going to be your complementary function now remember this is c2 this is c3 okay i that was a slip of pen this is c2 and this is c3 okay so three arbitrary constants will come now arbitrary constants they will only come in case of complementary function arbitrary constants will never come in case of a particular integral please keep that in your mind very very important okay now let us move on to particular integral so how to find particular integral for this so let me just name a subtopic particular integral so one by f d square now see you have to treat this as a function of d square all of you see how it works so what i will do this d cube i will break it up as a d square into d plus one okay so in place of d square you have to put a minus a square see here the formula this is the formula which we are going to follow here in place of d square you are going to put a minus a square remember a here is two so minus a square here would be minus four so what i'm going to do i'm going to write this like this so one by d square d plus one cos of two x minus one so this will become one by minus four d plus one of cos two x minus one okay now what to do next simple i will i will do one plus four d multiplication both in the numerator and denominator so it was one minus four d so i multiplied and divide with one plus four d okay oh can we do that yes very much we can do that so in the numerator it will remain one plus four d of cos two x minus one remember d is an operator don't treat it like it's a multiplication of two terms denominator you will have one minus 16 d square now again this d square you have to put as a minus four okay so wherever a d square is coming immediately have to put it as a minus four so it will become one plus four d by 65 okay now this operator will work on cos two x minus one now see what is happening on the numerator part all of you please pay attention one into this will remain the same function but when four when d is operating on it it will differentiate it remember d is a differentiation operator okay so the answer that you would get is one by 65 cos two x minus one so basically when one multiplies to this nothing will happen but four d what it will do it will differentiate it so it will give you something like let me put one by 65 outside so four times derivative of this will be minus eight sin two x minus eight sin two x and remember remember derivative four times the derivative of the whole thing derivative here will follow chain rule so minus sin two x into two so eight factor will come over here so this whole thing will actually become your particular integral are you getting my point so this is your complementary function this becomes your particular integral okay so your complete solution complete solution would be let me just copy first what is your particular complementary function c one c one e to the power minus x plus e to the power half x then c 2 cos of root 3 by 2x plus c 3 sin of root 3 by 2x plus one by 65 now I have started writing the complementary sorry I have started writing the particular integral one by 65 cos two x minus one minus eight times sin two x minus one okay this is your complete solution c 1 c 2 c 3 would be arbitrary constant that you need to find out from your boundary conditions now what will happen if your f of minus a square happened to be zero when such things will happen remember the formula will change like this so one by f d square sin or cos whatever you may take it will become x times one by f dash minus a square sin ax plus b now what is the meaning of f dash a square now whatever was your left hand side of the auxiliary equation you differentiate it once correct you differentiate it once and then put the value of d square as minus a square as I discussed with you in the previous problem okay let me take a small example of the same let us say I want to find out the particular integral I want to find out the particular integral for this function only the particular integral I will not do the complementary function part because of the time issues okay let's say I have this question and my question is find the particular integral for this okay so first of all if you see your auxiliary equation left hand side would be d q plus 4d right is it it's basically d times d square plus 4 okay so now now when I do the particular integral formula it is 1 by d d square plus 4 into sin of 2x correct okay now what will happen here if I put remember my a is 2 here so if I put my d square as minus 2 square that means 4 minus 4 will cancel off correct so I cannot put here so what I will do the formula suggests that you need to differentiate this guy so what are the derivative of d q plus 4d with respect to d so you have to do this so that will give you 3d square plus 4 right so now you have to now your formula will change to x by 1 by 3d square plus 4 sin 2x now you can put your d square as a minus 4 so when you do that you end up getting 3 into minus 4 plus 4 sin 2x in other words your particular integral will become x by 8 minus x by 8 sin 2x okay again this is subject to the fact this is subject to the fact I forgotten to write over here this is subject to the fact provided your f dash of minus a square is not 0 okay so a genuine question will arise in your mind what if f dash minus a square was 0 so if that is the case then you can continue the process further that means your particular integral so I'll write it down over here so if your f dash of negative a square was 0 then your particular integral will become x square 1 by f double dash of minus a square sin of ax plus b and this will continue on and on and on till you are getting f dash or f nth derivative of a square as 0 so this is also provided provided f double dash of negative a square is non-zero is this fine okay now we'll come to the last the third case okay third case is where your right hand side function is a polynomial in x let's say I take for general purpose x to the power m okay in such cases what do we do we use the binomial expansion of the given term how do we do that I'll explain this through an example okay this is best understood with an example let us say I want to get the particular integral for this ordinary linear differential equation d2 y by dx square plus dy by dx is equal to x square plus 2x plus 4 let's say I want to solve this okay so in this case what is your left hand side of the auxiliary equation d square plus d right okay now you want to find out the particular integral so particular integral will be what particular integral will be 1 by this of your x capital x this is your capital x term okay so x square plus 2x plus 4 you are applying the inverse of d square plus d see now what you do with this d square by d you'd write it like this okay now let 1 by d be as it is d plus 1 to the power minus 1 okay write the binomial expansion for this okay so it is like I am asking you to write what is the binomial expansion for 1 plus d inverse so you can treat it as 1 plus d inverse if you want for your convenience basically it's it'll be a geometric progression with infinitely big geometric progression with first term as one common ratio as minus d isn't it so it'll be 1 minus d plus d square minus d cube plus d4 dot dot dot up to infinity okay but don't worry you will not require to that extent see when you do d you are doing differentiation okay when somebody says d of x means you are doing d by dx of that term when somebody says 1 by d of x you are using you're doing the integration of that term okay please remember this fact please remember the simple thing it's very very important so what you're doing here you are let's say I keep this operator as of now okay one into this will give you x square plus 2x plus 4 minus d on this will give you the derivative of that so 2x plus 2 will come d square of this will further give you the derivative of this which will be 2 after that even if you do one more derivative it will become 0 so there's no point going further correct now the process is not completed I'll just simplify this so this is going to give you an x square plus 4 correct 1 by d means integration of this so it'll be x cube by 3 plus 4x now many people ask me sir after doing integration can I put a constant of integration no please don't do that I've already told you in particular integral no arbitrary constant will appear so this will become your particular integral in this case got the point okay so these are the three main cases that you will be getting for finding a particular integral okay so I would like to seal this up with a small question all right so let's say we have this differential equation and I want to find out the complete solution for this complete solution means I want complementary function plus particular integral okay now of course since you have something on the right side of this linear differential equation there would be some particular integral coming and purposely I have taken a mix of exponential trigonometric and a polynomial function in this case so that you can you know revise all the concepts so first of all I will go for auxiliary equation here auxiliary equation would be d square minus 4d plus 4 okay put it 0 because that is required for us to find the complementary function so the roots that come out from here is repeated roots from here because this is actually d minus 2 the whole square am I right okay so two repeated roots will come out from here 2 2 so complementary function we can write straight away from here it will be c 1 plus c 2 x e to the power 2 x remember when roots are repeated this is what we follow okay now what about particular integral particular integral will take them one by one okay so we'll break it up as 8 e to the power 2 x 8 sin 2 x and 8 x square so let us first cater to 1 by fd fd will be d minus 2 whole square e to the power 8 e to the power 2 x so 8 will be outside e to the power 2 x now the problem is when you put a 2 here it will become a 0 remember when fd when f a becomes 0 we have to differentiate once and put an x but the problem is even when you differentiate it once more and put d as 2 it will still become a 0 so what we'll do is we'll do 8 x square e to the power 2 x the double derivative of this term double derivative of this will leave you with a 2 correct so this answer will straight away become 8 8 by 2 which is 4 x square e to the power 2 x okay so let me call it as particular integral 1 particular integral 2 now I am calling it as particular integral 1 2 3 because there are three functions there so I'll add them all after we have done with all the three of them so next is 1 by d minus 2 the whole square of 8 sin 2 x right so the other function that I have is 8 sin 2 x okay actually 8 is just a constant that will have no role to play but here also you realize that it will become 1 by d square minus 4 d plus 4 sin 2 x put your d square as minus a square so this will become 8 1 by minus 4 minus 4 d plus 4 sin 2 x okay 4 and 4 will get cancelled off so this will give you minus 2 1 by d of sin 2 x now remember my dear students what did I tell you a little while ago 1 by d means you're doing the integration of this okay 1 by d means you're doing the integration of this are you getting my point okay so when you do the integration of this you'll get negative cos 2 x by 2 correct so this will just leave you with a cos 2 x so this is your particular integral obtained from this trigonometric term right now we'll have to go to the last term which is 8 x square so I'll do that also so 1 by f d that is x d minus 2 whole square of 8 x square so 8 I'll keep outside x square okay as I told you you can use binomial over here so what I'll do first I will take out I will write it as d times okay 1 by or you can write it like this 8 into 1 by d by 2 the whole square correct okay so this will give you 2 and if you take it in the numerator it will become 1 minus d by 2 to the power minus 2 of x square okay let us apply binomial expansion on this so this will become 2 times 1 plus d remember minus 2 into minus d by 2 will give you a d next term would be minus 2 minus 3 by 2 factorial d square by 4 and so on I don't need to go further because if I do it would be of no use because this is a second degree polynomial so d cube will all give you zero zero results right so don't go too far because it will not be of any use to you correct so if I just simplify this if I just simplify this I get 2 times 1 plus d plus 3 by 4 d square dot dot operator operated on x square so this will give me 2 x square so 1 into x square will give you x square d of x square will give you 2x because it will differentiate it d of x square means derivative of x square then 3 by 4 into derivative of this guy which is 2 okay so if I multiply I will get 2x square plus 4x plus plus plus 3 so this is your pi 3 okay so your final pi your final pi would be the sum of all the pi's that you have got so pi 1 pi 2 pi 3 see I've just named it pi 1 pi 2 pi 3 just for your convenience okay so if you don't want to name it like that it is up to you so your final particular integral will be 4x square e to the part 2x then cos 2x and then finally this 2x square plus 4x plus 3 okay so what's your complete solution complete solution is cf plus pi remember cf plus pi what was your cf your cf was c1 plus c2x e to the power of 2x and your pi is this fellow 4x square e to the part 2x cos 2x 2x square plus 4x plus 3 okay so dear all dear all students I hope this video session was useful for you to solve any linear differential equation with constant coefficient it will be very helpful for you when you're solving your simple harmonic motion and you're dealing with oscillatory electrical circuits whether with forced or non-forced functions on it thank you so much for watching