33 Linear Differential equation of higher order 1 Part 5

追加: 2026-04-29

[00:00:00]okay now this is very very important topic from this chapter linear differential equation of higher order with the constant coefficient so till now we were discussing our linear differential equation of first order now we are going to talk about linear differential equation of higher order so this will be the standard format of linear differential equation you have any derivative of Y with respect to X we in our earlier discussion we have seen what is the meaning of linear differential equation a differential equation was non linear if exponent of Y is not 1 exponent of dy by DX is not 1 or you have some tenth term containing product of dependent variable Y and it's derivative dy by DX so here you can see that exponent of all these terms is 1 you have Y to the power 1 dy by D X to the power 1 and similarly a method n minus 1 of the derivative the exponent of this is 1 here also 1 now regarding these coefficients here either you can have some constant numbers or you can have some function of X ok you will never have some function of Y here because in that case the differential equation will become nonlinear ok if I write Y here then you will have Y into derivative of Y with respect to X so it will become nonlinear so either you can have some constant here or some function of X here now if you have constant here then the differential equation is called a linear differential equation with constant coefficient and if you have some function of X here then it is called a linear differential equation with variable coefficient so for gate purpose mainly we have to talk about linear differential equation with constant coefficient there is one topic related with linear differential equation with variable coefficient so that will discuss after some time so for the present moment we have to talk about linear differential equation with constant coefficient so here I will have some constant say this constant is K and K n minus 1 n minus 1 or the derivative so anka fisticuff isiand ever I think - one similarly here k1 because the first derivative cannot why this equals 2x now we now we are going to introduce some symbols instead of writing this differential equation I will introduce the symbol capital D and this will mean a D by DX the first derivative if I am writing capital D Y so this will mean dy by DX similarly d square this will mean a second derivative okay if I am writing D square Y so this will mean second derivative of Y with respect to X so with the help of these symbols we are going to modify these equations so we can write this differential equation in this manner KN and D to the power n Y this will mean era the derivative of Y with respect to X plus next return KN minus 1 D to the power n minus 1 Y and so on finally we will get k1 a dy plus a k2 Y naught K - this was K naught K naught Y this equals to X now from all these terms if you take a Y common then you will get K n D to the power n K n minus 1 D to the power n minus 1 so on K 1 D plus a K naught Y this equals to X now you can see that this expression is nothing but some polynomial expression nd you have D to the power n D to the power n minus 1 and so on and these are some constant numbers so this expression you can write as some function of D y equals to X so from now on you can write the given differential equation in this manner f of dy equals to X where F of D is this polynomial function let me give you one basic example suppose if I have a differential equation say D 2 y by DX square may be 2 times D 2 y by DX square minus 3 dy by DX plus y and this equals 2 sine X say we have this differential equation then if we want to write this differential equation in this form then we can write this as 2 da square Y minus 3 dy plus y equals 2 sine X so 2 dias / - 3 D plus 1 y equals 2 sine X or you can write this as f of dy equals 2 sine X so the given differential equation can be written in this manner where this F of D will be F of T is nothing but this quadratic expression in D since it was a second-order differential equation so we are getting a quadratic expression so from now on instead of writing this differential equation I will write F of dy equals 2x and this will mean our differential equation 1 okay so this was basic introduction on this linear differential equation now we have to talk about this X so I am going to erase this portion and here I will explain this differential equation can be written in this manner we can write this as f of dy equals 2x now as far as this X is concerned there are two possibility for x one possibility is that this X will be 0 other possibility is that X is nonzero here you have some function of X or some constant so if x is 0 then the differential equation is called homogeneous linear differential equation okay and if X is nonzero then in that case it is called a non homogeneous non homogeneous linear differential equation for example in this example you can see that right hand side was sine X so here you have some function of X and therefore this is an example of non homogeneous linear differential equation but suppose if I had a 0 here okay suppose if this right-hand side was zero then in that case this differential equation will become homogeneous linear differential equation now in order to solve these type of differential equations there are two things that we need to calculate the first one is okay to solve this differential equation we have to find two things the first one is called complementary function in order to solve linear differential equation of our order with constant coefficient we need to calculate this complimentary function inside this complimentary function is represented by YC we will learn how to calculate complementary function but first let me give you the basic idea of how to solve this type of differential equation so first st up calculate this complimentary function next step calculate the particular integral first you have to calculate complimentary function YC then you have to calculate perfect particular integral of IEP now once you have calculated these two then your complete solution will be given by the formula complete solution Y this will be complementary function plus particular integral okay so using this formula you can find the complete solution first calculate complementary function then calculate particular integral then complete solution will be sum of complimentary function plus particular integral one more important thing if X is zero okay if this capital X is zero meaning that if the differential equation is homogeneous linear differential equation in that case the particular integral will be zero so for homogeneous linear difference and this you will find in most of the problems in gate exam okay in most of the problems in gate previous here paper you will find that this capital X is zero okay always you will be almost always you will be given this homogeneous linear differential equation so a particular integral is zero then your complete solution will be complete solution here will be y equals to complimentary function so for homogeneous linear differential you don't need to calculate a particular integral the particular integral will be always 0 so only you have to find the complementary function and that will be the complete solution so for homogeneous linear differential equation solution will be y equals to complimentary function for non homogeneous linear differential equation you need to calculate both the complimentary function and particular integral and then the solution will be sum of these two complimentary function plus particular integral so now we will learn how to calculate complimentary function for a different linear differential equation ok so now we are going to talk about rules for finding complimentary function so here the fastest step will be a step one we have to solve with this equation f of M equals to zero ok this equation is called auxiliary equation this is auxiliary equation so faster step will be you need to solve this equation f of M equals to 0 and from here you have to calculate the different value of M say that the values of M are M 1 M 2 M 3 and so on up to M n because for an eighth order differential equation this will be a polynomial equation of order and let me give you an example suppose if we had to solve this differential equation say D 2 y by DX square minus 4 dy by DX plus 3 y equals to 0 so say here since the right hand side is 0 particular integral will be 0 so we only need to calculate complimentary function now this equation can be written as d square minus 4 D plus 3 y equals to 0 so this is your F of D this function is f of D now the first step is you have to solve this auxiliary equation the auxiliary equation will be F of M equals to 0 so this means you here in a place of D you have to write M ok replace T by M then you a quadratic with an mm M s square minus 4 M plus 3 equals to 0 now you have to solve this equation so you can easily see that the factors will be 3 & 1 so this means the roads will be 1 & 3 so this will be the fastest step you have to solve this auxiliary equation f of M equals to 0 and from here you need to calculate different value of M here since we had only a second-order differential equation this auxiliary equation was quadratic equation and we have obtained a two different route if suppose this was a third order differential equation then this auxiliary equation will be a cubic equation and then we'll get three roots okay some of them might be imaginary but you have to calculate all the roots so for any thora differential equation you will get this in different roots and you will get n roots now depending on the nature of these roots total there are four possibility so the first possibility is this will be case 1 when all the roots are a distinct when all the roots are a distinct means you have obtained n roots and all those n numbers are different numbers for example here we have obtained the two roots and both of these roots are different numbers so this will come under this first case so if all the roots are distinct then in that case the complementary function will be given by this formula this you have to remember the formula will be complementary function will be C 1 e to the power M 1 X C 1 is some arbitrary constant plus C 2 e to the power M 2 X plus C 3 e to the power M 3 X and in this way up to C and E to the power M n X so this will be the formula to calculate complimentary function when all the roots are distinct the numbers and you can see that for an eighth order differential equation there has to be an arbitrary constant so these are the NRB constant okay so for example here if we want to find the complementary function so complementary function will be C 1 e to the power M 1 X M 1 is 1 plus C 2 e to the power M 2 X M 2 is 3 so this will be the complementary function for this given differential equation and since here the right-hand side is 0 the solution is also solution will be y equals to complimentary function which means the C 1 e to the power X C 2 e to the power 3 years since the particular integral was 0 here if the right hand side was nonzero some function of X then in that case we also have to calculate particular integral so that we'll see after the discuss our non complimentary function so this is first case when all the roots are distinct then this will be the formula to calculate complimentary function then the second case is when two roots are equal when M 1 is M 2 is M and all other roots all other roots are distinct when two roots are equal then what will happen so see now you cannot use the first formula because if two roots are equal then e to the power M 1 X and e to the power M 2 X these two will be same so if you take e to the power MX common C 1 plus C 2 will give us one arbitrary constant but we know that we have to have an arbitrary constant in a in authority NCL equation but if these two constant will give us 1 then you will get only n minus 1 arbitrary constant so here you cannot use the formula this formula here will get a slightly different formula here the formula will be in place of these first two terms you have to write C 1 X plus C 2 this will be corresponding to these two equal roots C 1 X plus C 2 e to the power MX plus now all the remaining terms will be same because all the other roots M 3 or fourth are distinct numbers so C 3 e to the power M 3 plus c4 e to the power M Forex and so on in the last term will be C and E to the power M and X okay so for example suppose if the roots of auxiliary equation are say the different value of M that you are getting is 1 1 & 2 these are the three values of M that you are getting so in this case the complimentary function will be C 1 X plus C 2 e to the power MX M is the equal root here equal root is 1 so e to the power 1 into X plus C 3 e to the power M 3 X M 3 is 2 so this will be the complimentary function if you have two equal roots now you can easily see what will happen if we have three equal roots let me take one more example here suppose if we have three equal roots say the roots of auxiliary equation are 2 2 2 & 3 now how to find the complementary function so now the formula for complementary function will be C 1 X square plus C 2 X plus C 3 this will be corresponding to these three equal loads e to the power 2 X plus C 4 e to the power M 4 X so this will be the complementary function if you have two equal roots then you will get a linear factor linear factor as coefficient of e to the power MX if you have three equal roots then you will get a quadratic coefficient for e to the power MX so in this way you can write the complementary function for these first two cases now we are going to see what happens when we get imaginary root so our next case is this third case when now we all know that imaginary roots are always in conjugate pair ok if you have any polynomial equation with real coefficients so here the coefficients are always going to be the yield numbers because this will be a real polynomial equation so if the coefficient are real numbers then imaginary roots are always in conjugate pair so if one wrote is alpha plus iota beta then other root has to be its complex alpha- iota beta and say the other roads are m3 m4 and so on up to em n say these are the routes that you are getting to imaginary roots these are complex conjugate and all the other routes are distinct so here you can apply the first formula because all the roots had a distinct a number and from this first formula will get one important result that you need to remember so here the complementary function Y will be C 1 e to the power M 1 X M 1 is alpha plus iota beta X plus C 2 e to the power M 2 X M 2 is alpha minus iota beta X plus all the other terms will be same those terms I am NOT writing here but you know what should be these terms right from here the remaining terms will remain will be same now we are going to simplify these faster two terms so as you can see that this expression you can write C 1 e to the power alpha X e to the power i aughtta beta X plus C 2 e to the power Alpha X e to the power minus I oughta be x plus these terms now if you take e to the power Alpha X common then you will get C 1 e to the power i aughtta beta X C 2 e to the power minus iota beta X plus the remaining terms now we all know that e to the power i aughtta theta is cos theta plus iota sine theta so similarly here also you can write this as e to the power Alpha X C 1 cos beta X plus y otras sine beta X plus C 2 cos beta X minus IATA sine beta X plus the remaining terms now you can see that if you take this cos beta X common from these two then in the coefficient that you will get is C 1 plus C 2 so that you can write as another arbitrary constant so you will get maybe I can write C 1 plus C 2 as simply C 1 so C 1 cos beta x plus now the coefficient of sine beta X that you are going to get is iota C 1 minus iota C 2 so that we can call that some other number C 2 so C 2 sine beta X so this will be corresponding to these first two imaginary roots plus the other terms so this is very important in case if you have one pair of imaginary roots then corresponding to that pair of imaginary roots so you will get e to the power Alpha X alpha is the real part of this complex of this imaginary root and C 1 cos beta X beta is the imaginary part so C 1 cos beta X plus C 2 sine beta X and the other terms will remain same other terms will be C 3 e to the power M 3 X and so on so let me give you one example here suppose if the roots of auxiliary equation are we need to write one pair of imaginary roots so suppose if one wrote is say for the third case suppose if one root M one is two plus three iota M though is two minus three Alta and the other roots are M 3 is say three okay or maybe some other number say m 3 is 5 now these are the three roots that you have boo plus 3i outer two minus three outer and the third root is five so here if you want to write the complementary function then complementary function will be e to the power alpha X ok X was independent variable so e to the power alpha X now C 1 cos beta X beta is 3 so C 1 cos 3x plus C 2 sine beta X plus C 3 e to the power M 3 X so this will be the complementary function if the roots of auxiliary equation are these three numbers corresponding to these two imaginary roots you have e to the power alpha X C 1 cos beta X beta is 3 plus C 2 sine beta X plus C 3 e to the power M 3 X and our last case is this will be case for when m1 is m2 is alpha plus iota beta you have one pair of equal imaginary roots and m3 is m4 is alpha minus iota beta okay suppose if these are the roots that you are getting then in that case the formula of complementary function will be almost similar except now the c1 will be replaced by c1 X plus c2 so now the complementary function will be e to the power Alpha X you have to replace that C 1 by C 1 X plus C 2 cos beta X plus C 3 X plus C 4 sine beta X so this will be the complementary function for for these roots of auxiliary equation now remember that in most of the cases you will get 1st and 2nd case only or the this case this fourth case will really occur because see here you need to here you must have at least 4 roots right now for the four root for roots the differential equation must be of fourth order and it is not always easy to solve fourth order polynomial equation a polynomial equation of fourth degree so therefore in gate problems you will really encounter these cases but still you should know this for example suppose here if I take the roots of auxiliary equation as 2 plus 3i out two plus three iota two minus three outer two minus three outer these are the four roots of auxiliary equation then how to write the complementary function the complementary function will be e to the power Alpha X then C 1 X plus C 2 cos beta X beta is 3 so cos 3x plus C 3 X plus C 4 sine beta X beta is 3 so sine 3 X so these were the 4 cases now we'll take some examples we'll make one table on one side right the root of news of a Greek person and on another side we'll see how to write the complementary function okay so now we are going to take some cases where the roots are causally equation are known okay I mean any actual problem the roots of our equation you have to calculate but we are going to learn once the roots or falls in equation are found then how to write the complementary function so these six different cases we are going to consider the first case is when the roots of auxiliary equation are 2 & 3 so here the complementary function will be this is our first case when you have we have all the distinct roots there are two roots and these roots are different so here the auxiliary equation will be C 1 here the complementary function will be C 1 e to the power M 1 X plus C 2 e to the power M 2 X now one more thing here I'm writing X because I am assuming that X is the independent variable and Y is the dependent variable so this will be the first case second suppose if the roads are 2 3 & 4 so again you have three roots and all the roots are distinct so here the complementary function will be C 1 e to the power M 1 X C 2 e to the power M 2 X plus C 3 e to the power M 3 X ok M 1 M 2 M 3 R 2 3 & 4 then we have this third case where you have two equal roots and one a distinct root so in this case the complementary function will be for equal roots you have to write one linear factor so C 1 X plus C 2 e to the power MX M is they equal root 2 2 and for this 4 we will get C 3 into e to the power M 3 X so this will be the complementary function for the third case the fourth one is where you have three equal roots and one distinct root so here the complementary function will be corresponding to equal roots we have to write one quadratic factor so C 1 X ax square plus C 2 X plus C 3 e to the power MX M is the equal root 4 plus C 4 e to the power M 4 X M 4 is the fourth root which is 5 then you have these two cases in 50s you have two roots and these roots are complex conjugate okay so this will be this was our third case when you have one pair of imaginary roots which are complex conjugate so here the complementary function will be e to the power Alpha X alpha is 2 so e to the power 2 X then C 1 cos beta X so C 1 cos 3x plus C 2 sine beta X so sine 3 X so this will be the complementary function if the roots of auxiliary equation are 2 plus 3 outer and 2 minus 3 out and finally the last one here the roots of auxiliary equation out of 5 iota minus 5 iota and 6 so here the complimentary function will be alpha is a 0 this you can write as 0 plus 5 iota and this is 0 minus 5 iota so if alpha is 0 then e to the power Alpha X will become 1 so you will simply get C 1 cos beta X beta is 5 plus C 2 sine beta X plus C 3 e to the power M 3 X M 3 is 6 so I am at this table only to explain all the four different cases that we had learned once the roots of auxiliary equation are known then you said you should be in a position to write the complementary function now we'll take some other problems which were asking get various your papers so these are our first few problems the first question is we have to saw solve this differential equation and the you can see that this right hand side is 0 so this means the differential equation is homogeneous differential equation so here we only have to calculate the complimentary function the particular integral will be 0 and the complete solution will be y equals to complementary function so here if we want to solve this differential equation you can see that this differential equation can be as d cube minus 6 d s / + 11 d minus 6y this equals to 0 so this is your f of D now in order to solve this differential equation we have to calculate the complementary function for which we have to solve this auxiliary equation because the roots of auxiliary equation will give us the complimentary function so auxiliary person will be F of M this equals to 0 so f of M equals to 0 this means M cube minus 6 M square plus 11 M minus x equals to 0 now there are a number of ways of solving this cubic equation one approach is that you have to find one of the roots by heat and trial method so if you substitute M equals to 1 here then you can easily see and always remember that you have to substitute the value of M which are a factor of this constant term ok so a factor of constant term six will be plus minus 1 plus minus 2 plus minus 3 and plus minus 6 so you can substitute only one of those numbers and generally at one or two you will get a zero so here you can see that if I substitute M equals to 1 then I will need 2 1 minus 6 plus 11 minus 6 which will be 12 minus 12 is 0 so this means M equals to 1 is one of the root which implies M minus 1 this will be a factor of this cubic polynomial so now you have to find the other quadratic factor so for that either you can divide this cubic by M minus 1 or there is another method which is directly division so here what we do that we try to create M minus 1 common factor so suppose if you have M cube here now forget about the next term what should be our next term so that we can get M minus 1 common in the next step so if I write minus M s were here then in the next step I can take ma square common from these two and I will get M minus 1 so now once you have written minus ma square now see what is the next step the next term is minus 6 M square so this means the remaining will be - 5ms we're now again forget about the next term what should be our next term here so that we can get em - one common so if i write plus 5m here then in the next step from these two i can take minus 5m common and then i will get a minus 1 as one linear factor now see what is the next term the next time is 11 so plus 5m plus 6 M minus x equals to 0 now you can see that how we are going to get the other quadratic factor from these two if I take M s square common I will get M minus 1 from these two if I take minus 5m common I will get M minus 1 and from the last two terms if I take 6 common then I will get M minus 1 equals to 0 now from all these three terms you can take M minus 1 common you will get M s square minus 5m plus 6 equals to 0 now you all know how to solve this quadratic expression quadratic equation so from here the value of M that you are going to get is 1 2 & 3 so once you have obtained the value of M now what should be our complimentary function the complementary since we have three roots and all the three roots are distinct so complementary function will be C 1 e to the power M 1 X C 2 e to the power M 2 X C 3 e to the power M 3 X and this will be the solution of a given differential equation because the particular integral is 0 as it is a homogeneous differential equation right hand side is 0 so this will be the solution of this given differential equation it was a third order differential equation so you have three arbitrary constants C 1 C 2 and C 3 now our next column is this question was asking to get 2018 so the problem is if YT is the solution of this differential equation then we have to calculate the value of y when T is pi and these two boundary conditions are given C there are two type of solution one is called general solution which has this arbitrary constant and other is called particular solution so for which these boundary conditions are given from very very calculated those arbitrary constant so here first we need to find the solution the general solution then using boundary condition we can find the particular solution so in order to solve this differential equation first let me write this differential equation s t - y by DX square plus dy by dt plus a 5 by 4 y this equals to 0 ok remember that here independent variable is T so our solution Y will be in form of T here in dependent variable was X so Y is a function of X now if you want to rewrite this differential equation this will be da square plus D plus a 5 by 4 y this equals to 0 and here this capital T is nothing but a D by DT T is the independent variable so if we have to solve this differential equation this is f of D so first we need to solve the auxiliary equation so all Cele a present will be F of M equals to 0 which means M s square plus M plus Phi by 4 this equals to 0 so from here I will get 4 M s square plus a 4 M plus a 5 equals to 0 so using quadratic formula the value of M will be minus B plus minus root over B Square minus 4 AC so this will be 8 T divided by 2a now you can see that this will be 60 minus 64 so I will get minus 4 plus minus 8 iota by 8 so finally the value of M that you are going to get is minus 1 by 2 plus minus iota this will be the value of M now once you have obtained the roots of auxiliary equation it is very easy to compute the complimentary function and once we know the complimentary function then that complimentary function will be the complete solution because the right-hand side is 0 here so here the complimentary function YC this will be e to the power now this is our third case when you have 1p imaginary route you have minus 1 by 2 + iota this is one route other wrote is minus 1 by 2 - I oughta so these are the two routes so complementary function will be e to the power alpha X C 1 cos beta X plus C 2 sine beta X ok this will be the complementary function and sorry not X here here we will get a T because T was the independent variable so this will be our complementary function and this is nothing but Y T the complete solution as the right-hand side is zero particular integral is zero now there is one very important concept that we have to learn here see the question is we have to calculate Y of Pi so this will be e to the power here we'll get alpha X not X but T here because T was independent variable so Y of Y this will be minus 1 by 2 + 2 pi C 1 cos PI plus c2 sine PI now we know that sine PI will be 0 cos PI will be minus 1 okay so this Y of Pi we need to calculate we need to calculate the value of y when T is PI so when T is PI sine PI will become 0 cos PI will be minus 1 so this will be the value of y of Pi so as you can see that we only need the value of this arbitrary constant C 1 we don't need the value of C 2 although we have two boundary conditions so using these two conditions we can calculate both C 1 and C 2 but we don't need the value of C 2 because we have to find the value of y when T is PI so this means we have to use Molly one of those boundary condition to find the value of C 1 so this is the solution that we have now if I use the first boundary condition first boundary condition is y 0 is 1 so this means when T is 0 the value of y is 1 why I am using this condition because this condition will give us the value of C 1 okay if I place T equals to 0 sine 0 will become 0 and we will directly get the value of C 1 here so this means value of Y is 1 and in place of T we have to substitute zero here so e to the power 0 C 1 costs 0 plus 3/2 sine 0 sine 0 will be 0 cos 0 will be 1 so the value of c1 that you are going to get is 1 okay so once you have obtained the value of C 1 value of y you of Y will be minus e to the power minus PI by 2 if you use the calculator this will be minus 0 point 1 so this is the answer to the given problem ok although the value or another boundary condition was given to calculate the value of C 2 but we don't need to calculate the value of C 2 as the value of y is asked at T equals to PI where CO 2 it will become 0 so this was one problem that was asking GAE 2018 we are going to take some other problems based on similar concept we have two more problems on the concept that we have learned homogeneous linear differential equation of higher order so the first question was asking a gate went in engineering services 2017 another problem is from gate 2018 in mechanical so first we are going to solve this problem here the question is we have to find the general solution of this differential equation as you can see this is fourth order differential equation and four options were given but I have not written the options here so here in order to find general solution since the right hand side is 0 particular integral will be 0 we only need to calculate the complimentary function so in order to calculate the complimentary function we have to solve the auxiliary question one more thing before writing auxiliary equation we can write this differential equation as D to the power of 4 minus 2 D cube plus 2 D square minus 2 D plus 1 why this equals to 0 okay each one of these term will operate individually on Y so D to the power 4 Y will give us this term minus 2 D cube Y will give us this term and so on so now we are only in a position to write the auxiliary equation auxiliary equation is F of M equals to 0 we are this expression is F of D so here if I place D buy em then I will get em to the power four minus two M cube minus two M s square sorry plus 2 m square minus 2 M plus one equal to zero now again there are a number of ways of solving this by quadratic equation it is a it is a fourth it is a polynomial equation of fourth degree so it is called by quadratic equation so in order to solve this by quadratic equation one approach will be calculate one linear factor another approach will be that we can write this as minus two M cube M four minus two M cube plus we can break this 2m s square as M s square plus M s square minus two M plus one now you will see why I am doing this from these first three terms if you take M square common then you will get M s square minus 2 m plus 1 plus M s square minus 2 m plus 1 this equals to 0 now this is nothing but M minus 1 whole square and if you from if you take M minus 1 whole square common then you will get M s square plus 1 equals to 0 and this will give us M s 1 1 and if you equate this to 0 you will get minus iota and plus iota so these are the four roots that you are going to get now once you have obtained the root you can easily write the complementary function for the two equal roots you will get C 1 X plus C 2 T to the power MX plus 4 this these two imaginary roots which have real part as 0 so e to the power Alpha X will become 1 you will simply get C 3 cos beta X plus C 4 sine beta X so this will be the solution of given differential equation ok so this was very simple motion you have to solve our a equation once you have obtained auxiliar roots of auxiliary equation then you can write the complementary function and the complementary function will be the complete solution as the right hand side was 0 here which means particular integral zero here so this was our first problem then we have one last problem from this concept this was asking gay 2018 in mechanical the problem has this differential equation is given and we have to first find the solution of the differential equation two boundary conditions are given to find the value of arbitrary constant and hence calculate the particular solution and then you have to calculate the value of y when x is 1 so first let me write the differential equation d square plus d minus 6y this equals to 0 this is the differential equation that we have so here if I want to write the auxiliary equation this will be F of M equals to 0 which means M s square plus M minus 6 equals to 0 M plus 3 into M minus 2 this equals to 0 which means the value of M is minus 3 and plus 2 these are the two values of M that you are going to get so therefore here the solution will be solution will be same as the complimentary function because particular integral will be 0 so solution will be C 1 e to the power M 1 X C 2 e to the power M 2 X this will be the solution of given differential equation now obviously they say general solution because you have these two arbitrary constant we need to calculate the particular solution using these boundary condition so it has given that Y is 0 this is 0 so this will mean C 1 plus C this is say 2 not C 3 so C 1 plus C 2 this will be 0 this is your first equation the second equation is wide and 0 is 1 so first we need to calculate a wide ass so why does will be 2 C 1 e to the power 2 X minus 3 C 2 e to the power minus 3 X now why does 0 why does 0 is 1 so this means 2 C 1 minus 3 C 2 this will be 1 so this is your second equation now if you solve these two equations the first equation is C 1 plus C 2 is 0 the second equation is 2 C 1 minus 3 C 2 is 1 so if you solve these two equations then you can easily see that the value of C 1 will be 1 by 5 and C 2 will be minus 1 by 5 these are the values of C 1 and C 2 C 1 is 1 by 5 and C 2 is minus 1 by 5 so therefore here your solution will be y equals to C 1 e to the power M 1 X here is the solution so in place of C 1 we have to write 1 by 5 e to the power 2 X minus 1 by 5 e to the power minus 3 X so this is no particular solution ok this is only one function whereas this is family of function for different value of C 1 and C 2 so now the question was what is the value of y of 1 so Y of 1 will be in place of X we have to write 1 so this will be e s square minus e to the power minus 3 by 5 if you do the calculation this will be 1 point 4 7 so the answer to our first problem is this was our solution for the second problem the answer will be 1 point 4 7 so now after this we'll talk about how to calculate particular integral and to solve non-homogeneous differential equations