 Namaste, Master, Mr. Biraj Dar Bala Saheb, Assistant Professor, Department of Humanities and Sciences, Walchand Institute of Technology, Solapur. In this session, we will discuss Linear Differential Equation of Higher Order with Constant Quipetients, Part 3, Learning Outcome. At the end of this session, students will be able to write complementary function of higher order linear differential equations and the solution of linear differential equation with constant coefficient of type f of d into y is equal to 0. Let us pause the video for a while and write answer to the question. Question is write the factors of standard polynomials a cube minus b cube, a cube plus b cube and a raise to 4 minus b raise to 4. Come back, I hope you return answer to this question. Let us see the solution. The factors of a cube minus b cube are a minus b into bracket a square plus a b plus b square. The factors of a cube plus b cube are a plus b into a square minus a b plus b square. The factors of a raise to 4 minus b raise to 4 are initially a square minus b square into a square plus b square. And then the factors of a square minus b square are a minus b into a plus b. Similarly, the factors of a square plus b square are a minus i b into a plus i b. So, these are the three standard polynomials with their factors, which we learned in lower classes. But these are now essential to find the roots of auxiliary equation for some examples. Now, let us start with example, because in previous session we discussed how to write a complementary function for the given linear differential equation. Now, example one solve the differential equation d cube plus d square minus 2 d bracket close into y is equal to 0, where d equal to d by d x. In this differential equation, y is a dependent variable and capital d equal to d by d x indicates that x is a independent variable. Now, solution of this differential equation is y equal to c f, because this differential equation is the type f of d into y equal to 0. It is a homogeneous linear differential equation and it is a solution containing only complementary function. Therefore, y equal to c f is it is a solution. Here f of d is d cube plus d square minus 2 d and to get auxiliary equation equate this f of d equal to 0. Therefore, we get d cube plus d square minus 2 d equal to 0 is auxiliary equation. Now, we have to solve this equation. In the left hand side, d is common taken out and in bracket d square plus d minus 2 equal to 0. Now, keep d as it is and again the factors of d square plus d minus 2 are d minus 1 into d plus 2 and it is equal to 0. Now, product of three factor equal to 0 implies d equal to 0, d minus 1 equal to 0, d plus 2 equal to 0. Therefore, d equal to 0 is one root and d minus 1 equal to 0 gives d equal to 1 and d plus 2 equal to 0 gives d equal to minus 2. Therefore, d equal to 0, d equal to 1, d equal to minus 2 are the roots of auxiliary equation. These roots are real and distinct roots. Let us suppose d equal to m 1 which is 0, d equal to m 2 which is equal to 1 and d equal to m 3 which is equal to minus 2. Now, by case one of a complementary function for real root, if the roots are real and distinct then we can write complementary function as here three roots, three constants we have to take c 1 e raise to m 1 x plus c 2 e raise to m 2 x plus c 3 e raise to m 3 x. x is a independent variable in the given equation only put the value of m 1, m 2 and m 3 in this formula, we get c f equal to c 1 e raise to 0 into x plus c 2 e raise to 1 x plus c 3 e raise to minus 2 x, but e raise to 0 into x is 1. Therefore, c f equal to c 1 plus c 2 into e raise to x plus c 3 into e raise to minus 2 x. Now, general solution of the given differential equation dependent variable that is here y which is equal to complementary function. Therefore, y equal to c 1 plus c 2 into e raise to x plus c 3 into e raise to minus 2 x is the required solution for the equation. Now, let us consider another example, example to solve the differential equation in bracket d square plus 6 d plus 9 bracket close into y equal to 0. In this equation y is dependent, but d is not mentioned then you can choose any variable as independent variable except y. In this equation f of d is d square plus 6 d plus 9 and equate this equal to 0, we get d square plus 6 d plus 9 equal to 0 which is called as auxiliary equation. Now, solve it by factorizing the left hand side, the factors of left hand side are d plus 3 d plus 3 that is why d plus 3 into d plus 3 equal to 0. Then separate into two equation d plus 3 equal to 0 d plus 3 equal to 0 which gives d equal to minus 3 d equal to minus 3. Here minus 3 is a real number repeated twice so that we can say roots are real and repeated twice that is here two roots m 1 which is equal to m 2 each equal to minus 3. And by using case 2 that is c f for repeated root if two roots are equal m 1 equal to m 2 then we know that two constant should be arranged as c 1 plus c 2 into x in complementary function. Therefore, the formula is c f equal to in bracket c 1 plus c 2 into x where x is independent variable chosen here bracket close e to power m 1 into x, but here m 1 and m 2 both are same that is minus 3. Therefore, c f equal to c 1 plus c 2 into x which is in bracket and into e raise to minus 3 x. So, this is the complementary function for the differential equation. Now general solution again it is a homogeneous equation so that c f will be it is a solution. Therefore, solution is y equal to c f which is equal to c 1 plus c 2 into x bracket close and into e raise to minus 3 x is the required solution. Now consider another example, example number 3. So, all the differential equation d raise to 5 y upon d t raise to 5 plus 8 into d square y by d t square which is equal to 0. In this differential equation y is a dependent variable, but t is a independent variable that is why while writing complementary function we have to use variable t. Now this differential equation is not in standard form or not in symbolic form. Hence first we write in symbolic form by replacing d by d t equal to capital D. So we have d raise to 5 y by d t raise to 5 can be written as d raise to 5 of y plus 8 into d square y by d t square can be written as d square of y which is equal to 0. Now in left side take y common hence in bracket d raise to 5 plus 8 d square bracket close into y equal to 0. Now it is of the type f of d into y equal to 0 here f of d is d raise to 5 plus 8 d square and to get auxiliary equation put f of d equal to 0 that is d raise to 5 plus 8 d square equal to 0 which is auxiliary equation. Now solve it by factorizing left side. In left side d square is common taken outside and in bracket d q plus 8 bracket close equal to 0. This is we can write as d into d d square we can write as d into d and in bracket d q plus 8 which is equal to 0. Product of 3 factor equal to 0 implies d equal to 0 d equal to 0 and d q plus 8 equal to 0. Now we have to solve d q plus 8 equal to 0 separately we will get another 3 roots. This d q plus 8 equal to 0 can be written as d q plus 2 cube equal to 0 and now d q plus 2 cube is of type a q plus b cube its factors are a plus b into bracket a square minus a b plus b square. Here a means d and b means 2 therefore its factors are d plus 2 into d square minus 2 d plus 4 bracket close equal to 0. Now separate into 2 equations d plus 2 equal to 0 d square minus 2 d plus 4 equal to 0 that gives d equal to minus 2 first equation and from second equation we can find the value of d using quadratic formula that is x equal to minus b plus or minus under root b square minus 4 ac upon 2 a it is a formula to find the root of quadratic equation. Using this formula here a is 1 b minus 2 c 4 hence d equal to minus of minus 2 plus or minus under root minus 2 its square minus 4 into 1 into 4 upon 2 into 1. Therefore d equal to minus 2 kept as it is and d equal to 2 plus or minus under root minus 12 whole divided by 2 that is equal to 2 plus or minus root of minus 12 can be written as 2 root 3 into i because root of minus sign is i and root of 12 we can write as 2 into root 3 whole divided by 2. Therefore d equal to minus 2 as it is and d equal to here 2 we can divide for the 2 terms of numerator we get 2 by 2 is 1 plus or minus 2 root 3 i upon 2 gives root 3 i. Hence total roots are d equal to 0 d equal to 0 d equal to minus 2 and d equal to 1 plus or minus root 3 i. Here root 0 is real repeated twice root minus 2 is real and distinct and root 1 plus or minus i into root 3 which is alpha plus or minus i beta type is a pair of imaginary root with alpha equal to 1 and beta equal to root 3. So this is the nature of roots according to this nature we have to write complementary function using independent variable t. Therefore C f equal to first I am writing C f for 2 equal root 0 0 2 constant I can arrange C 1 and C 2 as C 1 plus C 2 into independent variable t bracket close into e to power 0 into t now plus for root minus 2 C 3 constant I have taken C 3 into e raise to minus 2 t plus for root 1 plus or minus i root 3 using formula for 1 pair of imaginary root e raise to alpha t but alpha is 1. So that e raise to alpha t becomes e raise to 1 t and into bracket constant C 4 cos beta t but beta is root 3 therefore cos of root 3 into t plus C 5 into sign of root 3 into t curly bracket close. But e raise to 0 into t is 1 so that C f equal to C 1 plus C 2 into t plus C 3 into e raise to minus 2 t plus e raise to t in bracket C 4 cos root 3 t plus C 5 sin root 3 t curly bracket close this is the complementary function for the given differential equation. Now it is a general solution is again dependent variable that is here y equal to C f therefore y equal to C f is C 1 plus C 2 into t plus C 3 into e raise to minus 2 t plus e raise to t in bracket C 4 into cos of root 3 t plus C 5 into sin of root 3 t curly bracket close. So this is the required solution for the given differential equation. To prepare this session I refer this book as references thank you.