 Namaste. Myself, Mr. Viraj Dhar Bala Sahib, Assistant Professor, Department of Humanities and Science, Walchand Institute of Technology, Solaapur. In this session, we will discuss linear differential equation of higher order with constant coefficients part 2, Learning Outcome. At the end of this session, students will be able to write complementary function of higher order linear differential equations. Let us pause the video for a while and write the answer to the question. Question is write an auxiliary equation of the differential equation d raise to 4 y upon dx raise to 4 plus 8 d square y by dx square plus 16 y is equal to sin x. Come back, I hope you return answer to this question. Let us see the solution. Here a differential coefficient d by dx equal to capital D, so that this raise to 4 y by dx raise to 4 can be written as d raise to 4 of y, d square y by dx square can be written as d square y. With these two notation, the given equation reduces to d raise to 4 of y plus 8 d square y plus 16 y equal to sin x and y is taken common in the left hand side. Therefore, in bracket d raise to 4 plus 8 d square plus 16 bracket close into y equal to sin x. This is now in the form of f of d into y equal to x of x. Here f of d is d raise to 4 plus 8 d square plus 16 and when we equate f of d equal to 0, we get a auxiliary equation that is here d raise to 4 plus 8 d square plus 16 is equal to 0 is auxiliary equation for the given differential equation. Let us start with complementary function. In short, we can write C f. The definition, it is the solution of homogeneous differential equation f of d into y equal to 0. It depends on nature of roots of an auxiliary equation f of d equal to 0. It contains an arbitrary constants like C 1, C 2, C 3, etcetera. Number of such arbitrary constants equal to order of differential equation. Here f are two types of the roots namely real roots and imaginary roots. We have f of d equal to 0 auxiliary equation, but for nth order differential equation f of d is a naught d raise to n plus a 1 d raise to n minus 1 plus a 2 d raise to n minus 2 plus dash dash dash plus a n minus 1 into d plus a n equal to 0. To solve this equation, we have to factorize the left hand side. Here power of d is n, there are n factors are existing. Suppose, d minus m 1 into d minus m 2 into d minus m 3 and so on into d minus m n are n factors of the left hand side polynomial and it is equal to 0. When product of factor equal to 0, then equate each factor equal to 0, we get values of d. Therefore, d minus m 1 equal to 0 gives d equal to m 1, d minus m 2 equal to 0 gives d equal to m 2, d minus m 3 equal to 0 gives d equal to m 3 and so on. d minus m n equal to 0 gives d equal to m n, where m 1, m 2, m 3, m 4 and so on m n are constant. So, each value of d's are called as roots of an auxiliary equation. Now, rules to write complementary function, complementary function depends on nature of roots and we know that there are two type of the roots real roots and imaginary roots. Now, first of all see complementary function for real roots case 1, if all roots are real and distinct, then how to write a complementary function? Let us suppose m 1, m 2, m 3, m 4, etcetera m n, these n roots are real and distinct then while writing complementary function, the number of arbitrary constant c 1, c 2, etcetera must be equal to number of roots or order of differential equation. Here there are n roots, hence we choose n arbitrary constant c 1, c 2, etcetera c n and then write complementary function as you can take any root as a first root and any arbitrary constant here I can choose c 1 is the arbitrary constant for m 1 root, so that c 1 e raise to m 1 into independent variable x plus c 2 e raise to m 2 into x plus c 3 e raise to m 3 into x plus and so on plus c n into e raise to m n into x. So, this is the rule to write complementary function when roots are real and distinct. Now, case 2, sometimes the roots may be real and repeated. Suppose first point if suppose auxiliary equation f of d equal to 0 has the root m 1 repeated twice that means 2 roots m 1 and m 2 are real and equal that is m 1 equal to m 2 then how to write a complementary function. Here number of roots are 2, hence we choose 2 constant c 1 and c 2 when the roots are equal then we arrange c 1 and c 2 in a bracket in the form that is c f equal to in bracket c 1 into independent variable x plus c 2 bracket close e to power m 1 x because m 1 and m 2 are same. Or you can arrange that c 1 and c 2 as also c f equal to c 1 plus c 2 into x bracket close e to power m 1 x both are correct. Now, point b if 3 roots m 1 m 2 m 3 are real and equal that is m 1 equal to m 2 equal to m 3 then 3 constant c 1 c 2 c 3 can be arranged in the bracket as c f equal to in bracket c 1 plus c 2 into independent variable x plus c 3 into x square bracket close and e to power m 1 x and so on. Now complementary function for imaginary root as we know that imaginary roots are always in a pair that is of the type alpha plus or minus i beta where alpha is real part it may be positive negative or 0 and beta is imaginary part it is always positive. Now rules to write c f case 1 one pair of imaginary root that is if alpha plus or minus i beta be a pair of imaginary root for the auxiliary equation then in this pair there are two complex root alpha plus i beta and alpha minus i beta. So, that two constant c 1 and c 2 has taken therefore, c f equal to c 1 e raise to alpha plus i beta into x plus c 2 e raise to alpha minus i beta x. Now we have to simplify it to get in standard form in the right hand side e raise to alpha into x is common taken outside and in bracket c 1 e raise to i beta x plus c 2 e raise to minus i beta x. Now we have to express e raise to i beta x and e raise to minus i beta x in terms of cos and sin using Euler's formula e raise to i theta equal to cos theta plus i sin theta that is why c f equal to e raise to alpha x in bracket c 1 into bracket e raise to i beta x can be written as cos beta x plus i sin beta x bracket close plus c 2 in bracket e raise to minus i beta x can be written as cos beta x minus i sin beta x bracket close and big bracket close. This is due to Euler's formula e raise to plus or minus i theta can be written as cos theta plus or minus i sin theta. Now we have to simplify right hand side c f equal to e raise to alpha x as it is and in bracket collect in the coefficient of cos beta x we get in bracket c 1 plus c 2 into cos beta x plus now collecting the coefficient of sin beta x we get in bracket i c 1 minus i c 2 bracket close into sin beta x and then curly bracket close. Since c 1 c 2 are constant and i is imaginary constant so that c 1 plus c 2 is another constant denoted by a and i c 1 minus i c 2 is constant and it is denoted by b so that c f equal to e raise to alpha x and in bracket e into cos beta x plus b into sin beta x. Hence for one pair d equal to alpha plus or minus i beta of imaginary root the complementary function can be written in the form c f equal to e to power alpha x in bracket a cos beta x plus b sin beta x where a and b are arbitrary constant and treat this as a formula for c f for one pair of imaginary root. Now case 2 repeated pairs of imaginary root if alpha plus or minus i beta comma alpha plus or minus i beta are 2 equal pairs of imaginary roots for auxiliary equation then totally there are 4 roots hence we need 4 constant c 1 c 2 c 3 and c 4 but these 2 pairs are equal hence to write c f c f equal to e raise to alpha x in bracket now 2 constant we have to arrange in the coefficient of cos of beta x that is in bracket c 1 plus c 2 into x and bracket close into cos beta x plus remaining 2 constant c 3 and c 4 can be arranged in the coefficient of sin beta x that is in bracket c 3 plus c 4 into x bracket close into sin beta x and then curly bracket close. So, this is the complementary function for 2 equal pairs and so on. Now case 3 if 2 distinct pairs of imaginary root means if alpha plus or minus i beta and c plus or minus i d are 2 distinct pairs of imaginary roots then while writing complementary function write c f for first pair alpha plus or minus i beta as e raise to alpha x in bracket c 1 cos beta x plus c 2 sin beta x bracket close and plus c f for second pair c plus or minus i d as e raise to c into x and in bracket c 3 cos dx plus c 4 sin dx. So, this is the way to write complementary function for imaginary pairs. Now let us consider one example find complementary function of a differential equation 2 into d square y by dx square plus d y by dx minus 6 y equal to 0 here d by dx is capital D. So, that d y by dx equal to d y and d square y by dx square equal to d square y with these notations the given equation reduces to 2 d square into y plus d y minus 6 y equal to 0 and it is I can write as 2 d square plus d minus 6 it is in bracket and into y which is equal to 0. Now it is of the type f of d into y equal to x of x here f of d is the 2 d square plus d minus 6 to get auxiliary equation equate f of d equal to 0. So, we get 2 d square plus d minus 6 equal to 0. Now we have to solve this equation for solving we factorize it. So, that 2 d square plus d minus 6 equal to 0 can be written as 2 d square plus 4 d minus 3 d minus 6 equal to 0 because 2 into minus 6 is minus 12 and factors of minus 12 is a 4 into minus 3 sum of 4 and minus 3 is a plus d or plus 1 d and this is I can write 2 d I will take common from first 2 term and in bracket d plus 2 then minus 3 is common from last 2 term in bracket d plus 2 which is equal to 0 then d plus 2 is common in bracket 2 d minus 3 which is equal to 0 and 2 d minus 3 equal to 0 d plus 2 equal to 0 which gives d equal to 3 by 2 and d equal to minus 2 these are the 2 roots which are real and distinct. Hence let us suppose d equal to m 1 it is 3 by 2 and d equal to m 2 which is minus 2 and using case 1 of C f the C f is C 1 e raise to m 1 x plus C 2 e raise to m 2 x because 2 roots are there 2 constant C 1 and C 2 we have to take. Hence C f equal to C 1 e raise to 3 by 2 into x plus C 2 into e raise to minus 2 x is the complementary function. To prepare this video I refer this book.