 Namaste, Myself, Mr. Biraj Dar Bala Sahib, Assistant Professor, Department of Humanities and Science, Walchand Institute of Technology, Solapur. In this session, we will discuss linear differential equation of higher order with constant coefficients part 4. Learning outcome. At the end of this session, students will be able to write particular integral of higher order linear differential equation f of d into y is equal to x of x when x of x is equal to e to power a into x. Let us pause the video for a while and write answer to the question. Question is write complementary function of differential equation d minus 2 bracket cube into y equal to 0. Come back. I hope you have written answer to this question. Let us see the solution. This is the differential equation of the type f of d into y equal to 0. Here f of d is equal to d minus 2 bracket cube. Now, to get auxiliary equation, equate this f of d equal to 0. Therefore, we get d minus 2 bracket cube equal to 0. That is, we can write as d minus 2 into d minus 2 into d minus 2 is equal to 0, which gives d minus 2 equal to 0, d minus 2 equal to 0 and d minus 2 equal to 0. From this, we get d equal to 2, d equal to 2, d equal to 2. Here root 2 is real and repeated 3 times. And we know that when 3 roots are equal and real, m 1 equal to m 2 equal to m 3, then the complementary function c f can be written as in bracket c 1 plus c 2 into x plus c 3 into x square bracket close e 2 power m 1 into x. Here root is 2, which is repeated 3 times. Therefore, the complementary function c f equal to in bracket c 1 plus c 2 into x plus c 3 into x square bracket close e 2 power 2 into x, where x is independent variable. So, this is the required answer to the given differential equation. Now, let us start with a particular integral. As we know that the general solution of linear differential equation f of d into y equal to x of x is y equal to c f plus p i. C f means complementary function and we discussed this part in previous session. Now, today we can see how to find a particular integral for some standard functions. Particular integral is a part of solution of linear differential equation f of d into y equal to x of x. Here f of d is function of differential operator capital D, so that 1 upon f of d is known as inverse differential operator. Since f of d into y equal to x of x to remove f of d from the left hand side operate 1 upon f of d on both the side. Therefore, 1 upon f of d in bracket f of d into y is equal to 1 upon f of d of x of x. In the left hand side f of d gets cancelled, hence y equal to 1 upon f of d operating on x of x. The value of this y is nothing but a particular integral and it is denoted by p of y or y suffix p. Therefore, p i is equal to y suffix p is equal to 1 upon f of d operating on x of x. So, this is the formula we are using to obtaining particular integral for the given differential equation. Note that 0.1 meaning of 1 upon d operating on x of x. As we know that d means differential coefficient, so that 1 by d means inverse of differential coefficient that means integration. Therefore, 1 by d of x of x is equal to integration of x of x with respect to x. 0.2 meaning of 1 upon d minus a operating on x of x. It can be simplified as 1 upon d minus a of x of x is equal to e to power a x into integration e to power minus a x into x of x into d x. So, these are the two fundamental results are essential to find particular integral in further. Now rules to write particular integral. Rules to find particular integral depends on a nature of right hand side function x of x. Following are the standard types of functions x of x e raise to x e raise to x cos a x or sin a x x raise to m e raise to x into v and x into v. But in this session, how to find a particular integral when the function is e to power a x is discussing. Case 1, when x of x equal to e to power a x. As we know that p i equal to 1 upon f of d operating on x of x. Therefore, p i equal to 1 upon f of d, but x of x is e to power a x. Now to simplify this, we put every d equal to a in the denominator f of d. So, we get p i equal to 1 upon f of a into e to power a x if f of a not equal to 0. That means when the given function x of x is equal to an exponential function of the type e to power a x. Then to find particular integral, we put every d equal to constant coefficient a in the denominator f of d provided f of a not equal to 0. Now, case of failure. If suppose f of a equal to 0, then above step fails. In this case numerator by independent variable that is here x and differentiate denominator f of d with respect to d, we get f dash of d and then write as p i equal to x into 1 upon f dash of d and which is operating on e to power a x. That means I can put every d equal to a in the denominator f dash of d. We get p i equal to x into 1 upon f dash of a into e to power a x. This is the p i if f dash of a not equal to 0. If suppose again f dash of a equal to 0, then above step fails. Again repeat the above procedure and write as p i equal to 1 more x we have to multiply numerator. Hence x into x is x square into 1 upon differentiate f dash of d with respect to d. We get f double dash of d and into e to power a x. Now, to simplify write and said again put d equal to a in the denominator f double dash of d. We get p i equal to x square into 1 upon f double dash of a and into f double dash e to power a x. This is the particular integral if again f double dash of a not equal to 0 and so on. We may get answer in first step or in second step or may be in third step or may be in fourth step. We have to repeat this procedure till to get a answer. Note that when x of x is constant, then p i equal to 1 upon f of d operating on k which is equal to 1 upon f of d. k we can write as k into e raise to 0 x which is equal to k we can take outside the operator and to operate 1 upon f of d on e raise to 0 x. We put d equal to 0 by case 1. Hence we get p i equal to k into 1 upon f of 0 into e raise to 0 x if f of 0 not equal to 0. So, this is we can write as p i equal to 1 upon f of d operating on k is nothing but write k as it is simply put d equal to 0 in the denominator. We get k upon f of 0 if f of 0 not equal to 0. That means for function x of x which is equal to constant, then to find particular integral put every d equal to 0 in f of d provided f of 0 not equal to 0. Now, let us consider one example. So, all the differential equation d square minus 2 d plus 1 into y is equal to e raise to 3 x plus 8. This differential equation is belongs to type f of d into y equal to x of x. So, here f of d equal to d square minus 2 d plus 1 and x of x equal to e raise to 3 x plus 8. Now, to find complementary function we write auxiliary equation by equating f of d equal to 0. Therefore, d square minus 2 d plus 1 equal to 0 is auxiliary equation. Now, solve it by factorizing its factors are d minus 1 into d minus 1 which is equal to 0 which gives d minus 1 equal to 0 and d minus 1 equal to 0. Therefore, d equal to 1 and d equal to 1. Roots are real and equal. Therefore, by formula when 2 roots are equal m 1 equal to m 2 c f equal to c 1 plus c 2 into x bracket close e raise to m 1 x. Using this formula we can write c f equal to in bracket c 1 plus c 2 into x bracket close e to power 1 x that is e raise to x. This is the complementary function. Now, to find particular integral we have the formula p i equal to 1 upon f of d operating on x of x. Now, put f of d value and x of x here we get p i equal to 1 upon d square minus 2 d plus 1 operating on e raise to 3 x plus 8. Now, separate into two terms in the right hand side. Hence, we get p i equal to 1 upon d square minus 2 d plus 1 operating on e raise to 3 x plus 1 upon d square minus 2 d plus 1 operating on 8. Now, using case 1 when the function is of type e raise to a x put d equal to a and when the function is a constant then put d equal to 0. Therefore, in first term function is e raise to 3 x a is 3 that is why put d equal to 3 and in the second term the function is 8 which is constant. Therefore, we put d equal to 0 hence p i equal to 1 upon 3 square minus 2 into 3 plus 1 into e raise to 3 x plus 1 upon 0 square minus 0 plus 1 into 8. After simplification we get p i equal to 1 upon 4 into e raise to 3 x plus 8 by 1 that is 8. Hence, general solution is dependent variable here it is y is equal to c a plus p i. Therefore, y equal to c f is c 1 plus c 2 into x bracket close into e raise to x plus p i is 1 upon 4 into e raise to 3 x plus 8. So, this is the required general solution for the given equation. To prepare this video session I refer this book as a reference. Thank you.