 Namaste, Myself, Mr. Birajdar Bala Saheb, Assistant Professor, Department of Humanities and Sciences, Walchand Institute of Technology, Sholapur. In this session, we will discuss linear differential equation of higher order with constant coefficients part 5. Learning outcome. At the end of this session, students will be able to solve higher order linear differential equation of the type f of d into y equal to x of x when x of x not equal to 0. Let us pause the video for a while and write answer to the question. Question is write particular integral of differential equation in bracket d minus 5 bracket square into y is equal to e raised to 2 x. Come back, I hope you have written answer to the question. Let us see the solution. The given differential equation is of the type f of d into y equal to x of x here f of d is a d minus 5 bracket square and x of x is equal to e raised to 2 x. We know that formula for particular integral is p i equal to 1 upon f of d operating on x of x. Therefore, p i equal to 1 upon f of d is in bracket d minus 5 bracket square and into x of x is e raised to 2 x. The function is exponential function e raised to a x type therefore, by case 1 to find p i put every d equal to e in the denominator. Here e is a 2 coefficient of x therefore, put d equal to 2 in the denominator. Therefore, p i equal to 1 upon in bracket 2 minus 5 bracket square into e raised to 2 x. Therefore, p i equal to 1 upon 2 minus 5 is minus 300 square is 9 therefore, 1 upon 9 into e raised to 2 x is the required particular integral of the differential equation. Now, let us start with the case 2. When the right hand side function x of x is equal to e raised to x then how to find a particular integral? 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 operating on a raised to x and that is we can write as 1 upon f of d operating on e to power x into log a because a raised to x we can write as e raised to log of a raised to x that is we can write as e to power x into log a. Here coefficient of x is log a which is a constant so that it is belongs to case 1 hence here coefficient of x is log a that is why put d equal to log a in the denominator. Therefore, p i equal to 1 upon f of d operating on a raised to x is equal to 1 upon f of log a into a raised to x. It means that for function x of x equal to a raised to x then to find particular integral we put every d equal to log a in the denominator f of d. Note that when x of x is a hyperbolic cos a x or hyperbolic sin a x first express these functions in terms of a exponential form by definition of hyperbolic function and then use case 1. So, hyperbolic cos a x we can write e raised to x plus e raised to minus a x upon 2 hyperbolic sin a x we can write as e raised to x minus e raised to minus a x upon 2. Now, let us see example 1 solve the differential equation d q plus 4 d into y equal to 2 raised to x. This differential equation is in the form of f of d into y equal to x of x here f of d is d q plus 4 d and x of x is equal to 2 raised to x. Now, first we have to find complementary function for that we write the auxiliary equation by equating f of d equal to 0. Therefore, d q plus 4 d equal to 0 is auxiliary equation now we have to solve it. d is common taken outside. So, that in bracket d square plus 4 which is equal to 0 this is we can write d equal to 0 and d square plus 4 equal to 0. Now, we have to solve d square plus 4 equal to 0 separately that is I can write d square plus 2 square equal to 0, but the factors of d square plus 2 square are d minus 2 i into d plus 2 i equal to 0. Hence, we get d equal to 0 as it is, but d minus 2 i equal to 0 gives d equal to 2 i and d plus 2 i equal to 0 gives d equal to minus 2 i. Here 2 i and minus 2 i are complex roots conjugate to each other write them together, but real part is absent assume it as 0. Therefore, d equal to 0 as it is and d equal to 0 plus or minus 2 i and it is of the type alpha plus or minus i beta. So, one root is real and distinct and another two roots are imaginary roots. Hence, complementary function C f equal to for d equal to 0 C 1 constant and e raise to 0 into x for pair 0 plus or minus 2 i alpha 0 beta is 2. So, that using formula C f equal to e raise to alpha x in bracket C 1 cos beta x plus C 2 sin beta x, but already C 1 is taken then take C 2 and C 3 while writing C f for the pair 0 plus 2 i that is plus e raise to 0 into x and in bracket C 2 into cos of beta x, but beta is 2 therefore, cos 2 x plus C 3 into sin 2 x, but e raise to 0 into x is 1. Therefore, C f equal to C 1 plus C 2 into cos 2 x plus C 3 into sin 2 x now particular integral p i equal to 1 upon f of d operating on x of x. Therefore, p i equal to 1 upon d q plus 4 d operating on 2 raise to x. Now, function 2 raise to x is of the type a raise to x therefore, when the function is a raise to x type then put d equal to log 2, but here e raise to show that put d equal to log 2 hence p i equal to 1 upon log 2 bracket cube plus 4 into log 2 into 2 raise to x. Therefore, the general solution is y equal to C f plus p i hence y equal to C f is C 1 plus C 2 into cos 2 x plus C 3 into sin 2 x and plus p i is 1 upon log 2 bracket cube plus 4 into log 2 into 2 raise to x it is the general solution of the given differential equation. Now, let us consider example 2. So, all the differential equation d square minus 1 into y equal to hyperbolic cos of 2 x again it is of the type f of d into y is equal to x of x here f of d is a d square minus 1 and x of x is hyperbolic cos 2 x. Now, for C f we write auxiliary equation that is equate f of d equal to 0 that is d square minus 1 equal to 0. Now, factorize the left hand side factors are d minus 1 into d plus 1 equal to 0 that gives d minus 1 equal to 0 and d plus 1 equal to 0 from this we get d equal to 1 and d equal to minus 1. Roots are real and distinct hence complementary function C f equal to C 1 into e raise to 1 x that is e raise to x plus C 2 into e raise to minus 1 x that is e raise to minus x this is the required complementary function. Now, for p i formula is p i equal to 1 upon f of d operating on x of x therefore, p i equal to 1 upon f of d is d square minus 1 and x of x is hyperbolic cos 2 x. Now, we have to express that hyperbolic cos 2 x in terms of exponential form by using definition of hyperbolic function. Therefore, p i equal to 1 upon d square minus 1 as it is and hyperbolic cos 2 x you can write as e raise to 2 x plus e raise to minus 2 x upon 2. Therefore, p i equal to 1 by 2 constant coefficient taken outside and in bracket 1 upon d square minus 1 operating on e raise to 2 x and plus 1 upon d square minus 1 operating on e raise to minus 2 x. Both the function are belongs to the type e to power x. So, that put every d equal to a this is by case 1, but in first term e raise to 2 x that is why e is 2 therefore, d equal to 2 and in second term e is minus 2. So, that put d equal to minus 2 hence p i equal to 1 by 2 in bracket 1 upon 2 square minus 1 into e raise to 2 x plus 1 upon minus 2 it is a square minus 1 into e raise to minus 2 x. So, after simplification we get p i equal to 1 upon 2 in bracket 1 by 3 into e raise to 2 x plus 1 by 3 into e raise to minus 2 x. Now, 1 by 3 is common taken outside hence p i equal to 1 by 6 into bracket e raise to 2 x plus e raise to minus 2 x, but hyperbolic cos of a x is equal to e raise to x plus e raise to minus a x upon 2. Therefore, using this formula e raise to 2 x plus e raise to minus 2 x you can write as p i equal to 1 by 6 in bracket 2 into a hyperbolic cos 2 x and 2 by 6 is 1 by 3 therefore, p i equal to 1 by 3 into a hyperbolic cos 2 x. Now, general solution y equal to c i plus p i therefore, y equal to c f is c 1 e raise to x plus e raise to minus x and plus p i is 1 by 3 hyperbolic cos 2 x. To prepare this session I use this book as a references. Thank you.