 Myself, Mr. Birajdar 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 Coefficient Part 1. Learning Outcomes At the end of this session, students will be able to, first one, define linear differential equation of higher order, second, write higher order linear differential equations in short form using differential operator capital D. Let us pause the video for a while and write answer to the question. Question is, write an order and degree of differential equation d square y by dx square minus 5 into dy by dx square plus 6 into y is equal to e raised to x. Come back, I hope you return answer to this question. Let us see the solution. Here, the highest order derivative in the equation is 2, therefore order of this differential equation is 2. And the power of highest order derivative is 1 in the equation so that the degree of differential equation is 1. This is according to the definition of order and degree of differential equation which we learn in lower classes. Now, let us start with linear differential equation. It is denoted by LDE shortly. Definition. A differential equation is said to be linear differential equation if the degree of dependent variable and its derivatives is 1 and they are not multiplied each other in the equation. If coefficient of derivative terms and the coefficient of dependent variable are constants, then the differential equation is called linear differential equation with constant coefficient. Let us consider an example 1. The differential equation d square y by dx square plus 3 into dy by dx plus 2 into y equal to e raised to x. In this differential equation is y is a dependent and x is independent variable. Here a degree of y and its derivatives are dy by dx and d square y by dx square. The degree of these three is 1 and they are not multiplied each other in the equation so that this differential equation is belongs to linear differential equation. Also the coefficient of derivative term are constant as well as the coefficient of dependent variable y is also constant. Hence, this differential equation is also called as linear differential equation with constant coefficient. Another example number 2. 2 into dq y by dtq plus 3 into d square y by dt square plus dy by dt equal to 0. In this again y is a dependent variable and t independent variable and derivative of y are dy by dt d square y by dt square dq y by dtq. The degree of all these three differential coefficient is 1 and they are not multiplied each other in the equation so that this differential equation is also belongs to linear differential equation. And again the coefficient of these derivative term is constant so that it is also called linear differential equation with constant coefficient. In general an ordinary differential equation a naught d raise to n y by dx raise to n plus a 1 d raise to n minus 1 y by dx raise to n minus 1 plus a 2 d raise to n minus 2 y by dx raise to n minus 2 plus and so on plus a n minus 2 into d square y by dx square plus a n minus 1 into dy by dx plus a n into y equal to x of x denoted by 1 where a naught a 1 a 2 etcetera a n are constant coefficients and x of x is a function of x or may be constant is known as linear differential equation of nth order with constant coefficients. Now operator d the differential coefficient d by dx is called differential operator and it is denoted by symbol capital D so that dy by dx can be written as d of y and d square y by dx square can be written as d square of y and so on d raise to n y by dx raise to n can be written as d raise to n up y. With these notation the above differential equation 1 can be written as a naught d raise to n up y plus a 1 d raise to n minus 1 up y plus and so on plus a n minus 2 d square up y plus a n minus 1 d up y plus a n into y equal to x of x. In the left hand side every term operating on y so that y is taken outside so in bracket 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 and so on plus a n minus 2 d square plus a n minus 1 d plus a n bracket close into y which is equal to x of x. Now expression in the bracket is a function of d and it is denoted by f of d. Therefore f of d into y is equal to x of x denoted by equation 2. Now equation 2 is called a symbolic form of equation 1. Both equations are identical but way of writing is different. Here 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 and so on plus a n minus 2 d square plus a n minus 1 d plus a n which is called as polynomial in a d. Now general solution of linear differential equation. The general solution is the functional relation between dependent variable and independent variable which satisfy the differential equation and it is containing an arbitrary constants. The number of such constant must be equal to order of differential equation. So that in general the general solution of linear differential equation of the type f of d into y equal to x of x with constant coefficient when x of x is not equal to 0 is given in two parts as dependent variable here y which is equal to complementary function plus particular integral. In short we write it as y equal to C f plus p i where C f is the solution of a homogeneous differential equation f of d into y equal to 0 and p i is the particular solution of equation f of d into y equal to x of x when x of x not equal to 0. Now auxiliary equation of a differential equation. An auxiliary equation of a linear differential equation f of d into y equal to x of x is obtained by equating 0 to polynomial f of d that means f of d equal to 0 gives an auxiliary equation of the differential equation. But for this 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 and so on plus a n minus 1 into d plus a n. And when we equate it equal to 0 we get 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 and so on plus a n minus 1 into d plus a n which is equal to 0. It is called as a auxiliary equation of a given nth order differential equation. It is an algebraic equation in d the values of d which satisfy this equation are called roots of an auxiliary equation. Now let us consider example one write an auxiliary equation of differential equation d cube y by dx cube plus d square y by dx square minus 2 into dy by dx which is equal to e raise to x and find roots of an auxiliary equation. Here d by dx is capital D so that dy by dx equal to dy d square y by dx square equal to d square y and d cube y by dx cube equal to d cube y. With this notation the given equation can be written as d cube y plus d square y minus 2 dy equal to e raise to x that is when y is taken common in bracket d cube plus d square minus 2 d bracket close into y equal to e raise to x. This is of type f of d into y which is equal to x of x here f of d is d cube plus d square minus 2 d and when we equate this f of d equal to 0 we get auxiliary equation therefore d cube plus d square minus 2 d equal to 0 is the auxiliary equation of q 1 differential equation. Now, we have to solve this equation here d cube plus d square minus 2 d equal to 0 to solve this equation we will we factorize the left hand side for that d is common taken in bracket d square plus d minus 2 equal to 0 and keeping d as it is are equate d equal to 0 one equation and d square plus d minus 2 equal to 0 another equation and now we have to solve equation d square plus d minus 2 equal to 0 separately by factorizing here d square plus 2 d minus d minus 2 d can be written as 2 d minus d and minus 2 which is equal to 0 from first 2 term d is common in bracket d plus 2 and from last 2 term minus 1 is common in bracket d plus 2 which is equal to 0 and when we take d plus 2 is common we left d minus 1 in bracket which is equal to 0 then d plus 2 equal to 0 gives d equal to minus 2 and d minus 1 equal to 0 gives d equal to 1 hence total roots of auxiliary equations are d equal to 0 d equal to minus 2 and d equal to 1 to prepare this video session I use this book as a reference thank you