 Hello everyone, myself, Mrs. Mayuri Kangre, Assistant Professor of Mathematics from the Department of Humanities and Sciences, Valchan Institute of Technology, SolarPur. Today we are going to see higher order linear differential equations. The learning outcome is at the end of this session, the students will be able to solve the examples of higher order linear differential equations. Let f of d of y equal to x be given higher order linear differential equation. Its general solution is given as y equals to cf plus pi that is yc plus yp, where this cf or yc is the complementary function and pi or yp is the particular integral. In this video, we are going to see the shortcut method to find the particular integral for the case of capital X that is the right hand side of the equation as x raise to m, where m is a positive integer. Let f of d of y equal to x raise to m, where m is a positive integer be given higher order linear differential equation. Then the particular integral pi or yp it is equals to 1 upon f of d into x that is 1 upon f of d into x raise to m, we will express this f of d as 1 plus or minus phi of d. So that yp equals to 1 upon 1 plus or minus phi of d into x raise to m. Where f of d equal to 1 plus or minus phi of d form is obtained from a given polynomial in d by taking lowest degree term as common factor. Therefore, we can write this 1 upon 1 plus or minus phi of d as 1 plus or minus phi of d bracket raise to minus 1 multiplied by x raise to m. After expressing it in this form, use the following results to expand the bracket. The results are 1 plus x bracket raise to minus 1 equals to 1 minus x plus x square minus x cube and so on and 1 minus x bracket raise to minus 1 equals to 1 plus x plus x square plus x cube and so on. Operating each of the expansion on x raise to m. Before we solve the examples, please pause the video for a minute and give the answer of this example. Expand 1 plus 3x by 2 bracket raise to minus 1. I hope all of you have expanded this function. So let us check the solution. We know that 1 plus x bracket raise to minus 1 equal to 1 minus x plus x square minus x cube and so on. Therefore, 1 plus 3x by 2 bracket raise to minus 1 will be expanded as. Now here at the place of x, we will treat it as 3x by 2. Wherever the x is there, we will replace it by 3x by 2. Therefore, we can expand it as 1 minus 3x by 2 plus 3x by 2 bracket square minus 3x by 2 bracket cube plus 3x by 2 bracket raise to 4 minus 3x by 2 bracket raise to 5 and so on. So it gives us 1 minus 3x by 2 plus 9x square by 4 minus 27x cube by 8 plus 81x raise to 4 upon 16 and so on. Now let us move to the examples. We call d cube plus 3d square plus 2d bracket closed y equal to x square. Here the given equation is d cube plus 3d square plus 2d bracket closed y equal to x square. Therefore f of d is d cube plus 3d square plus 2d and capital X is x square. Here the case of x raise to m to be applied as capital X is x square. Now first of all, we will find out the complementary function. So the auxiliary equation is d cube plus 3d square plus 2d equal to 0 taking d as common. So d into the bracket d square plus 3d plus 2 equal to 0. We can factorize this gives us d into d plus 1 into d plus 2 equal to 0. Therefore we get d equal to 0 minus 1 minus 2. Now here the roots are real and distinct. Therefore the complementary function yc will be equals to c1 raise to 0x plus c2 raise to minus x plus c3 e raise to minus 2x. Now e raise to 0x is 1 so we can write the cf that is yc equals to c1 plus c2 e raise to minus x plus c3 e raise to minus 2x and we will call it as equation number 1. Now let us move to the particular integral Pi which is defined as 1 upon f of d into x. So we will get the yp here as 1 upon d cube plus 3d square plus 2d into x square. Here to apply the case of x raise to m, we have to express this f of d as 1 plus or minus phi of d and for that purpose we have to take or we have to select the smallest degree term as common. Here the smallest degree term is 2d. So take it as common, therefore we can write the yp as 1 upon 2d into the bracket. Now this 2d is divided to d cube plus 3d square, 1d from the numerator and denominator get cancelled and we get the bracket value as d square plus 3d upon 2 plus 1 operated on x square. Now it can be written as 1 upon 2d as it is addition we can change the position 1 plus d square plus 3d by 2 into x square. Now it can be written as 1 upon 2d into 1 plus d square plus 3d upon 2 bracket raise to minus 1 into x square. Now expanding it using the relation 1 plus x bracket raise to minus 1. In this expansion we will treat d square plus 3d by 2 as x. So we get the expansion as 1 upon 2d into the bracket 1 minus d square plus 3d by 2 plus d square plus 3d by 2 bracket square and continuing the same way multiplied by x square. So the power of x if the power of x is 2 expand the function up to the power 2. If the power of x is 3 expand the function up to the power of 3 here the power of x is 2 so we will expand the function up to the power 2. Now we will simplify it therefore yp equals to 1 upon 2d into the bracket 1 minus d square plus 3d upon 2 plus now expanding this bracket square. So d raise to 4 plus 6d cube plus 9d square upon 4 using the formula a plus b bracket square equals to a square plus 2ab plus b square. Now we will multiply this x square to the bracket so we can write it as yp equal to 1 upon 2d into the bracket x square minus d square x square plus 3d x square upon 2 plus d raise to 4x square plus 6d cube x square plus 9d square x square upon 4 and so on. Now here as we know that d stands for d by dx for x square d of x square will be the first derivative of x square will be 2x d square it means the second derivative of x square will be 2 the third derivative of x square will be 0 and third onwards all will be 0. So this d square x square see here d square x square is 2 dx square it will be 2x d raise to 4x square it will be 0 d cube x square it will be 0 d square x square it will be 2. So substituting these values in these expansions we will get yp equal to 1 upon 2d into the bracket x square minus 2 plus 3 into 2x upon 2 plus 0 plus 6 into 0 plus 9 into 2 upon 4. Simplify all these values we will get yp equals to 1 upon 2d into the bracket x square minus 2 plus 3 into 2 gives us 6x upon 2 plus 0 plus 18 upon 4. Now it can be written as 1 upon 2d into the bracket x square as it is minus 2 upon 2 gives us minus 1 minus 6 upon 2 gives us minus 3 into x. Now plus 18 upon 4 can be written as 9 by 2. Now minus 1 plus 9 by 2 is 7 by 2 therefore we get yp as 1 upon 2d into the bracket x square minus 3x plus 7 by 2. As we know that 1 upon d of x is nothing but the integration of x dx therefore yp will be equals to 1 upon 2 integration of x square minus 3x plus 7 by 2 into dx. The integration gives us 1 by 2 into x cube by 3 minus 3x square by 2 plus 7x by 2 we will call it as equation number 2. Using equation 1 and 2 we can write the solution of the given equation as y equals to c1 plus c2 e raise to minus x plus c3 e raise to minus 2x plus 1 by 2 into the bracket x cube by 3 minus 3x square by 2 plus 7x by 2. Thank you.