 Hello everyone, myself, Mrs. Mayuri Kamre, assistant professor of mathematics from the Department of Humanities and Sciences, Valshan 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, 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 Cf is the complementary function and Pi is the particular integral, denoted as yc and yp, respectively. Now, in this video, we are going to see the shortcut method to find the particular integral. For the case, capital X is equal to x raise to m, where m is a positive integer. Now, let F of d of y equal to x raise to m, where m is positive integer, be given higher-order linear differential equation. Then, by definition yp is equal to 1 upon F of d into capital X, therefore, we get yp equal to 1 upon F of d into x raise to m. Here we get yp as 1 upon 1 plus or minus phi of d into x raise to m. F of d equal to 1 plus or minus phi of d form is obtained from a given polynomial in D by taking the lowest degree term as common factor. Here, we will take the lowest degree term from the F of d as common, which will convert this F of d as 1 plus or minus phi of d. So, we can write it as 1 plus or minus phi of d bracket raise to minus 1 into x raise to m. Now, to expand it, we will use the result 1 plus x bracket raise to minus 1 equal to 1 minus x plus x square minus x cube and so on and 1 minus x bracket raise to minus 1 equal to 1 plus x plus x square plus x cube and so on. So, operating each of the expansion on x raise to m will give us the yp. Before we solve the example, please pause the video for a minute and give the answer of this example. Expand 1 minus 5x upon 3 bracket raise to minus 1. I hope all of you have obtained the solution. Let us check the solution. As we know that 1 minus x bracket raise to minus 1 equal to 1 plus x plus x square plus x cube and so on. Expand 1 minus 5x upon 3 bracket raise to minus 1 as 1 plus 5x upon 3 plus 5x upon 3 bracket square plus 5x upon 3 bracket cube and so on. Here, while expanding, this 5x upon 3 will be treated as x in this expansion of 1 minus x. So, we get 1 plus 5x upon 3 plus 25x square upon 9 plus 125x cube upon 27 plus 625x raise to 4 upon 81 and so on. Now, let us solve the examples. Solve d square plus d minus 2 bracket closed y equal to 1 plus x minus x square. The given equation is of this form. Therefore, f of d is d square plus d minus 2 and capital X is 1 plus x minus x square. If you observe this capital X, it will be under the case of x raise to m. Now, first of all, we will find out the complementary function. The auxiliary equation is d square plus d minus 2 equal to 0 gives us d plus 2 into d minus 1 0 which gives us d equals to minus 2, so 1. Here the roots are real and distinct. Therefore, the complementary function yc is equals to c1 e raise to minus rho 2x plus c2 e raise to x, we will call it as equation number 1. Now, let us go to obtain the particular integral yp. By definition, yp is equals to 1 upon f of d into x. So, here yp will be equal to 1 upon d square plus d minus 2 into 1 plus x minus x square. Now, to convert this f of d in the form of 1 plus or minus 5 of d, we will take the lowest degree term as common. Here the lowest degree term is minus 2. Therefore, take minus 2 as common, so we get yp equal to 1 upon minus 2 into the bracket d square plus d upon minus 2 plus 1 into 1 plus x minus x square which can be written as minus of 1 upon 2 into the bracket 1 minus d square plus d upon 2. This minus sign for 2 is written for the term. So, we will get 1 minus d square plus d upon 2 into 1 plus x minus x square. So we can write it as minus 1 by 2 into 1 minus d square plus d upon 2 bracket raise to minus 1 into 1 plus x minus x square. Now, we will expand it by using the relation 1 minus x bracket raise to minus 1 equal to 1 plus x plus x square and so on. Now, here this d square plus d upon 2 will be treated as x. So, the expansion will be now yp equal to minus 1 by 2 into the bracket 1 plus d square plus d by 2 plus d square plus d by 2 bracket square and so on multiplied by 1 plus x minus x square. Therefore, yp will be equals to minus 1 by 2 into the bracket 1 plus d square plus d by 2 plus now expanding this square gives us d raise to 4 plus 2 d cube plus d square upon 4 so on multiplied by 1 plus x minus x square. This 1 plus x minus x square is now multiplied to the bracket so we get yp equals to minus 1 by 2 into the bracket 1 plus x minus x square plus d square into 1 plus x minus x square plus d into 1 plus x minus x square divided by 2 plus d raise to 4 into 1 plus x minus x square plus 2 d cube into 1 plus x minus x square plus d square into 1 plus x minus x square divided by 4. Now, here the d stands for d by dx so for 1 plus x minus x square the derivative of this 1 plus x minus x square is 1 minus 2x d square of 1 plus x minus x square is minus 2 and the third derivative onwards all are 0. So, here d square of 1 plus x minus x square will be minus 2, it will be 1 minus 2x, it will be 0, it will be 0 and this will be minus 2. So, replacing all by these values we get the yp as minus 1 by 2 into the bracket 1 plus x minus x square plus into the bracket minus 2 plus 1 minus 2x completely divided by 2 close the bracket plus next bracket 0 plus 2 into 0 plus minus 2 divided by 4 close the bracket and the next terms will be 0. Therefore, yp will be equals to minus 1 by 2 into the bracket 1 plus x minus x square plus minus 2 plus 1 minus 2x upon 2 plus 0 plus 2 times 0 minus 2 upon 4 plus 0 that is minus 1 by 2 into the bracket 1 plus x minus x square plus now minus 2 plus 1 gives us minus 1 minus 2x upon 2 here minus 2 upon 4 simplifying all the terms gives us minus 1 by 2 into the bracket 1 plus x minus x square minus 1 by 2 minus 2x upon 2 minus 2 by 4 here 2 get cancelled and 2 by 4 is 1 by 2. So, can be written as yp equal to minus 1 by 2 into the bracket 1 plus x minus x square minus 1 by 2 minus x minus 1 by 2 plus x minus x get cancelled plus 1 minus 1 by 2 and 1 more minus 1 by 2 get cancelled as we know that minus 1 by 2 minus 1 by 2 gives us minus 1 and plus 1 and minus 1 will be 0. So, we get yp equals to minus 1 by 2 into minus of x square that is x square 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 c 1 e raise to minus 2x plus c 2 e raise to x plus x square by 2. Thank you.