 Hello everyone, myself, Mrs. Mayuri Kangre, assistant professor of mathematics from the Department of Humanities and Sciences, Valchian Institute of Technology, Soulapur. 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 cf or yc is the complementary function and pi or yp is the particular integral. In the previous videos, we have seen the shortcut method to find the particular integral, where the capital X is sin x or cos x. If the capital X is sin x or cos x, we have seen that to find the yp, we will rewrite it as 1 upon f of d into sin x or cos x, where this f of d will be rearranged so that it will be in terms of f of d square. Suppose you write this yp as 1 upon f of d square into sin x or cos x, replace d square by minus of a square, where a is the coefficient of x in sin x or cos x. While substituting, do not take the minus a bracket square, it should be minus of a square. Therefore, we get the yp as 1 upon f of minus of a square into sin x or cos x, provided that this denominator should be non-zero. Then we have seen that if the denominator is equal to 0, we multiply by x and take the derivative of f of d and repeat the same procedure which gives us yp equal to x into 1 upon f dash of minus of a square into sin x or cos x, provided again the denominator is non-zero. And if this denominator is 0, repeat the same procedure till you get the non-zero denominator. Now, before we start the examples, please pause the video for a minute and give the answer of this example. Solve d2y by dx square plus y equal to 0. I hope all of you have solved the example. Please check the answer. Here the equation is d square plus 1 bracket closed y equal to 0. Therefore, f of d is d square plus 1 and the auxiliary equation is d square plus 1 equal to 0, which gives us d equals to plus or minus i, which is a complex and distinct root. This plus or minus i can be written as 0 plus or minus 1 i. Therefore, the complementary function is nothing but the solution can be written as e raise to 0x into C1 cos x plus C2 sin x. Therefore, the solution y is C1 cos x plus C2 sin x. Now, let us solve the examples. Solve d square plus d plus 1 bracket closed y equal to sin square x. The given equation is of this form, which gives us f of d as d square plus d plus 1 and capital X as sin square x. First of all, we will find out the complementary function. So, the auxiliary equation is d square plus d plus 1 equal to 0. Solve this equation d equal to minus 1 plus or minus i into root 3 upon 2. The roots are complex and distinct and can be written as minus 1 by 2 plus or minus i into root 3 by 2. Therefore, yc equals to e raise to minus 1 by 2x into the bracket C1 cos root 3 by 2 into x plus C2 sin root 3 by 2x. We will call it as equation number 1. Now, let us find out the particular integral yp. By definition, it is 1 upon f of d into x. So, here we get 1 upon d square plus d plus 1 into sin square x. Now, we will shift this sin square to sin r cos form. So, we can write it as 1 upon d square plus d plus 1 into the bracket 1 minus cos 2x upon 2. This 1 by 2 is a constant can be taken outside the operator. Therefore, the yp will be 1 by 2 into the bracket 1 upon d square plus d plus 1 into 1 minus cos 2x. Separating the terms gives us 1 by 2 into 1 upon d square plus d plus 1 into 1 minus 1 upon d square plus d plus 1 into cos 2x. Now, this 1 can be written as e raise to 0x. Therefore, yp will be equal to 1 by 2 into the bracket 1 upon d square plus d plus 1 into e raise to 0x minus 1 upon d square plus d plus 1 into cos 2x. Now, using the case of e raise to ax for the first term and using the case of cos ax for the second term. By the case of e raise to ax, we will replace d by a that is d by 0. And here by using the case of cos ax, we will replace d square by minus of a square. Here a is 2. Therefore, minus of a square will be minus 4. So, replace d square by minus 4. So, we can write yp equal to 1 by 2 into the bracket 1 upon 0 plus 0 plus 1 into e raise to 0x minus 1 upon minus 4 plus d plus 1 into cos 2x. Now, here 1 upon 1 into e raise to 0 is also 1. So, 1 upon 1 into 1 is 1. So, we get the next step as yp equal to 1 by 2 into the bracket 1 minus 1 upon d minus 3 into cos 2x. See here minus 4 plus 1 gives us minus 3. So, we get d minus 3 here. To apply the case of cos ax, the denominator should be in terms of d square. But it is free from d square. So, to bring the term of d square here, we will use the method of rationalization. Therefore, yp can be written as 1 by 2 into the bracket 1 minus 1 upon d minus 3 into d plus 3 upon d plus 3 into cos 2x. We can write the yp as 1 by 2 into the bracket 1 minus d plus 3 into cos 2x upon d square minus 9. Now, replace a d square by minus 4. So, we get yp as 1 by 2 into the bracket 1 minus d into cos 2x plus 3 into cos 2x upon minus 4 minus 9. Now, d into cos 2x is the derivative of cos 2x which is minus sin 2x into 2 gives us minus 2 sin 2x plus 3 cos 2x as it is. And in the denominator minus 4 minus 9 gives us minus 13 with one more negative sign here gives us plus sign. So, yp will be equal to 1 by 2 into the bracket 1 plus minus 2 sin 2x plus 3 cos 2x upon 13. We will call it as equation number 2. Therefore, using equation 1 and 2, the solution of the given equation can be written as y equals to e raise to minus 1 by 2x into the bracket C 1 cos root 3 by 2x plus C 2 sin root 3 by 2 into x plus. Now, this 1 by 2 is multiplied to the bracket. So, we get 1 by 2 plus 1 by 26 into the bracket minus 2 sin 2x plus 3 cos 2x. Now, let us solve the next example d square plus 1 bracket closed y equal to sin x into sin 2x. The given equation is of this form. Therefore, f of d is d square plus 1 and capital X is sin x into sin 2x. First of all, we will find out the complementary function. Therefore, the auxiliary equation is d square plus 1 equal to 0 which gives this d square equals to minus 1. Therefore, d equal to plus or minus i. The roots are complex and distinct and can be written as 0 plus or minus 1i. Therefore, the yc, the complementary function will be e raise to 0x into the bracket C 1 cos x plus C 2 sin x. Now, e raise to 0x is 1. Therefore, the yc will be equal to C 1 cos x plus C 2 sin x and we will call it as equation number 1. Now, let us go for the yp that is particular integral. It is equals to 1 upon d square plus 1 into sin x into sin 2x. Now, we will use the result sin a sin b equal to 1 by 2 cos of a minus b minus cos of a plus b. Therefore, the sin x into sin 2x will be equal to, now a is x and b is 2x. So, it will be equal to 1 by 2 into the bracket cos of x minus 2x minus cos of x plus 2x that is 1 by 2 into cos of minus x minus cos of 3x. Now, we know that the cos of minus theta is cos theta. Therefore, we can write the yp as 1 upon d square plus 1 into 1 by 2 cos x minus cos of 3x. This 1 by 2 is a constant in the multiplication form can be taken outside the operator. Therefore, we can write it as 1 by 2 into the bracket 1 upon d square plus 1 into cos x minus 1 upon d square plus 1 into cos 3x. Now, using the case of cos ax for the first term a is 1. So, d square is replaced by minus 1 and for the case of second term a is 3 here. Therefore, the d square will be replaced by minus 9. Therefore, we get yp equal to 1 by 2 into the bracket 1 upon minus 1 plus 1 into cos x minus 1 upon minus 9 plus 1 into cos 3x. Now, minus 1 plus 1 gives us 0. Therefore, case fails. Here f of d is d square plus 1 gives us f dash of d as 2d. Therefore, we can write yp as 1 upon 2 into the bracket x into 1 upon 2d into cos x minus 1 upon minus 8 cos 3x minus 9 plus 1 gives us minus 8. Then upon d cos x will be integration of cos x or dx that is sin x. Therefore, yp will be equals to 1 by 2 into the bracket x by 2 sin x. Now, minus minus becomes plus 1 by 8 cos 3x which gives this yp as x by 4 into sin x plus 1 by 16 into cos 3x. We will call it as equation number 2. From 1 and 2, we can write the solution of the given equation as y equals to c 1 cos x plus c 2 sin x plus x upon 4 sin x plus 1 by 16 cos 3x. Thank you.