 Hello everyone. Myself, Mrs. Mayuri Kangre, assistant professor of mathematics from the Department of Humanities and Sciences, Valchan Institute of Technology, Solaapur. 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 equals to capital 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 which can be denoted as yc and yp respectively. In this video we are going to see how to find the particular integral if the capital X is in standard form that is the shortcut method to find out the particular integral. So let us begin the shortcut method to find the particular integral when the capital X is sin x or cos x. Let f of d of y equals to sin x or cos x be given higher order linear differential equation. Then the particular integral yp is equals to 1 upon f of d into sin x or cos x. We will rearrange this f of d as f of d square that means the terms of f of d are rearranged so that it will contain d square. So we can write yp as 1 upon f of d square into sin x or cos x. After writing it in this form we will put d square as minus of a square where a is the coefficient of x in sin x or cos x. Here do not take this minus of a square as minus a bracket square both are different because minus a bracket square will give us plus a square where here we are going to take the a square first with the negative sign and this d square will be replaced by minus of a square. Therefore yp will be equals to 1 upon f of minus of a square into sin x or cos x where this denominator f of minus of a square is not equal to 0. That means the denominator should be non-zero value. Suppose if this denominator is equal to 0 then yp will be x into 1 upon f dash of minus of a square into sin x or cos x. That means here we will multiply by x to the yp and we will take the derivative of f of d which gives us f dash of d. In that f dash of d we will rearrange for d square that means we will express this f dash of d as f dash of d square and again we will replace a d square by minus of a square which gives us f dash of minus of a square and again this denominator should be non-zero. If it is again 0 then yp will be again multiplied by one more time by x so we will get x square into 1 upon f double dash of minus of a square into sin x or cos x again provided that the denominator f double dash of minus of a square should not be equal to 0. And we can continue the process till we get the non-zero denominator if you will get the denominator as 0. Now before going to start the examples please pause the video for a minute and give the answer of this example. The example is solve d2y by dx square minus 5dy by dx plus 6y equal to 0. I hope you all are getting the solution, let us check the solution. The given equation is d2y by dx square minus 5dy by dx plus 6y equal to 0 expressing it with operator d gives us d square minus 5d plus 6 bracket closed y equal to 0 where this f of d is d square minus 5d plus 6 therefore we get the auxiliary equation as d square minus 5d plus 6 equal to 0 that is d minus 3 into d minus 2 equal to 0 which gives us d equals to 3 comma 2 where the roots are real and distinct. So by the case of real and distinct roots we can write the complementary function which is the solution of this example as c1 e raise to 3x plus c2 e raise to 2x. Now let us go for the examples. For example, d cube minus d square minus 6d bracket closed y equal to sin x. Here the given equation is of this form therefore f of d is d cube minus d square minus 6d and the capital X right hand side is sin x. Therefore the auxiliary equation is d cube minus d square minus 6d equal to 0 taking d as common gives us d into the bracket d square minus d minus 6 equal to 0 that is d into the bracket d minus 3 into d plus 2 equal to 0 which gives us d equal to 0 3 minus 2 which are the roots. The roots are real and distinct therefore we can write the complementary function yc equals to c1 into e raise to 0x can be written as c1 only as we know that e raise to 0x is 1. Therefore yc will be equal to c1 plus c2 e raise to 3x plus c3 e raise to minus 2x we will call it as equation number 1. Now let us find out the particular integral pi that is yp which is defined as 1 upon f of d into capital X. So here it will be 1 upon d cube minus d square minus 6d into sin x. First of all we will rearrange the f of d in terms of d square. Here we have d cube which can be expressed as d into d square. So we can write yp as 1 upon d into d square minus d square minus 6d into sin x. Now comparing this sin x with the sin x so we get a as 1 minus of a square is minus 1. Therefore replacing a d square by minus 1 which gives us yp equals to 1 upon d into the bracket minus 1 minus into the bracket again minus 1 minus 6d into sin x. Therefore yp will be equals to 1 upon minus d minus of minus becomes plus 1 minus 6d into sin x. Now plus 1 minus d minus 6d gives us minus 7d therefore we get yp equal to 1 upon 1 minus 7d into sin x. Now to apply the case of sin ax the denominator should contain a d square so that the d square will be substituted by minus of a square. Now to get the d square in the denominator here we will follow the rule of rationalization. Therefore yp will be equal to 1 upon 1 minus 7d into 1 plus 7d upon 1 plus 7d into sin x. If you observe the denominator it is in the form of a minus b into a plus b which gives us a square minus b square. Therefore we can write it as the numerator will be 1 plus 7d only divided by a square minus b square that is 1 square minus 7d bracket square which gives us 1 minus 49d square into sin x. Now if you observe here the denominator is now containing a d square so we can use the shortcut method of sin ax. Here d square will be replaced by minus 1 and sin x will be multiplied to the numerator. Therefore we can write the yp as sin x plus 7d into sin x upon 1 minus 49 into the d square will be replaced by minus 1. Now sin x will remain as it is plus 7 as it is the derivative of sin x is cos x divided by 1 minus of minus becomes plus 49 gives us 50. Therefore yp will be equals to sin x plus 7 cos x divided by 50 which will be called as the equation number 2. Using equation 1 and 2 the solution of the given equation will be y equals to yc plus yp that is y equals to c1 plus c2 e raise to 3x plus c3 e raise to minus 2x plus sin x plus 7 cos x upon 50. Thank you.