 Hello everyone, myself, Mrs. Mayuri Kangri, Assistant Professor of Mathematics from the Department of Humanities and Sciences, Vulture Institute of Technology, Sallabur. 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 this yc or cf is the complementary function and yp or pi is the particular integral. We want to find the solution of this differential equation f of d of y equals to capital X. Already we have seen how to find the complementary function. Now we will see how to find the particular integral. When we are going to find the particular integral, it depends upon the nature of f of d and nature of capital X. If the capital X, the right hand side of the equation is in any of the standard form for shortcut method, we will go for shortcut methods. So today, we are going to see the shortcut method to find the particular integral. For the case, capital X that is the right hand side is sin x or cos x. That means the equation is of the form f of d of y is equals to either sin x or cos x. For those examples, we are going to see how to find the particular integral. Let f of d of y equals to sin x or cos x be given higher order linear differential equation. Then the particular integral that is yp is given as 1 upon f of d into sin x or cos x. Now we want to evaluate this yp. For that, first of all we will rewrite this f of d as f of d square. That means the terms which are involved in f of d will be expressed in terms of d square wherever possible. Therefore, the yp will be now 1 upon f of d square into sin x or cos x. Once we express it in the form of f of d square, we will observe coefficient of x in sin x or cos x and we will replace this d square by minus of a square. Therefore, put d square equals to minus of a square where a is the coefficient of x in sin x or cos x. Now remember that here we are going to consider minus of a square. It is not the minus a bracket square. So don't take minus a bracket square. Both are different. Therefore, this yp will be now 1 upon f of minus of a square into sin x or cos x where this f of minus of a square is not equal to 0. That means this denominator should be non-zero value. Suppose if this denominator is equal to 0, that means if this f of minus a square is equal to 0, then we will take the derivative of f of d which will be f dash of d and in that f dash of d, again we will rearrange it in the form of f dash of d square and again by the previous way, we will replace a d square by minus of a square. So we will get the yp as x into 1 upon f dash of minus of a square into sin x or cos x where this f dash of minus a square that is the denominator is not equal to 0. Suppose if this denominator is again 0, that means if f dash of minus of a square is also equal to 0, then yp will be obtained again multiplying by x. So we get x square into 1 upon f double dash of minus of a square into sin x or cos x where this f double dash of minus a square is not equal to 0. Now here this f double dash of minus of a square is obtained just like the previous steps. Now suppose again if it is equal to 0, we will continue the same process till we get the denominator as non-zero quantity. Before we solve the examples of this case, pause the video for a minute and give the answer of this example. The question is solve d2y by dx square minus 6 into dy by dx plus 9y equal to 0. I hope you all have solved the example. So let us check the solution. The given equation is d2y by dx square minus 6dy by dx plus 9y equal to 0. We will rewrite the equation using the operator d. So we can get d square minus 6d plus 9 into y equal to 0, that is f of d equal to d square minus 6d plus 9. The auxiliary equation is d square minus 6d plus 9 equal to 0 which gives us d equals to 3,3 where these roots are real and repeated. Therefore the solution of the given example is c1 plus c2x into e raise to 3x. Now let us solve the examples. First example, solve d2y by dx square minus 4dy by dx plus 4y equal to cos 2x. The given equation is d2y by dx square minus 4dy by dx plus 4y equals to cos 2x. We will rewrite it using the operator d gives us d square minus 4d plus 4 bracket closed y equals to cos 2x. Now here f of d is d square minus 4d plus 4 and capital X the right hand side is cos 2x. Therefore the auxiliary equation is d square minus 4d plus 4 equal to 0 which gives us d equals to 2,2 where the roots are real and repeated. So we can write the complementary function yc as c1 plus c2x into e raise to 2x. We will call it as equation number 1. Now we will find out the particular integral Pi which is defined as 1 upon f of d into capital X that is 1 upon d square minus 4d plus 4 into cos 2x. Compare this cos x with given example cos 2x which gives us a equals to 2 therefore minus of a square will be minus 4. Therefore yp will be equals to 1 upon d square is replaced by minus 4. So minus 4 minus 4d plus 4 into cos 2x. Here minus 4 and plus 4 get cancelled and we get the yp as 1 upon minus 4d cos 2x that is yp equals to minus 1 by 4 as it is into 1 upon d of cos 2x. As we know that 1 upon d of x is nothing but the integration of x dx. Therefore 1 upon d of cos 2x is equals to integration of cos 2x dx and we know that the integration of cos x is sin x upon a therefore we can write it as sin 2x upon 2. So we get the yp as minus 1 by 4 into sin 2x upon 2 which gives us yp equals to minus of sin 2x upon 8 we will call it as equation number 2. The solution of the given equation is now y equals to yc plus yp using the equations 1 and 2 we get the solution of the given example y as c1 plus c2x e raise to 2x minus sin 2x upon 8. Now let us go for the next example solve d cube y by dx cube plus dy by dx equals to sin x. The given equation is d cube y by dx cube plus dy by dx equals to sin x that is d cube plus d bracket close to y equals to sin x. Now here f of d is d cube plus d and the capital X that is the right hand side is the sin x. The auxiliary equation is d cube plus d equal to 0 taking d common which gives us d into d square plus 1 equal to 0. Therefore we get the roots as d equal to 0 and d equals to plus or minus i where this 0 is the real root and plus or minus i is complex and distinct root which can be expressed as 0 plus or minus 1i. That is why we can write the complementary function yc equals to c1 e raise to 0x plus now the real part is 0. So we can write it as e raise to 0x into the bracket c2 cos x plus c3 sin x as we know that e raise to 0x is 1 therefore this yc will be equals to c1 plus c2 cos x plus c3 sin x. We will call it as equation number 1. Now particular integral yp is defined as 1 upon f of d into capital X. So here it will be 1 upon d cube plus d into sin x. To express it in the form of d square we will take d common therefore yp will be equals to 1 upon d into the bracket d square plus 1 into sin x. Comparing this sin x with the sin ax therefore we will get a equals to 1. So a square will be 1 with the negative sin gives us minus of a square equals to minus 1. Therefore the yp will be equal to 1 upon d into the bracket. Now d square is replaced by minus 1. So minus 1 plus 1 into sin x. Here minus 1 plus 1 is 0 therefore the case fails. When we get the denominator as 0 we will go for method if f of minus of a square is 0 then yp equals to x into 1 upon f dash of minus of a square into capital X where this denominator f dash of minus of a square should not be equal to 0. Since f of d is d cube plus d f dash of d will be equal to 3 d square plus 1. Therefore yp will be equals to x into 1 upon 3 d square plus 1 into sin x. Again replace d square by minus 1. So yp will be equals to x into 1 upon 3 into minus 1 plus 1 into sin x which gives us x into 1 upon minus 2 into sin x that is minus of x sin x upon 2. We will call it as equation number 2. Now using equation 1 and 2 we can write the solution of the given equation as y equals to yc plus yp that is y equals to c1 plus c2 cos x plus c3 sin x minus x sin x upon 2. Thank you.