 So, we will continue with the ah second order differential equations. This is a part of the quantum chemistry course and in the last lecture on second order differential equations we looked at the homogeneous equations homogeneous linear second order differential equations, we were looking at ah those equations with constant coefficients. So, we shall continue that and also probably ah extend to that to some simple examples of non homogeneous equations. The idea is ah that you familiarize yourself with the methods of solving simple second order differential equations, but in the second part of today's lecture and probably also later I would follow this up with the ah power series method. We shall just see a few simple examples of non homogeneous ah second order differential equations the same thing non homogeneous linear second order ah DE ah with constant coefficients ok. We shall see that and then follow this by the power series method, but the power series method is the one that ah we would ah go on and on for the next few lectures particularly for ah those of you who encounter this for the first time. I hope that I provide you sufficient amount of details ah so that you feel comfortable with it and that you would try other equations in physics using this method, but we will keep our objectives very straight in that ah it is ah all of these things are being done with the idea of solving the Schrodinger equation model problems particularly the problems of ah the particle ah as a harmonic oscillator and the two dimensional motion on a sphere as well as the full three dimensional motion of the electron in a hydrogen atom and the energy levels of the hydrogen atom the electron in the hydrogen atom so ah that is the purpose. So, let us get back to a little more general aspects of the second order differential equations. You remember that when we write down a differential equation a d square y by d x square plus b d y by d x plus c y is equal to 0 that is with constant coefficients. We know that this has two solutions namely ah y of x is given by a exponential something like k 1 of x plus b exponential k 2 of x the two constants k 1 and k 2 are obtained by solving or substituting one of the exponentials in here and solving the corresponding quadratic equation. So, you remember that k 1 or 2 is given by minus b plus or minus square root of b square minus 4 ac by 2 a and then we looked at the special cases where k 1 is equal to k 2 and also the case where k 1 and k 2 are both complex that is this is imaginary this is negative and therefore, you have an imaginary solution which brought to us the details of the exponential multiplied by trigonometric functions sin or cosine ok. So, given that there is one more thing that we all should keep in mind that two solutions to a second order differential equation u 1 of in this case let me call this as u 1 of x and call this as u 2 of x ok they are independent if certain criterion is fulfilled and the independence of the solution is through a criterion known as non zero Ronskian. Let me define that in a minute two solutions u 1 of x and u 2 of x are independent linearly independent solution that is one cannot be expressed in terms of the other, but any combination of these two is equal to 0 with constant coefficients mean that the coefficients have to be 0 ok. I will write that down or linearly independent only if the Ronskian is non zero. What is a Ronskian? It is defined in terms of the functions and the derivatives. So, in this case the Ronskian is written as w u 1 u 2 it is a function of those two solutions and it is a determinant of u 1 of x u 2 of x u 1 prime of x u 2 prime of x and this should not be 0 ok. That is the product u 1 of x u 2 prime of x minus u 1 prime of x u 2 of x is not equal to 0 then this implies that the solutions u 1 and u 2 are independent general solutions to the second order differential equation. Now what if we extend this to the homogeneous non homogeneous case and also may be instead of second order, third order and higher order differential equations in general if ok. What is the linear dependence? Linear independence mean means the following c 1 u 1 of x plus c 2 u 2 of x is equal to 0 means c 1 is equal to c 2 is equal to 0 that is called the linear independence of the two functions that is this one cannot be expressed in terms of the other in terms of constants that is what this means. And in general if you extend to this two n functions it is a linear combinations of the n functions please remember earlier we talked about basis functions in quantum mechanics at some point and mentioned the linear independence of the wave functions and so on. If you have seen that all of this follow the same definition ok. So, if you have more than two functions for a non homogeneous differential equation we will see that in a minute. If you have that what is meant by them being linearly independent the Ronskian is determined by that all of these functions let me write this for 3 in the first place and then write the general one. Again the determinant should be u 1 of x u 2 of x u 3 of x u 1 prime of x u 2 prime of x and u 3 prime of x and the second derivative u 1 double prime of x u 2 double prime of x and u 3 double prime of x. Therefore, for three functions we take the determinant comprising of the functions the first derivative and the second derivative and we say that this determinant is nonzero and if you find that it is nonzero then these three are the solutions to the and if these three are solutions to the differential equation and the determinant is nonzero then these three are three independent solutions to the second to the second order third order or other differential equations. So, essentially linear independence means writing down the corresponding Ronskian and verifying that the Ronskian is nonzero. So, what is it for the general n functions? It is the same thing it will contain n minus 1 derivatives of all the functions that is n minus 1 derivatives. So, if I write w in terms of u 1 u 2 u n then the Ronskian is u 1 of x u 2 of x all the way up to u n of x and then you have u 1 prime of x u 2 prime of x u n prime of x that is the first derivative down to n minus 1 derivative of x n minus 1 prime if you want to write that and some people even write this but anyway I will keep this as explicit as I can this is u 1 and this is u 2 the n minus 1 derivative and likewise the u n n minus 1 derivative of x this determinant should be nonzero if the functions u 1 u 2 u n or solutions of a differential equation and or independent this condition will be satisfied ok. So, this is something to remember now let us check some of the solutions that we have had for both the homogeneous and also the non-homogeneous differential equations and verify whether they are actually nonzero and so on ok. So, let me just give you a couple of examples I think we had probably talked about the differential equation u y double prime minus 4 y prime plus 3 y is equal to 0 ok. You know that when it is factored with the solution u of x is equal to a e to the k x you are going to get k squared minus 4 k plus 3 times this whole thing because the derivatives will give you that that is going to be 0 and therefore, this is factored out as k minus 3 into k minus 1 is equal to 0 therefore, k is equal to 3 or 1. So, we have the general solution y of x is equal to a e to the 3 x plus b e to the x are these linearly independent just determine the derivatives w the functions are e to the 3 x e to the x and the derivative of e to the 3 x is 3 e to the 3 x and the derivative of e to the x is e to the x itself and if you look at the determinant of this determinant then what you have is e to the 4 x minus 3 e to the 4 x the answer is minus 2 e to the 4 x which is not 0 and therefore, these two solutions are independent general solutions to the differential equation and a linear combination of these two is obviously, the general solution and it is arbitrary because we do not know what a and b are that we can determine if we know the two conditions on y or one condition on y at a particular point and it is derivative at the same point if we know then we can in principle find out a and b. Therefore, a and b are arbitrary until you put the boundary conditions or you put the initial conditions whatever that may be you have you need two conditions to determine a and b, but the solution is linearly independent ok. So, the next is consider the same thing for a non homogeneous, but simple second order differential equations and we can see that the solutions are obtained specific solutions are obtained by solving both the homogeneous equation and the non homogeneous part independently and then adding the solutions to the differential equation as the overall solutions ok. So, let us take a look at ok before we do that you will want to verify on the same manner that we have this differential equation 4 y prime plus 4 y is equal to 0 you remember we wrote the solutions u 1 of x as e to the 2 x and u 2 of x as x e to the 2 x and you can verify that these two are indeed independent by writing down the run scheme by taking the derivatives of the exponential 2 x and the exponential of x e raise to 2 x putting them and calculating that I will leave it to you as an exercise ok. So, let us look at the homogeneous non homogeneous part and let us consider some simple equations. So, let us take a simple example of a differential equation y double prime minus 2 y prime minus 3 y is equal to cos x ok. Let me see I have actually solved that yeah I guess I guess I have ok. So, the solution to this is expressed by the two solutions that will be a particular solution with cos x being non 0 and the homogeneous part of this minus 2 y prime minus 3 y is equal to 0 will have the homogeneous solution y of x and the particular solution you can write it as y part of x then the overall solution for this differential equation y of x is the y homogeneous of x plus the y particular solution of x and this we already know how to calculate you know you can factor this out and get the two exponentials, but it is this the particular solution that we will see because we have a cosine x as a function and we have second derivatives connecting second derivatives of y connecting the cosine x you know that the cos and sin alternate when you take the derivatives. Therefore, the differential equation that we have gives us an an indication that the function y may be expressed as a linear combination of cosine and sin. So, that between first derivative and second derivative the most we will get is the cosine and sin and then match the result with the conditions of the differential equation and determine the coefficients. So, if you do not know what it means let me write the solution y of x as some constant c or let me let me see in my notes I have used a and b yeah a and b. So, what I would do is I will write y is a cosine x whatever is here plus b sin x and then substitute this its derivative its second derivative and obtain two equations involving a and b because there are two coefficients here and those two equations will give you the values of a and b and therefore, that is the particular solution because a and b is fixed in that process and what you will have is the solution satisfying precisely this identity and then this is what we have as y particular and that needs to be added to the y homogeneous. So, let us just do that the derivative y prime of x is minus a sin x plus b cos x the second derivative y double prime of x is minus a cos x minus b sin x and so, if we substitute that in the differential equation the non homogeneous part with the cosine on the right hand side what you will have is minus a cos x minus b sin x that is y double prime and then you have minus 2 y prime. So, it is minus 2 times y prime is we have minus a sin x plus b cos x and then we have minus 3 y which is minus 3 times a cos x plus b sin x is equal to cos x. So, now, you have two independent functions cosine and sin linearly independent of each other because a cos x plus b sin x is equal to 0 you know cannot be solved ok except for a is equal to 0 and b is equal to 0 ok. So, what you have is terms connecting the cosine x and terms connecting the sin x. So, if you collect all the cosine x coefficients it is minus a and it is minus 2 b and it is minus 3 a ok and then you have sin x which is minus b you have minus plus 2 a and minus 3 b and that is equal to cos x times cos x sorry that is sin cos is equal to cos x sorry ok. Therefore, you see that the coefficients all of this is equal to 1 and all of this is equal to 0 which tells you that 2 a is equal to 4 b or a is equal to 2 b and if you put a is equal to 2 b it is minus a is minus 2 b minus 3.