 Welcome back to the lectures on the Principles of Quantum Mechanics. The series of lectures is being offered to you through the national program on Technology Enhanced Learning NPTEL and my name is Mangala Sundar Krishnan. I am a professor of chemistry in the department of chemistry Indian Institute of Technology, METROS. In the series we have been looking at in the first part of the series we have been we have been looking at some of the elementary mathematical aspects related to the Schrodinger equation and also basic quantum mechanics and in that context in the last lecture we looked at first order differential equations. So, this lecture is purely a lecture for solving problems some simple problems on first order differential equations. Therefore, I shall state the problems for those of you who want to do the solutions by themselves and not listen to the lecture. You can always refer to the lecture for details when you need to, but you can skip this lecture and go to the next one for some details on the second order differential equations and the power series method for solving mathematical problems in quantum mechanics. So, this is more like a hand holding or a problem solving lecture and I shall state a few problems that we want to consider in this hour. We will split this into two 20, 25 minute sessions. So, that you can have a break at some convenient point that I have decided that that is a convenient point ok. The first set of problems very elementary to slightly more involved. Let me write a simple differential equation d y by d x minus 10 y is equal to 0 and another problem d y by d x minus say phi y is equal to 6. These are given as problems for the background quiz on the website. So, they are not with any different from what is already there and the third that we will look at in this group is d y by d t minus y by t is equal to 0 ok. So, there are 3 yeah these 3 problems first let us look at that. Obviously, these are trivial I think you must know that this is a simple exponential solution. So, you can write that d y by y is equal to minus 10 d x which gives you immediately y of x is equal to a e to the d y by d x is plus 10 x. So, it gives you a e to the 10 x where a is the arbitrary constant. So, one condition will tell us what the value of a is that condition can be on the y or on the y prime or whatever it is at a particular value of x. So, there is one arbitrary constant since it is a first order differential equation. The second problem d y by d t d x d y by second problem d y by d x minus phi y is equal to 6 is amenable by a simple method of finding this as a single function which you can immediately see that it is d by d x of e to the minus phi x y of x. The differential of this will give you this part the on the left hand side and since you have multiplied the whole equation by e to the minus phi x in getting this form this is 6 e to the minus phi x ok. All that I said was multiply by e to the minus phi x and that gives you this ok. So, the solution would be y of x is equal to e to the plus phi x integral e to the minus phi x d x 6 times and the answer that you would get is y of x is equal to minus 6 by 5 plus a constant times e to the phi x ok which if you substitute in the differential equation you will get the equation back by phi. The third problem d y by d t minus y by t is equal to 0 is a separable problem because it is d y by y is equal to d t by t and then integration gives you a constant. So, you will get y is equal to a t it is essentially l and y is equal to l and t plus c. So, it is a constant it is a straight line in t ok. So, this is the first set of problems. Let us look at a slightly more involved set of problems involving the other methods. Let us look at the other set being d y by d x is equal to x square plus y square by x y solve for y. Second problem is d y by d x is equal to y by x plus square root of x y and one more d y by d x is equal to 2 plus y e to the x y y 2 y minus x e to the x y. So, in all these three cases solve for. So, let us take the first one this you recognize is a homogeneous of order 2 of degree 2 the equation d y by d x is equal to x square plus y square by x y. Therefore, the standard method that we learnt in the last lecture is that let y by x be u and y is equal to u of x d y by d x is given by u plus x d u by d x. Then you have d y by d x given as u plus x d u by d x is equal to dividing by x square we get 1 plus u square because it is y by x whole square and then you get y by x which is u and so x d u by d x is 1 plus u square by u minus u which is 1 minus u square. So, you get 1 by u therefore, the solution is x d u by d x is equal to 1 by u or u d u is equal to d x by x integrate to get u square by 2 is equal to l n x plus c and u itself is y square by x square therefore, you have the final answer y of x is equal to square root of 2 x square l n x plus c prime c x square that should be a 2 perhaps and u square by 2. So, that should be a 2 here also. If you have an initial condition or if you have a condition for y given the value of x you can obtain the value of the constant c is a constant. The second this is the first of the three problems yes. So, the second one d y by d x is equal to y by x plus square root of x y this also has the same form it is homogeneous and the degree is 1. Therefore, immediately you can write u plus x d u by d x is equal to y by x which is u and then you have by 1 plus root u or x d u by d x is equal to minus that is a u which you have to cancel. So, it is minus u root u by 1 plus root u. Therefore, the differential equation is minus d u by u root u minus d u by u is equal to d x by x and the answer turns out to be you can solve this immediately. This is square root of u the answer is 2 by root u for this and the answer here is minus l n u this is equal to l n x plus c giving you the answer that 2 root x by y is equal to l n y plus c c a constant. The third problem because the way it looks like d y by d x with it appears non-linear y e to the x y 2 y minus x e to the x y. So, we try this by writing it as m d x plus n d y is equal to 0 and the c if dou m by dou y is equal to dou n by dou x in which case we have a function of x and y which is a constant as a solution ok. So, if you rewrite this differential equation you have 2 y minus x e to the x y d y plus minus 2 minus y e to the x y d x is equal to 0. So, this is assuming this to be n and this to be m let us look at the derivatives is d y. So, I have to look at dou n by dou x and that is the partial derivative with respect to x here which will give me minus e to the x y minus x y e to the x y since differentiating this only with respect to x only ok. And dou m by dou y you look at the derivative here is minus e to the x y minus the derivative of the exponential with respect to y is x y e to the x y and therefore, we have this equation dou n by dou x is equal to dou m by dou y and therefore, there is a function f of x of y which is a constant such that d f is dou f by dou x given y d x plus dou f by dou y given x constant d y and that is 0 and this is your m this is your n and all that we did was dou squared f by dou y dou x is equal to dou squared f by dou x dou y for well behaved functions ok that is what we used. Therefore, the immediately we know that integral m d x with respect to x will give you the integral of minus 2 minus y e to the x y d x only and the answer will be not a constant, but any function of any arbitrary function of y because we are only integrating with respect to x. So, the answer we can write as minus 2 x minus e to the x y plus a function of y plus other constants, but that we do not want to include right now and how do we determine this we have to find out what n d y gives which will give you a function of y and an arbitrary function of x you have to match these two in order to get the final answer. So, the n d y integral is integral 2 y minus x e to the x y d y which is given as y squared minus e to the x y plus g of x. Now, you can see that here is one solution and here is the other solution and the function f of x y is the correct combination of these two such that f of x y satisfies the above equation. So, you can see that there is a y squared here and which is the h of y and e to the minus x y is common to both of them there is a g of x which is minus 2 x and therefore, the answer is f of x y is y squared minus e to the x y e raise to x y minus 2 x and of course, any arbitrary constant to that because you can always add a constant here you can always add a constant here and then well make it c double prime some arbitrary constant. So, the solution should be y squared minus e x raise to x y minus 2 x plus a constant. So, this is one way of solving the differential equation. So, let me complete this exercise with one more set of problems namelyverifying the integration integrating factor or these exact equations. Question given is y d x minus x d y and the factor minus 1 by x square is it an integrating factor. You can see that if this is m d x and n d y you can see that dou m by dou y is 1 dou n by dou x is minus 1 and therefore, you see that this is not equal to each other I mean they are not equal to one one other and therefore, you see that this is not an exact equation. However, if we multiply this equation y the if we multiply this quantity by minus 1 by x square you have minus y by x square d x minus plus 1 by x d y then you can see that this is a perfect differential. Therefore, if you have an equation like this is equal to 0 then what you do is you multiply this equation by minus 1 by x square and then make this into a perfect differential and solve this. This is easy to verify because if this is m and this is n dou m by dou x is dou y I am sorry dou y is minus 1 by x square and dou n by dou x is again minus 1 by x square. So, that is an that is the integrating factor. Likewise,there are two other problems is 1 by x y an integrating factor for the quantity y d x plus x d y actually y d x plus x d y itself is a perfect differential it is d of x y therefore, dividing it by x y obviously, will make it an integrating factor anyway. So, again make it perfect differential. So, we do not need it, but it is definitely an integrating factor for this problem because what you have is 1 by x d x plus 1 by y d y and you see that the derivatives in both cases will be 0 which is obvious because when you divide x y by x y you have a constant therefore, the derivative of a constant is 0. So, if f of x y is x y then you can see immediately d f is given by this quantity right. However, if you have a minus sign as you have here you do need an integrating factor. So, these are some simple examples of thefirst ordered differential equations. There are many many suchexamples and I probably will givean assignment on the websiteplus some handwritten solutions of those. So, those of you who want toget familiar with a little bit more of these operational methods of elementary first order differential equations please refer to the NPTEL website under my course the Principles of Quantum Chemistry or Quantum Mechanics ok. You can search for myname and you will see the courses that are there on the website. So, let me stop this part of today's lecture here have a break and then what we would do is to start with elementary second order differential equations and then continue that with somesimple examples in the next lecture. So, that the three lectures the onethe the preceding one this one and the next one hour lecture one and a half hours whatever together will form an elementary introduction to differential equations solution for those who want to study quantum chemistry and whose mathematical backgrounds are not I mean formally well defined in the areas of mathematics. However, if you want to learn these things more thoroughly it is important for you to go back to the books and read some of the proofs the theorems and the existence uniqueness of these solutions etcetera. A lot of these is also done in the NPTEL program under the mathematics and under the physics department. This is a hands on help to the chemistry graduate and undergraduate students who would come across quantum mechanics for the first time and are often disturbed by the fact that they have to solvesomewhat difficult differential equations and then never use them in practice later on because almost all the learningmethods and the practice is computational and numeric. However, it is important to appreciate some of the elementary solutions and particularly exact solutions of quantum mechanical problems. So, let me stop here we will continue this after a break.Welcome back we shall continue this with the second order differential equations. Again we will only do what is minimally needed to take us through the solutions ofthe quantum mechanical problems of harmonic oscillator and the hydrogen atom. So, most of you are familiar with these elementary differential equations second order d square by e x square y plus some constant k squared y of x is equal to 0 or d square y by d x square minus k squared y is equal to 0. Well, the sign difference is notthat simple, but it leads to basically different solutions I mean different the solutions are of completely different nature. This gives you oscillatory solution and this gives decaying, growing solutions. So, exponentials both, but very different nature ok. Then the other types of second order differential equations the general linear second order differential equation d square y by d x square plus a function p of x d y by d x plus another function q of x y is equal to 0. This is homogeneous linear second order d e and the allied inhomogeneous equation d square y by d x square plus some function p of x d y by d x plus q of x y is equal to g of x inhomogeneous or non-homogeneous this right this is non-homogeneous equations. When p and q are constants in p q or p q or constants then there are simple methods for solving them. When p and q are not constants, but functions and also sometimes a function preceding the second derivative itself. For example, well l of x d square y by d x square plus p of x d y by d x plus q of x y is equal to some other function g of x. We have far more general and much more detailed methods of solving them. One way is to look at this l x by not looking at it, but by bringing it as a divisor, but then there could be problems if l of x goes to 0 at some particular value of x which is in the domain of the solution of the equation. And those things are called the singular points and solutions of differential equations with singular points is one of the most beautiful and most thoroughly studied branches of mathematics. And a lot of that is also useful in quantum physics and in of course the entire branch of physics and engineering. We will not do any of those things at the moment. I mean you can take them up later, but probably we will do a little bit of these. And then proceed directly to what is known as an infinite series method or power series method for solving the differential equations, particularly second order differential equations. First simple solutions d square y by d x square plus k square y is equal to 0. It should be immediately obvious that y of x has the solution a cos k x plus b sin k x twice differentiating this you will get this you will get that equation ok. Another way of writing the solution is instead of a cos k x to write to this as e to the i k x plus e to the minus i k x by 2 plus b by 2 i e to the i k x minus e to the minus i k x which gives you two new constants c exponential i k x plus d exponential minus i k x. So, this is the complex representation for the solutions to the differential except that that c and d now can be complex constants or real constants you can see immediately b by 2 i plus a by 2 or minus whatever is defined by c and d. So, it is possible to write this. This is an important equation that we will see in a particle in a box and particle in a 2 d box and so on. So, we will have to keep this in mind ok. We will see this again. Another basic differential equation is d square y by d x square minus k squared y is equal to 0 and this has clearly only one solution y of x is equal to a e to the k x some arbitrary constant b e to the minus k x ok. Differentiating this twice you will get k squared and you will also get minus k whole square. So, you will get exactly this equation, but this is exponential growth and exponential decay this is not oscillatory as you have in the previous case ok as you have in this case ok. So, the sign is quite important and the fact is that k square is real if k square is imaginary. Obviously, this will become negative and so you have solutions if k is imaginary. Obviously, this is e to the i some real number and the solution goes back to this elementary. Now, the type of equations that we will study for in this course or the following equations ok. One Legendre differential equation of order alpha and this is a special equation solved by obviously, the legendary Legendre one minus x square d square y by d x square minus 2 x d y by d x plus alpha into alpha plus 1 y is equal to 0. In this course I will not solve this equation, but it is important in physics and in mathematics it is known as Bessel's equation of order n given by x square d square y by d x square plus x d y by d x plus d y by d x plus d y by d x plus x square minus n square y is equal to 0. Please note that it is multiplied by the second derivative is multiplied by x square. Therefore, x is equal to 0 is a singularity and one has to be careful about writing the solutions of this or solving this equation ok. We will not do this in this course, but it is important to remember that it is an extremely important equation in physics and mathematics. In this course we will see this equation log air differential equation for the solution of the radial part of hydrogen atom electron problem. The log gas equation is given by x square d square y by d x square plus 1 minus x d y by d x plus alpha y is equal to 0 and associated log air differential equation x square d square y by d x square plus 1 plus k minus x where k is non 0 d y by d x plus alpha y is equal to 0 and lastly oh there are two more equations. We have might differential equation d square y by d x square minus 2 x d y by d x plus 2 alpha y is equal to 0 ok. So, this we will solve explicitly in this course, this we will solve explicitly and also the Legendre there is also an associated Legendre equation associated given by 1 minus x square d square y by d x square minus 2 x d y by d x plus alpha into alpha minus alpha plus 1 minus m square y is equal to 0. So, we shall solve these types of equations in detail later on, but let us look at simple theory of second order differential equation. Well, simple examples I would not call it theory, simple examples for second order differential equation. If we consider homogeneous equations with constant coefficients these are easy to solve meaning the equation a d square y by d x square plus b d y by d x plus c is equal to 0 homogeneous constant coefficients a b c then the and a is not equal to 0. So, if you have merely d square y by d x square a is equal to 1 otherwise any number that goes with it then it is easy to solve this equation by having this operator defined d as d by d x and the operator d square as d square by d x squared and you can see that a function which does not change through differentiation to any order is an exponential function. Therefore, we might propose as a simple solution to this differential equation an exponential function and then if we substitute that exponential function with some unknown constants in there we can probably transform this equation into a linear equation not a linear differential equation in terms of those coefficients and then determine this coefficients and since you have second derivative here and first derivative here when you take the derivative of exponentials with coefficients the most you will get or the powers of the second powers of the coefficient and the first power of the coefficient and so on. So, this will turn out to be a simple quadratic equation in terms of those coefficients and quadratic equations have analytic solutions in terms of the roots of the quadratic equations. So, it is very easy and very quick to understand the solutions as to why they are important. So, if I have an equation like a y double prime as I have written down b y prime plus c c y is equal to 0 and if I propose a solution y is equal to a constant times an exponential say k x we could choose minus k, but it does not matter because when you solve for the y using this form you will get different values for k. So, if you do that d y by d x is k y and d square y by d x square is k squared y and so you will immediately get the equation a k squared plus b k plus c times y is equal to 0 and since y of x is not 0 everywhere arbitrarily it is non 0. You have the quantity a k square plus b k plus c is equal to 0 and this gives you the solution for k as minus b plus or minus square root of b squared minus 4 k c by 2 a. So, you get two values for k k which you can call it as k 1 and k 2, k 1 is obviously one of the roots minus b plus square root of b square minus 4 a c by 2 a and k 2 is minus b minus square root of b square minus 4 a c by 2 a and given these two roots the general solution for the differential equation will be a e to the is equal to a e to the let me just get this right e to the k 1 x plus b e to the k 2 x where k 1 and k 2 are given by this. So, that becomes the solution of the differential equation. So, when you have constant coefficients for a second order differential equation it is nothing other than a simple quadratic form, but interesting thing is when k 1 and k 2 are equal or when k 1 and k 2 are not real, but complex then you have to worry about some special forms. If k 1 and k 2 are not equal to each other real or complex the solution is easy to write down. So, the best would be to illustrate this using a simple example. Let me take an example for solving such an equation. The examples are there in the assignments in the website, but let me see if I have one here. Yeah. So, let us take two examples and we will illustrate the differences between them. For example, one problem that we have is y double prime minus 4 y prime minus 5 y is equal to 0 ok. You have just seen that this with the substitution y is equal to a e to the k x this essentially goes down to k squared minus 4 k minus 5 is equal to 0 which is easily factored into k minus 5 times k plus 1 is equal to 0. Therefore, k is equal to minus 5 k is equal to 5 or minus 1 mean that is a quadratic equation solution anyway. So, you have immediately the solution of the differential equation is y of x is a e to the 5 x plus b e to the minus x easy real different values for k. Now, an interesting variation of this would be to solve this differential equation y double prime minus 4 y prime instead of minus 5 let us put plus 5 y is equal to 0. Then if you factorize this what you would get is k squared minus 4 k plus 5 is equal to 0 and this the roots have to be solved through this discriminant form. So, you will get k is equal to 2 plus or minus 2 i b minus b plus or minus b square root of b square minus 4 a c all those things you get 2 roots. Therefore, you have the solution y of x is equal to a e to the 2 plus 2 i of x and b e to the 2 minus 2 i of x which you can rewrite as e raise to 2 x a e to the 2 i of x plus b e to the minus 2 i of x and this you know is cos 2 x plus i sin 2 x and this is cos 2 x minus i sin 2 x and therefore, the solution y of x becomes e raise to 2 x you collect all the cosine terms and the sin terms you have complex coefficients c is equal to times cos 2 x plus d times sin 2 x and you get cos 2 x plus b e to the you know that c is a plus b and b is i times a minus b i a minus i b that is what you have. But you have an oscillatory term with the growing exponential that is because the leading term or the real part is positive. If the real part is negative you will have a decaying exponential but oscillatory. So, this is what happens if the coefficients k 1 and k 2 are imaginary or complex. What happens if they are both the same? For example, y double prime minus 4 y prime plus 4 y is equal to 0 which when you factor out it is k square minus 4 k plus 4 is equal to 0 or it is k minus 2 square. Therefore, k is equal to 2 is the only solution is the only answer. Does that mean that y of x is equal to a e to the 2 i of x minus 2 x is the only solution? No. Does that mean? No. Because it is a second order differential equation we expect two general solutions or general solutions with two arbitrary constants and therefore, there is another form that we need to worry about to in instead of explaining why that form is let me give you that particular form namely we write this y of x as a e to the 2 x plus a general form b times x e to the 2 x that it will turn out to be only x and not x square or x cube etcetera you can easily see that. But if we have x to the e to the 2 x also will be a solution to this equation. So, let me just do this in a slightly different way y it is x that is a sorry write this y double prime minus 4 y prime plus 4 is equal to 0. We will write this as d minus 2 introducing the notation that I put in in the very beginning d f 1 minus 2 f 1 is equal to 0. Where f 1 is a solution is a solution ok. So, we have already found out f 1 to be a e to the 2 x. So, if there is another solution f 2 what we need to do is to try out f 2 is equal to g of x e raise to 2 x we will keep that because exponential is always there for all the second order constant coefficients differential equations. But we need to now determine the arbitrary function g of x and in a minute in a few minutes you will find out that this is nothing other than x ok. So, let us look at this if f 2 is a solution then d by d x sorry d is the operator d d minus 2 times d f 2 minus 2 f 2 is equal to 0 because that is the differential equation d square minus 4 d plus 4 f 2 is equal to 0 which is factored into these operator forms. Now, the derivative if you substitute for d f 2 using the g of x e raise to 2 x what you will have is d by d x minus 2 d g by d x minus 2 d g e to the 2 x is equal to 0 ok or d square g by d x square d g by d x of e to the 2 x minus 2 d g by d x of e to the 2 x is equal to 0. And therefore, what you have is d g by d square g by d x square times e to the 2 x is equal to 0 because this 2 will cancel with this 2. I have done only differentiate this whole thing using d by d x this one. So, what you have is d square g by d x square is a is equal to 0 which means d g by d x is a constant or g is a constant times x. So, you see that it comes out immediately when you substitute a general form a general form f of x this is the general form that we used sorry the general form that we used was f 2 of x g f x. If you substitute that in the differential equation you immediately get this answer ok. Therefore, the general solution for the differential equation is one is the a e raise to 2 x the other is the x e raise to 2 x. And this is the form for the homogeneous differential equation with constant coefficients such that you get a perfect square for the differential forms like this. So, there are many tricks of the trade here and the job of a theoretician is of course to keep as one of my post doctoral colleagues senior colleagues when I was under training when I was associated with him kept on telling that you have to collect the tools and your toolbox should keep on piling up mathematical techniques for theoreticians. One should continue to to learn and keep them ready and use the necessary ones. So, differential equations are are many methods the homogeneous linear second order differential equation has some very simple solutions as we have seen. In the next lecture I will take this a little bit further and talk about the general solutions for the differential equations with the coefficients which are not constants, but functions like the p and the q that we wrote down. And then probably given some time we will also look at one important method for chemists namely an infinite series representation of the solution and then solving the differential equations using an infinite series. We will assume the infinite series to be convergent all those details will not be discussed. We will come back to the differential equations in the next lecture. Until then thank you very much.