 Good morning. So, in this lecture, we start with second order linear differential equation. First the homogeneous cases and then we will go to the non-homogeneous cases. As outlined in the previous lecture, we will start with the simplest case, which is this. Homogeneous differential equation with constant coefficient and then we will consider the homogeneous differential equation with variable coefficient. And then one by one, we will consider these more difficult cases. So, first the simplest differential equation of second order. So, for this differential equation, we try to find the solution and the search is for a function y with upon differentiation produces such functions with in an additive manner can cancel one another. So, we look for the type of function with upon differentiation will produce its own kind. So, that a sum of such functions can vanish together. So, what kind of a function we are talking about? Exponential function fits the description, because the derivative of e to the power x is e to the power x itself. Derivative of e to the power k x will be k into that is constant into e to the power k x itself. So, then the function itself and its derivative with certain coefficients here can be added together to produce the same kind of function. And as the total, if the total coefficient can be made to vanish, then as a sum we can get 0. So, we know what kind of solution we will expect. So, we assume y equal to e to the power lambda x and then simply differentiate it twice. So, the first derivative we get as lambda into e to the power lambda x and the second derivative as lambda square into e to the power lambda x. Now, these three expressions if we insert in this, then we get this equation. And now, we say that since this part cannot be 0. So, for this equation to be satisfied, we must have this equal to 0. And this equation is called the auxiliary equation of the differential equation. So, from here directly we can say that if we are looking for this coefficient lambda in the exponent, then from here we write lambda square from here we write a lambda plus b. And that equal to 0 gives us the auxiliary equation for this differential equation. And since this is a quadratic equation. So, we will expect two roots from here. That is two solutions of this quadratic equation. And that is very easy. So, we solve for the two solutions of this quadratic equation. Let us call them lambda 1 and lambda 2. And then, putting those values in turn here we will get two solutions. And we expected that because it is a second order differential equation. So, e to the power lambda 1 x and e to the power lambda 2 x are the two solutions that will satisfy this. And now, if e to the power lambda 1 x satisfies, then any multiple also will satisfy similarly for this. However, a quadratic equation can yield three types of solutions. And according to that there will be certain variations in the form of the solution that we will get. First case is the real and distinct solutions. That is when lambda 1 and lambda 2 are both real, but they are unequal. That is the case when the discriminant of this quadratic equation is positive. That is when a square is greater than 4 b. Then we will have lambda 1 and lambda 2 to distinct real solutions. And in that case, these two will be linearly independent. And these two in a linear combination will provide us the complete solution. That is this. Now, if the two roots of this quadratic polynomial turns out to be same. That is real and equal. Then e to the power lambda x, e to the power lambda x, they will be the same solution repeated. And therefore, they will not be two linearly independent solutions. It will be the same solution. So, therefore, we cannot linearly combine like this to get the general solution or the complete solution. So, what to do with that? That is in this case if a square is equal to 4 b. That is the discriminant is 0. Then we get both roots same. That is simply if here a square is equal to 4 b. Then what we have here is lambda square plus root over 4 b lambda plus b. That means, we will get the common value of lambda as minus a by 2. So, in that case, we will have both the values of same as minus a by 2. And the only solution in hand is this. So, we need a method to develop another solution. Towards the later part of this lesson of this lecture, we will see a methodical way to find that second solution when we have one solution in hand. But currently, let us simply verify that x into e to the power lambda x is another solution of the same differential equation. So, let us try to insert y 2 that is x into e to the power lambda x into this equation. Where lambda is minus a by 2 and a square is equal to 4 b in this particular case. So, we try to insert this solution into the differential equation and see whether it satisfies the differential equation. And then the second derivative will be the derivative of this. So, we will get lambda from here and another lambda from here. So, here the derivative of this has been included and the derivative of this, the first part of it in which this is differentiated is included. And the second one in which this part is differentiated that will give us this. Now, we will multiply this with a and this with b and add up. So, e to the power lambda x terms we will get from these two cases and x into e to the power lambda x terms we will get from here, here and here. So, let us put all of them together. So, this plus a times this plus. Now, x e to the power lambda x with that we will get lambda square from here a times this and b times this. Now, we have already seen that lambda is a solution of this. That means lambda square plus a lambda plus b is 0. So, this is 0 and in this particular case lambda happens to be minus a by 2. So, twice lambda plus a this is also 0. So, that means these two terms independently vanish and therefore, the sum is 0. That means this sum which we have constructed the left hand side of the differential equation that is satisfied. So, here we just verify that this is another solution and obviously, this solution x into e to the power lambda x is linearly independent from e to the power lambda x because the ratio of the two solutions y 1 and y 2 is not constant it is x. So, then like this we can construct two linearly independent solutions and then we can get the general solution as a linear combination of these two linearly independent solutions. This is this. The third possible case of the solutions of that quadratic equation is when the discriminant is negative and a square is less than 4 b then we get the two roots as complex conjugates like this minus a by 2 plus minus i omega and in that case these are certainly the corresponding solutions e to the power minus a by 2 plus i omega x and e to the power minus i omega a by 2 minus i omega x they are certainly linearly independent. However, this is not a very nice form of writing this solution. So, we reorganize the solution a little bit e to the power minus a by 2 x we take outside and then inside we will have c 1 into e to the power i omega x which is this and on this side we will have plus c 2 into e to the power minus i omega x which is this and then we club together cosine terms which we get as c 1 plus c 2 let us call it a and we club together the sin terms which will be i into c 1 minus c 2 let us call that b then this becomes the more nice looking form more elegant form of the same solution in terms of two new constants a and b. There is a third form also which is quite useful in many situations that is in terms of the phase. So, what we can say is that rather than having a here and b here if we say that let us call a square plus b square under root as c then we can make the sin cos substitution here and call a equal to c cos alpha and b equal to c sin alpha then getting c outside we can put together cos omega x cos alpha plus sin omega x sin alpha which is this. So, this is a third form of the same solution which is also quite often found useful. So, in this in all the three cases we can find the complete solution for this differential equation. Now, apart from this differential equation this one there is another differential equation linear, but not with constant coefficient. Currently we are discussing the solutions of homogeneous equations with constant coefficient, but then there is a particular kind of differential equation which is not with constant coefficient, but in many aspects it resembles this differential equation and that is the famous other differential equation in which the coefficient of y double prime is x square coefficient of y prime is x and coefficient of y is 1 that is apart from constant coefficient a and b. So, constant coefficient can come there and apart from that the variable part of the coefficient has whatever is the power of that for the coefficient of y there is one more power in y prime and another more power in y double prime and so on. So, in this particular case in order to have a sum of these terms in such a manner that together they can vanish that is one can compensate for the other we will need all of them to be same type of functions and that means that we should have a function here which upon differentiation gives another function which is 1 degree less and that such that when that multiplied with x will produce something which will be similar to y. So, that it can be together added similarly here another differentiation should involve another degree less. So, such that another multiplication with x will again produce similar things we know that x to the power k is a function like that which as many times as you differentiate it degree will go on reducing. So, if we substitute a function of this kind in the case of other equation then very quickly a similar auxiliary equation we can develop in that other case it was lambda square plus a lambda plus b but in this case it will be k square plus a minus 1 k plus b that can be verified easily by just two differentiations and substitutions in the same manner as we handle the previous equation. Anyway then we get an auxiliary equation or initial equation and we know that those values of k which satisfy this equation if inserted here and then x to the power k is taken then it will satisfy this differential equation. So, our first job is to solve this quadratic equation and again similar to the last case we get three cases if the roots are real and distinct then we will get two different values of k 1 k 2 both real immediately then we put x k x to the power k 1 and x to the power k 2 and this gives the general solution. On the other hand if roots are real and equal then this will be this will give us one solution as x to the power k but the other solution other linearly independent solution similarly we can develop as logarithm of x into x to the power k that will give the second linearly independent solution. The I would advise that this particular case you should verify the way we verify the case in the previous example previous differential equation in the case of roots being complex conjugate you have k 1 and k 2 which are like this. And again in this case also rather than having the solutions in terms of the complex numbers we can club together the cosine terms and the sin terms and develop the solution in this manner. The real part of this k is taken outside and the imaginary part is put inside so that it can be clubbed together in terms of cosine and sin. And again through sin cos substitution another form can be obtained in the with the help of the phase. So, this entire set of solutions can also be obtained very easily by making a substitution of the independent variable rather than x being here if we insert x equal to e to the power t that is a very elegant approach because if we insert x equal to e to the power t in which case t is log of x and then dx by dt turns out to be x itself and dt by dx is 1 by x and all these substitutions with further derivatives. If we insert here then we will find that this differential equation which is in terms of x can get reduced to a differential equation in terms of t and that differential equation will be this. And therefore, this particular differential equation Euler-Costier equation has a very close relationship with the differential equation with constant coefficient which we have studied earlier. Now, this much for the case of homogeneous differential equations with constant coefficient and it is immediate cousin which is the Euler-Costier equation. Now, let us go back and see what we planned to do after that for this the entire complete solution is quite simple. Now, our next issue is to solve a similar differential equation with variable coefficient that is where coefficients are functions of x. Now, note that all through this discussion we will consider these coefficient functions p x and q x and in this case r x also in this cases r x also as functions which are continuous and bounded. And therefore, we will get the advantage of the existence and uniqueness theorem which we discussed in an earlier lecture. So, now we consider the fundamental theory of homogeneous equations that is the second order homogeneous equation we take like this. And first we use the well posedness of its IVP. Now, we already established that a linear differential equation with coefficient functions which are continuous and bounded is well posed with an arbitrary set of initial conditions. So, the initial value problem of this ordinary differential equation with arbitrary initial conditions any position and any speed any y and any y prime at the initial point x 0 as a unique solution and it depends continuously on the initial conditions as long as p x and q x are continuous and bounded also in the interval in which the solution is being studied. So, this result this particular fact we will use in establishing some of the results in a quite straight forward manner. Now, one issue can be very easily notice that at least two linearly independent solutions we can see very clearly. In one case we consider these initial conditions that is at x 0 y is 1 and y prime is 0. And let us call that solution as y 1 that is that solution of this differential equation which has value 1 at x 0 and rate 0 at x 0. Another solution we can consider as that solution of this differential equation with initial conditions y at x 0 x 0 and y prime is 1. So, these are certainly two linearly independent solutions because if you consider a linear combination of this to vanish that is this plus this is equal to 0. Then just simply by putting one of the value x 0 we find that y 2 of x 0 is 0. So, this goes to be this goes to 0 and y 1 of x 0 is 1. So, here we get only c 1 the right side left side is c 1 which is 0. Similarly, if we differentiate it and then insert the initial condition x 0 then we find that the c 2 coefficient is also 0. That means a linear combination of these two solutions being 0 necessarily means that the two contributions are independently 0 and that is the definition of linear independence. So, if you can if we can think of two solutions of this differential equation which have these initial conditions we can see very easily that the two solutions are linearly independent that shows that at least two linearly independent solutions of this we can find. Now, we said at least two linearly independent solutions which we can see very clearly. Can we say that there will be at most two linearly independent solutions also that is other than these two we can find no other solution which will be linearly independent to both of them the answer is yes. We can say also this that is two and exactly two linearly independent solutions we can find out for this differential equation. And to establish that let us consider one or two important points first is the definition of Ronskian. Ronskian function of two solutions y 1 and y 2 two functions is defined in this manner the determinant of this 2 by 2 matrix y 1 y 2 y 1 prime y 2 prime. So, that will be y 1 y 2 prime minus y 2 y 1 prime this is defined as the Ronskian of these two solutions. Now, the important result is that two solutions y 1 and y 2 are linearly independent if and only if there is some value x 0 where the Ronskian vanishes. Now, if we want to establish this result then we need to establish two points. One is that if the Ronskian vanishes at some point then they are linearly dependent and other is that if they are linearly dependent then the Ronskian will vanish. So, let us first consider that if the two solutions are linearly dependent then there is some point where the Ronskian will vanish. So, in order to establish that we take this y 1 and y 2 and consider them to be linearly dependent. Two functions linearly dependent means that one will be k times the other that is the two should be proportional. If we take that then y 2 prime will be k y prime. So, in the expression for the Ronskian we simply insert y 2 as k y 1 and y 2 prime as k y 1 prime and we find that k will be common and we will have y 1 y 1 prime minus y 1 y 1 prime which is 0. So, that shows that the Ronskian is 0 everywhere in particular at some point x 0 it will be 0. So, forward part of the result we have found very easily that is if y 1 and y 2 are linearly dependent then the Ronskian vanishes everywhere not only at x 0. So, in particular at x 0 it vanishes. Now, we want to show the converse that is if at some value x 0 the Ronskian vanishes then the two solutions are linearly dependent. So, say if there is a value x 0 for the Ronskian vanishes like this till now we have not shown that it vanishes everywhere. We have just in the converse proof that is we have taken some value x 0 where the Ronskian vanishes that is the premise. Now, if this is 0 this determinant is 0 that means the corresponding matrix is singular that is this matrix now if this matrix is singular then it will have a null space which means that there can be non-zero vector c 1 c 2 which will be in the null space of this matrix that is which will be the solution of this linear system homogenous linear system of equations. Now, we choose such a non-zero vector c 1 c 2 which is a solution of this that is we choose a vector in a null space of this matrix. Since this matrix is singular we will always have a null space. So, we choose a vector in that null space a non-zero vector in that null space and construct this function with these c 1 c 2 values. Now, we claim that this function so constructed with these values c 1 and c 2 satisfies this initial value problem the same differential equation and 0 initial condition. You can see very easily that this is true because y 1 is a different is a solution of this differential equation. So, c 1 y 1 is also a solution similarly c 2 y 2 and when you insert this you will have c 1 into y 1 double prime plus p y 1 prime plus q y 1 which will vanish plus c 2 into y 2 double prime plus p y 2 prime plus q y 2 which will vanish. So, this function satisfies the differential equation. So, far as satisfying the initial conditions is concerned you can see that you evaluate this at x 0 and you get c 1 y 1 at x 0 plus c 2 y 2 at x 0 that is 0 that is immediately satisfied because c 1 c 2 is a solution of this system of linear equation. So, c 1 y 1 at x 0 plus c 2 y 2 at x 0 equal to 0 that is the first row of this vector equation. Similarly, the second row of this vector equation the c 1 y 1 prime plus c 2 y 2 prime at x 0 is 0 tells us that this function also satisfies the second condition. Now, you see that this function satisfies the differential equation and both the initial conditions that means that this is a solution of this initial value problem. Till now we are saying this is a solution of this initial value problem, but remember that in a previous lecture we established here that a linear differential equation with continuous and bounded coefficient functions has unique solution with any arbitrary set of initial conditions. That means IVP of this differential equation for any initial conditions any set of initial conditions is unique. So, that means that this is a solution of this IVP means this is the unique solution of this IVP, but then we can also see that y equal to 0 is certainly a solution of this y equal to 0 satisfies this differential equation trivially this initial condition trivially and derivative equal to 0 trivially. So, y equal to 0 satisfies this initial condition initial value problem trivially and just now we have seen that this is the unique solution how can that be that can be possible only in one way in which this solution is nothing but 0, because this is the unique solution of this IVP and we can see that y equal to 0 is certainly a solution of this. So, that can be the unique solution if it is that same solution y equal to 0. So, that means this solution that we have so constructed is actually equal to 0 that shows what that shows that we can find two numbers C 1 and C 2 not both 0 such that this vector is a non-zero vector such that this sum vanishes without C 1 and C 2 being both 0 that means that the two functions y 1 and y 2 are linearly dependent. Now, we have shown the converse also that means if there is a value at x 0 value x 0 where the Ronskian vanishes then that will imply that the two solutions y 1 and y 2 are linearly dependent and that will mean from here you see that the Ronskian vanishes everywhere in a circular manner that will mean that if the Ronskian vanishes that vanishes at some point then it will vanish everywhere. So, as a consequence of this we find that if the Ronskian is 0 then not only it means that the two solutions are linearly independent linearly dependent it also means that the Ronskian will vanish everywhere for all x. And now if the Ronskian is found to be non-zero at some point for a situation then that also will mean that it will be non-zero everywhere because if somewhere else it is 0 that will imply that it is 0 everywhere which will contradict with this that means at one point if you find the Ronskian to be non-zero from there you can directly claim that it will be non-zero always and therefore the Ronskian function will never change sign. If it is positive then it will remain positive for all values of x if it is negative then it will remain negative everywhere because due to continuity from for going from positive to negative or vice versa it has to cross 0 which it cannot do. So, we find that if the Ronskian can be shown to be non-zero at one point immediately we can conclude that it is non-zero everywhere and y 1 and y 2 are linearly independent solutions. And now the general solution you will get in that case as a linear combination of the two and we can say that this general solution is also the complete solution. What does it mean? It means that no third solution is possible which is linearly independent to both of these and that means this is the complete solution that is any solution for the differential equation that you can find can be certainly put in the form of a linear combination of these two which means that there is no single solution for the linear O D. If we want to show that then what we will do? We will pick a solution candidate solution suppose capital Y is a solution of the differential equation and then we will try to put it in the form of a linear combination of y 1 and y 2 which are two linearly independent solutions. If we succeed that means that for any solution of the differential equation we can always put it in this manner. So, what we do is that for this y of x we choose a point x 0 and evaluate the two basis members two solutions y 1 and y 2 which we found earlier and this new solution also. For all these three solutions we find the value at x 0 and the values of their derivative also at that point and then the values of y 1 y 2 we put here values of their derivatives we put here and the y and y prime values at the same point we put here and then we construct this linear system of equations and ask for the values of c 1 and c 2. Now, see y 1 and y 2 are two linearly independent solutions so that means their Ronskian is non zero which means that the determinant of this is non zero which means this is a non singular matrix. If this is a non singular matrix then when we ask for values of c 1 c 2 satisfying this we get a unique solution that is unique values of c 1 and c 2 we get. Now, as we get that unique values of c 1 and c 2 and then we construct with the help of this c 1 and c 2 this particular solution y 1 and y 2 are linearly independent solutions of this differential equation. So, now with the help of these coefficients which we found from the solution of this we develop this particular solution y star and then we know that this y star satisfies this differential equation and this y star must satisfy these two initial conditions also why so because c 1 into y 1 c 1 y 1 plus c 2 y 2 at x 0 that is the first equation in this system c 1 y 1 plus c 2 y 2 at x 0 that is equal to this the solution of this system of equation c c 1 c 2. So, this one is evaluated this function when evaluated at x 0 gives us the left side of this first equation which is equal to this. So, it satisfies this similarly from the second line second row of this equation we find that this condition also is satisfied. So, that means this function y star that we have constructed out of the solution of this linear system of equations that satisfies this initial value problem that satisfies this differential equation and that satisfies this initial condition that means it is a solution of the initial value problem. Now, again based on the uniqueness theorem we know that if this is a solution if this is a solution of this initial value problem then it is the unique solution of the initial value problem. But then the way we defined capital Y and its derivative, capital Y itself the original candidate solution itself is also a solution of this. That means this solution that we have constructed with the help of this T 1 and T 2 happens to be exactly same as capital Y, because of the uniqueness of this solution. So, that is the candidate function itself. So, that means that candidate function which we picked up from where we started turns out to be expressible as a linear combination of the two solutions Y 1 and Y 2 which we started with and that shows that there is no third solution possible which is linearly independent of Y 1 and Y 2. So, that is why we say that for completely describing the solutions of the second order differential equation we need two and two only linearly independent solutions of it and that gives us the complete solution. However, based on the differential equation itself if we want to find out two linearly independent solutions then there is no guaranteed procedure to identify two such solutions analytically in general. And that is a something block there is however a way to find a second solution if we have already in hand one solution that is if we have one solution in hand that is if Y 1 is available which is already known to satisfy the differential equation and we want to find another solution say Y 2 which is linearly independent of Y 1 then there is a method to do that and that is called reduction of order. That means if we want to solve a general second order differential equation homogenous differential equation then in the most difficult case we may not be able to identify analytically two linearly independent solutions. However if from somewhere through some consideration we can identify one solution then onwards we can completely solve the problem. That means we can find out a second solution which is linearly independent to this and we can combine the two in a linear manner C 1 Y 1 plus C 2 Y 2 and get the complete solution. The way we do that is the method called reduction of orders. So, suppose Y 1 is a solution of the differential equation Y double prime plus P y prime plus Q y equal to 0 and we want to find a second solution which is linearly independent of the first. So, we assume the second solution as u x into y 1 now as long as u x is variable depends on x this will turn out to be linearly independent of y 1. So, then taking u y u y 1 as a second solution we force this to satisfy the differential equation. So, as we insert its second derivative first derivative and the function itself in the differential equation. So, the second derivative of this is u double prime y 1 plus twice u prime y 1 plus u y 1 double prime this is y 2 double prime plus P y prime y prime y 2 prime. So, u prime y 1 plus u y 1 prime. So, this is first derivative plus Q into y 2. So, this is the result of substitution of this y 2 into the differential equation. Now, here we club together this term this term and this term in all these three terms you find u is common. So, take this term here then this term here and this term here u is taken outside the other three terms that is this is here this is here and from here this term is here. Now, y 1 is already available as a solution of the differential equation. So, this is 0 and therefore, we have this equal to 0. This part turns out to be 0 and now you see that here in this differential equation in terms of u, u double prime is appearing u prime is appearing u itself is not appearing because all the terms containing u have together vanished. Then what we can do is that this u prime we can call as something else. Let us call it as capital U and then we find that here we have y 1 into capital U prime plus this thing into capital U. Now, if you divide it with y 1 then here you will get capital U prime plus this expression divided by y 1 into capital U. So, we get this is the differential equation that we get in capital U which is actually small u prime derivative of small u. Now, this is the reduction of order. Now, this is the first order differential equation and as we rearrange this that is as we call this u prime as d u by d x then multiply overall with d x divide overall with u then we get d u by u here which is this and plus d 2 plus twice d y 1 by d x. So, multiplication with d x we will take away that d x. So, we will have d y 1 by y 1 plus p into d x this u has gone here. So, we will get this. Now, we can see that as we integrate this differential this will be l n u this will be twice l n y 1 plus this will be simply integral p d x equal to constant or taking exponential all over this sum we will get converted to product and therefore, we will get u into y 1 square into e to the power integral p x p d x. So, this whole thing will be constant whatever constant we choose we reach the same final result. So, we can choose as 1 and as we do that we can expression we can get an expression of u in terms of everything else that is we take all these things and divide here. So, this capital u turns out to be 1 by y 1 square into e to the power this negative because it is going on the other side. So, this is capital u which is actually u prime. So, we simply integrate it. So, the integral gives us u x the coefficient function which needs to be multiplied with u 1 to get the second solution y 2. So, this is the way u into y 1 that gives us the second independent solution second linearly independent solution of the same differential equation and this is linearly independent because u x is not constant and see that u x the factor u x can never be constant because for it to be constant this integrand must be 0 and that will not be possible because the way the solutions we are constructing y 1 will be continuous and bounded. So, this cannot be 0. Similarly, this p is continuous and bounded. So, its integral will not be minus infinity or something like that. So, infinite infinity and minus infinity this will not be. So, therefore, this cannot be 0 and similarly y 1 cannot be infinite. So, this factor also cannot be 0. So, product cannot be 0 and therefore, its integral cannot be constant. So, therefore, u will be always variable and therefore, y 1 into u will be linearly independent of y 1. So, this way if we have one solution in hand from there we can work out another solution which is linearly independent. Now, you will remember that in this particular case I told you that currently for the time being let us just verify this. If we wanted to find the second linearly independent solution with this y 1 known we could have done that with the use of this reduction of order and in that case after the necessary steps we would get u is equal to x and that would show that x into e to the power lambda x is another solution which is linearly independent of the first solution. So, now that we have got this. So, let us summarize the findings till this point and see a particular function space for the perspective of this solution of differential equations. Operator D that is D by D x means differentiation and in the context of function space it operates on an infinite dimensional function space as a linear transformation. That means it maps all functions in the infinite dimensional function space to other functions in particular it maps all constant functions to 0 and that one dimensional subspace of the function space is its null space. The second derivative or d square is another operator that has a two dimensional space that is q 1 plus t 2 x with a basis which is this. So, all linear combinations of 1 and x when operated upon by this second order operator. So, that vanishes that means this has a two dimensional null space. You can think of composite operators like this D plus A that is D by D x operated over something plus A into that something A is a constant number. This again will have a null space like this x D plus A is another first order operator which will have a null which is this. Similarly, a second order linear operator which is this now this operated over y equal to 0 gives us the second order differential equation which we have been studying till now. Now, this itself apart from the y part of it is a second order linear operator and it possesses a two dimensional null space. So, the differential equation which we have been discussing till now that the solution of that basically turns out to be the finding of the null space of this second order linear operator or in particular we try to find a basis for it. That means basis means two linearly independent members of that null space and this issue is very analogous to the solution of this homogeneous system of linear equations and that is the development of a basis for the null space of A. Now, as we have seen that this solution y is a member of the infinite dimensional vector space of continuous functions. Here this x the solution is a member of a finite dimensional vector space and apart from that many things are common. Now, to take the analogy further we consider the right hand side also and complete this analogy in ordinary vector space of finite dimension and in the function space of infinite dimension. This A x equal to b a system of non homogeneous equation. The corresponding differential equation is this here the unknown is x which is a vector of finite dimension here the unknown is a function y which is a vector of infinite dimension. Now, for this saying that system is consistent corresponding statement here is p x q x and r x are continuous and bounded functions. Now, a solution x star here is analogous to a solution a particular solution y p of this differential equation. For this linear system of equations apart from x star if there is an alternative solution x bar then similarly in this case consider another alternative solution y bar. Now, the way in the case of ordinary linear systems of equations two different particular solutions give a difference which satisfy A x equal to 0. The corresponding homogeneous system of equations similarly here y bar minus y p will be another function which will satisfy this differential equation. The same left hand side, but in the right hand side it is 0 and here it will mean that the corresponding difference is in the null space of the coefficient matrix A and here it will mean that the corresponding difference is in the null space of the linear operator d 2 d square plus p d plus q. Now, here also we found the complete solution by adding a complete basis member of the null space to a particular solution same thing we will do here when we need to solve this non homogeneous equation that is one particular solution added to a general member of the null space will give the complete solution of this non homogeneous equation. Till now we were solving the homogeneous equation now we get into the non homogeneous equation also. What is the methodology to solve this and find the most general solution? We find the null space of A that is basis members of the null space and then one particular solution of this different this system of equations we find and compose in this manner. Similarly here we will first find the null space of this which is the solution of the corresponding complete solution of the corresponding homogeneous equation which we were doing just till now this equal to 0 and then find out one particular solution of this which is y p and compose in this manner. One particular solution of the non homogeneous equation added to a general member a general solution of the corresponding homogeneous equation gives us a general solution or the complete solution of this non homogeneous equation. So, this is the way we will now next take up the problem of solving this differential equation which is non homogeneous. First we will start with the case of constant coefficient that is in this manner we will consider the first case with constant coefficient and see one simple method. So, what we will do? We will first solve the corresponding homogeneous equation in any case as this equal to 0 and then obtain the basis with two solutions and construct this. Now, note that this is not a solution of this differential equation this is the solution of the corresponding homogeneous equation and it is shown as y a this is sometimes called the complementary function. Next we will find one solution of this differential equation non homogeneous equation and then construct the complete solution of this and finally, if there are some initial or boundary conditions known then those conditions can be now imposed to determine the particular solution with the determination of the values of c 1 and c 2. Now, as I just now noted that y 1 and y 2 and therefore, this is not a solution of this equation, but they are solutions of the corresponding homogeneous equation and therefore, and in another manner if you find two solutions which are solutions of this you do not expect the linear combination of that to satisfy this differential equation as it did in the case of the homogeneous equation. Here the impact the implication of linearity of the differential equation is in the sense of super position that is with zero initial condition if y 1 and y 2 are responses for two different functions r 1 and r 2 then the response or the solution corresponding to c 1 r 1 plus c 2 r 2 will be c 1 y 1 plus c 2 y 2 in that sense linearity operate on this non homogeneous differential Now, we first consider the case with constant coefficient in which a particular very simple method works and that is the method of undetermined coefficient. Let us take this non homogeneous equation with constant coefficient. Now, if r x is a function of a few particular kinds and if the coefficients are constant as shown here then this method of undetermined coefficient will work very easily. In that what we do we choose certain type of function for y and then substitute to find out the undetermined coefficient. So, for example, let us consider if r x is x to the power n then what kind of y p we should choose which will satisfy this. Now, understand that before we take up this question we have already found the complete solution of the corresponding homogeneous equation y double prime plus a y prime plus b y equal to 0 for that we have got y 1 and y 2 and therefore, c 1 y 1 plus c 2 y 2 is the complementary function that is y a solution complete solution of the corresponding homogeneous equation that we have already got. Now, we are looking for a particular solution of this non homogeneous equation. So, if r x is x to the power n then we are what kind of y p we are looking for. Now, in order to satisfy this x n x to the power n part something here should give a similar term which will cancel with this in order to satisfy this equation. So, if y y p we take as x n then that can be managed with this part, but then its derivative will have x to the power n minus 1 there is nothing on this side to cancel that. So, what we have to do is that we have to cancel that x n minus 1 and that x n minus 2 that will come here with the help of these fellows only. So, apart from choosing a term containing x to the power n we must include a term containing x to the power n minus 1 as well which in turn will produce x to the power n minus 2 x to the power n minus 3 and so on. Now, that will mean that 1 by 1 we will end up including x to the power n minus 2 then n minus 3 n n minus 4 till we find that we are at the end x to the power 1 x to the power 0 which is constant and that is it. So, that means that for positive integer value n here x to the power n whatever is n choosing y p to be a function of k function of type k into x to the power n will not suffice, but an entire n s degree polynomial we have to include and apart from x to the power n if in r x many other terms are also present x to the power n minus 2 n minus 4 n minus 3 n minus 1 for all of that you will have the same thing. Now similarly, if r x is an exponential function then the similar exponential function e to the power lambda x we can include and that through derivatives can cancel this part this side. So, what we will do for this we will choose k into e to the power lambda x and try to fit the function here and then equate both sides to determine the value of k. Now, if we have a sum like this appearing here then we use the superposition that is k 1 into r 1 plus k 2 into r 2 if that is here then we will choose the function in the form k 1 into this plus k 2 into this time. Now, here we use the principle of superposition to summarize if the right hand side function is a polynomial of degree n then the candidate solution also should be chosen as a polynomial of degree n. Now note that some of the terms missing here will not help us in reducing the terms here we as long as the term x to the power n is present here we have to start with a complete polynomial of n s degree and coefficients we have to determine it may be a full polynomial as a result which has all the terms. If we have the RHS function as e to the power lambda x then we choose this as the candidate function candidate solution and try to find out the value of k. If the right hand side function is cosine or sin or a linear combination of the two then in all these cases we have to choose this and determine k 1 and k 2 through substitution. Now, even if the right hand side function has only cosine or only sin till here in the candidate solution we must include both because through differentiation sin and cosine will produce the other kind also. Now, if there is a product of these two e to the power lambda x cosine omega x or e to the power lambda x sin omega x. Now, you will notice that e to the power lambda x cos or sin or this all of these are actually same kind of functions because through a quick reference to complex algebra you can convert these into this itself. So, therefore, here also the same rule applies if this is involved in R x or this or a combination of both then we will need to include a term like this and further polynomial into e to the power lambda x also will give the same kind of candidate solution. Now, all these we can do in the case of constant coefficient cases and this particular few candidate RHS functions right hand side functions here also there is a situation in which we may need to modify the rule. For example, in this case with the homogeneous equation giving solutions e to the power x and e to the power 5 x in the first case we can choose the candidate function as k into e to the power 3 x and that will give us the solution which is this. On the other hand in this second case there is a problem. In the second case the homogeneous equation itself has a solution which is e to the power 3 x and therefore, choosing k into e to the power 3 x as the candidate solution for the non homogeneous part will not succeed. Because this is already included here and it is evaluated on the left side as 0 that is this fellow when inserted here evaluates to 0. So, there is no way to satisfy this and in this case the candidate is not k into e to the power 3 x, but k x e to the power 3 x and the value of k turns out to be 1. In this case the homogeneous equation itself has not only e to the power 3 x, but x into e to the power 3 x as well and in that case choosing even this will not help and in this case we try to choose k x square e to the power 3 x and that suffices and the coefficient turns out to be half. So, this is a particular modification rule that is if the candidate function this or this or this is already a solution of the corresponding homogeneous equation with this or this satisfying the auxiliary equation then we need to modify the candidate solution by multiplying with x. Now, if that multiplied with x version along with the original version both are sitting as y 1 and y 2 here then we have to further modify it and take k into x square into e to the power lambda x. Now, this method succeeds with only constant coefficients and with only a selected few right hand side functions r x. In the next lecture we will consider the general method which is the method of variation of parameter which will succeed in all cases and after that in the next lecture we will continue to generalize the findings till now the discussion still now for second order differential equations in the case of the higher order differential equations without any limit to the order of the differential equations. So, these two things we will do in the next lecture. Thank you.