 Welcome back to this lecture. In the last lecture, we were introduced the second order linear equations. We have introduced the concept of Ronskian, and we also seen the solution space of homogeneous equation is two dimensional, which is in some sense a remarkable fact. But the only difficulty compared to your first order equations, where the solution could be easily found by a integral calculus problem. There is no general method to find the two independent solutions. So, if you have a procedure to find independent solution or if anybody can prove that independent solutions, then the n-th air solutions space can be understood as a linear combination of two independent solutions. Since there are no recipes, I will begin with two methods to how to find the independent solution. There is not very general method and still we do not have a complete procedure. So, we will start with one easy method. It may not work all the time. It depends on particular problems you are dealing with. It might work or might not work. So, let me recall the equation, general equation. Recall the general equation. That is nothing but L of y. For shorter notation, we will put it or you can also put L of y, y prime, y double prime. That is equal to y double prime plus p t, y prime plus q t, y equal to 0. This is the general equation, which we are going to introduce. So, the idea is to look for a solution of the phone. Look for a solution of the phone. There is no reason why such a solution we can look for, but then for certain equations, it will work out. We are not climbing U and V are solutions. We are looking for a solution in the product form U and V. We are profutivate enough. We will be able to find a solution, find U and V by solving some other easy equations. So, do a computation to that one. So, you can do the computation. I will not do the entire computation, but y equal to U prime V plus U V prime. Then, y double prime. You can compute y double prime. When you compute y double prime, you get U double prime V. Then, U prime V prime 2 times will come to U prime V prime plus U V double prime. Substitute in the equation. Let me call this equation to be 1. Substitute in 1. You get a bigger equation. You will give you a large equation. You have to substitute all that. You will get U double prime V plus 2 U prime V prime plus U V double prime plus p t y prime U substitute. You get U prime V plus U V prime plus q t U V equal to 0. That is all. The next step is to make the equation. Choose U in such a way that you try to remove the V prime term from the equation. You put in such a way that summation of this equation not this one. The V prime term, this one, this combined term V prime is absent. So, step choose U so that V prime term is absent. V prime term is absent. As I said, the idea is to look for get equations for U and V, which are simpler. That is, if U prime term is absent, you want to choose this way 2 U prime V prime plus p t U p t U this V prime equal to 0. That gives an equation for U prime is equal to 2 U prime plus p t U equal to 0. So, this is an easy equation. This is an easy equation for U prime. So, you solve for U and that is an easy part. Solving U for is an easy part because it is a first order equation. So, that is no difficulty absolutely. But the next result is that it is possible that because there is no V prime term, but then U is already determined. So, computing U double prime U etcetera is already known to you, but then there are no V prime term. So, you will have an equation. It is possible that the equation for V is not always true. If it is always true, it is an equation for V is easy to solve. So, let us see an example. If you are lucky, you will be able to do that. So, let us see a simple example. Again, y double prime plus 2 t y prime plus 1 plus t square y. There will be plenty of equations to solve equal to 0. So, what is P t here? P t is equal to 2 t. Q t is equal to… You should also get familiarized with how to solve equations. That is important, 1 plus t square. So, what do you get it? If you solve for U t, you can write down the equation U t is equal to… You can write down the earlier equation. So, equation for U t is this one. So, this is what your equation P t is easy. So, you can write down the equation for U and you can solve it. So, let me not spend time here. So, let me write. This will imply your U t is nothing but e power minus t square by 2. Now, substitute that is an easy part. Substituting and computing, let me avoid a computation here because that one can do it easily. Computing can write down equation. You can actually see that it will imply v double prime will be 0. You see, so you have an equation. This happens for this particular equation. So, looking for a solution of the form y equal to U v and this will give you v is equal to… C 1 t plus C 2 and that will imply your y t is of the form C 1 t plus C 2 into e power minus t square by 2. So, you have your solution easily in this case. So, many examples can be done that way. So, I will just show that example. Now, I will go to the next method. Method 2 is called the reduction of order. This method is quite good, but then you have to pay a price. The idea of this method, you need to know one solution. So, if you know one solution, it gives you a method, a systematic method to find the second independent solution. So, assume a solution y 1 is known. So, this is the price you have to pay. So, still we do not have this. So, if you are able to find one solution, this method will give you a systematic procedure by reducing the order to find the second independent solution. What will happen? If y 1 is a solution, a constant multiple cannot be a solution. Constant multiple will also be a solution, but then the constant multiple of y 1 cannot be an independent solution. So, that is the first observation one has to make it. And no constant multiple of y 1 will give you a second solution. So, the possible way to find a second solution instead of multiplying by a constant, you multiply by a function itself. So, look for a solution of the form y 2 equal to c t by t, c 1 by y t, you see, because if you take c as a constant multiple, you are not going to get an independent solution. Any constant multiple will always be a solution, but if you want to have a solution y 2, which is independent of the solution y 1. So, now do the computation. So, you have to do a bit of computation. I will not do the computation. So, the exercise is just to do the computation. What are the computations? You have to use it to find y 1 prime, y 2 prime, y 2 double prime and substitute in the equation and substitute in l y 2. So, you want l y 2 to be a solution to your equation. So, you l y u 2. You also use the fact that use the fact l y 1 equal to 0, because y 1 is already given to be a solution to your homogeneous equation. So, choose l y 1 equal to 0, then you want y 2 be a solution. So, use l y 2 equal to 0. If you do the computation, this will imply you will get an equation c prime of t y 1. y 1 is known to you plus you get c prime of t into 2 y 1 prime plus p is a 1 of y 1 is equal to 0. So, this is known to you, because y 1 is given to you. This y 1 is known to you. Though it looks like a second order for c prime, since second prime for c that is not c present here. So, you can write down this in principle c double prime of t by c prime of t is equal to minus 2 y 1 prime by y 1 minus p u c. So, you have an equation even though it looks like a second order equation, but if you put. So, let me use the color. If you put v is equal to c prime will imply v prime by v is equal to minus 2 y 1 prime minus y 1 minus p u c. So, that will give you, you can solve it. So, if you put v, solve for v first. So, that is why it is called an order reduction method of reduction of order. So, you are actually solving a first order equation here. So, you have a first order equation solve for v, which you can write it as explicitly, because y 1 is known to you. So, you can write for solve for v t and, but this implies once you get solve for v that is nothing but c prime of t is equal to v, again you are solving. So, you have a second equation to solve for c. So, you see first you solve for v here using the first order equation and then v is given to you and you can just write c t and c t is nothing but integral of v t d t. So, and v can be obtained from here, which is a function of t, which you can be solved for that which will give you. So, that is a method of reduction. So, let me give you an example again to begin with an example for you. So, you have an example, a simple example I am trying to give you t y double prime plus t square y double prime t y prime minus y equal to 0. As I told you again there is still it does not give how to determine the one solution that you have to do by trial and error or looking at the equation you may have a some way of finding that. For example here, so immediately looking at this equation properly, if I choose y equal to t y prime will be 1 and this will get vanished and since y equal to t this will get vanished. So, this immediately gives you this may be easy if that is easy we have nothing more to do it here. So, y t equal to t is a solution you see immediately we can see that y t equal to t is a solution and then how do the order reduction you look for another look. So, let me call it y 1 equal to look for y 2 t is equal to c t t look for y t c t t. So, how do you do that one? So, one more point I want to do here. So, we have determined a solution y 1 and y 2 you can actually show that this y 1 and y 2 are actually y 1 and y 2 are independent solutions. You have can compute the Ronskian from here all that you can see that you can compute the Ronskian of y 1 and y 2 and see that the Ronskian is not so wrong and. So, can actually prove that the solutions y 1 and y 2 are independent. So, which you can do which you can do it as an exercise. So, if you compute this y 2 prime and if you do the computation a bit here do the computation here you can actually see that c double prime by c prime is equal to probably minus 3 by t you can do the computation. So, that is implies finally do the computation first to compute c prime compute solve for c prime solve for c prime then solve for c 1 and do if you do this a small exercise you can actually get that your c t is you can take c t or any constant multiple you can see that c t is equal to 1 over t you see. So, you will have another solution. So, the general solution can be written as a solution c not c t c t you will get it as 1 over t square that will imply your y 2 t is of the form 1 over t. You can take constant that does not matter for independent. So, you get solution. So, the other general solution is your c 1 t plus c 2 y t. So, you see a general solution. You expect this singularity because if you look at this equation this is actually a singular equation. So, the solution singular have the problem. So, if t not equal to 0 and if you are solving only on the positive side of t you have no problem. As long as you have a singularity you will expect such kind of singular problem to show some singular behavior which we have seen already in the previous examples. So, we will go to the here you can see that here you can see that t and 1 by t are independent solutions. It is the same situation in the earlier case also for general this is a very general method to solve that. Now, we will go to a special class of equations where a general method is available to the find the independent solution. These are second order equations second order equations with constant coefficients. We will do that second order constant equations. So, how does a l y look like now? Your l y will be of the form. I can put a constant here a not equal to 0. So, that is singularity a y double prime plus b y plus c equal to 0. You want to see that one solutions that one. So, here how do you proceed with that one? Again observe the equation c y p y prime plus c y equal to 0. So, look here is a factor this again we are trying to get some idea from using this equation a not equal to 0 a b c are constants now. Since a not equal to 0 one can divide by a and you can write in the p q form itself that is something. So, that is a singularity, but let me put it a here because this is the form you have already seen when you study the spring mass system where a equal to m be equal to c the damping coefficients and c equal to k your spring constant. Now, you have also recall what we have done in the beginning in the module one when we have studied the spring mass system. Time permits may be able to do a very little about it recall what we have done there in the module one. So, let us do it this equation the one observation is that if y prime is proportional to y that is y prime is equal to some k y with for some constant k for some constant k that will immediately tell you y double prime is equal to k y prime, but y prime is k y. So, you can immediately write instead of k we already used it for some spring constant in that example. So, I want to preserve that. So, I will use here r y prime this is equal to r square y and it shows that if y is proportional to y y prime is proportional to y y double prime is also proportional to y with proportionality constant r square. So, the immediately shows that if y prime is proportional to y y double prime is also proportional to y. So, if the y comes out. So, if you substitute that if immediately tells you that you get a r square plus b r plus c of y equal to 0. Definitely this is a constant a r square plus b r plus c is a constant and you want y of course, if y is identically 0 this will satisfy and y identically 0 is a solution to your homogeneous system, but what we are interested is a non trivial solution. So, if you want to get a non trivial solution for a non trivial solution we need a r square plus b r plus c equal to 0 and this is called the characteristic equation. So, immediately if you are just starting with a solution of the form. So, what we have a thing is that you started with the solution of the not started with a solution of the form if you are looking for a assuming that y prime equal to alpha y that is nothing, but y prime is equal to r y. So, if y is a solution to this equation y prime equal to r y then r will satisfy this one that is if y satisfies y prime equal to r y that will imply r should satisfy satisfy a r square plus b r plus c equal to 0 the that is why we study always the characteristic equation with constant coefficients. So, the moment this has a solution a r square plus b r plus c equal to 0 is a solution and then if you look for a solution to this form that will be a solution to your second order equation. So, what are the solutions to this equation y prime equal to r. So, that is a reason. So, look for a solution of the form look for a solution what you study of the form y equal to e power r t you see. So, this is just to motivate most of the time it is introduced that look for a solution of the form y equal to r t this gives you a some motivation why we are looking for the solution and that immediately implies a r square plus b r plus c equal to 0. So, you look for a solution of this form and of course, this has two roots this gives you two roots r 1 r 2 solve it what are the roots you get it. R is equal to minus b plus or minus square root of b square minus 4 a c by 2 a c. So, this gives you the different situations. So, the cases. So, you will start with a different case. Case 1 is the case which you are easier case for a c plus c. So, this is the case of a positive this is the first case. In this case two real distinct roots that is very important that if it is not distinct you do not get a two solution distinct roots. Therefore, y 1 equal to e power r 1 t what is r 1 with a plus sign here with a plus sign here and y 2 equal to e power r 2 t with a minus sign here you get that e power r 2 t. So, r 1 is with plus r 2 t equal to thing. So, you have r solutions are the independent are the independent are the independent symbol exercise for you are the independent a symbol exercise show. How do you show we have a good method to show independent show w of y 1 y 2 is at t simple computation because you know y 1 you know y 2 you compute y 1 y 2 prime minus y 1 prime y 2 you can see that it is 0 if r 1 and r 2 are distinct two distinct roots r 1 and r 2 that is important r 1 r 2 and r 1 not equal to r 2 it is just a computation for any anybody can do it. So, there is absolutely no problem. So, you have. So, what would be the general solution in this case general solution general solution y is equal to a a is equal to a e power r 1 t plus b e power r 2 t see. So, you have that now case two case two b square minus 4 a c equal to 0 this is a double root. So, therefore, r is equal to minus b by a minus b by 2 a is a double root double root. Therefore, y 1 we get only one solution double root does not give you two independent solution e power r t is one solution you see the moment you get to one solution now you know the method of reduction of order to find a second solution. So, use here is a symbol again exercise for you I do not have to worry about that that is an easy thing and it is also nice to do it. So, that method of reduction can be learned properly find a second root find a second solution y 2 second solution y 2 using y 2 of the form c 1 c t solution c t into e power r t. So, you have a solution c t into e power r t using the method of reduction using the method of order reduction. Very simple computation absolutely no difficult use the fact it is a double root the only thing that use r is a double root you have to use this. What is the meaning of double root that is a means a r square plus b r plus c equal to 0 since it is a double root is derivative is also 0. So, you have this condition 2 a r plus b is also equal to 0. So, to use this both the facts. So, use this fact and you get it again we can get y 2 of the form p e power r t. So, you can solve c t to see that it is nothing but t. So, what is the general solution general solution. So, again so this t is not coming in a very arbitrary 1 it comes as a second root. So, it will be general solution y t is equal to c 1 plus. So, you have c 1 into e power r t and c 2 into this one. So, that c 1 c 2 t or let me use a b. So, that you do not get confused with that c a plus b t e power. So, you have a nice thing. So, now we will go to the case 3 with complex root case 3 complex roots. What are the complex roots here b square minus 4 a c is negative. So, you have 2 complex roots r 1. So, what are the let me go there I think minus b r is equal to minus b plus or minus square root of b square minus 4 a c by 2 a. So, it is a negative. So, you will have r 1 is of the form something like alpha plus i beta 2 complex root r 2 is equal to alpha minus i beta. So, what is alpha alpha is nothing but minus b by 2 a and beta is of the form square root of b square minus 4 a c minus b square. So, this is the minus we take it out by 2 a. So, that will give you e power r 1 t e power r 2 t are complex solutions. But it is an easy factor to see that when you have complex solutions it will immediately give you real and imaginary parts are all real solutions real and imaginary parts imaginary parts are real solutions. So, and that will immediately give you e power alpha t into cos beta t e power alpha t sin beta t are 2 solutions. Just write down e power r 1 t you get it these 2 things e power r alpha t and e power r 2 solutions what about their independence are they independent yes they are independent the 2 solutions which are independent. How do you compute the same trick there is no problem independent a simple a computation is not our real exercises, but it is a simple computation which one have to vary. So, if you compute. So, this is your y 1 and this is your y 2 here. So, you will have you compute e power w of y 1 w of y 2 if you compute you will get nothing but you compute this you get beta into e power 2 alpha t may be. So, you will have a computation that is a not a problem because you can just compute y 1 y 2 prime and y 2 y 1 prime y 2 you get it this is not equal to 0 y y that is easy because this is non 0 which is exponential and beta is non 0 because beta is a we are in the situation of complex roots and b square minus 4 a is negative. So, this is non 0 you see this is a non 0 commodity. So, if it is a complex roots it is a complex roots your beta should be non 0 because beta equal to 0 is the case where you have real roots you see. So, that proves immediately they are independent. So, this completely gives you n the complete analysis of your equations with the constant coefficient you can always determine if the equation with constant coefficients that 2 roots with this thing. Now, we will go to the general case of non homogeneous equation and today first in the first lecture in this lecture homogenous equation in this particular lecture we will give you the thing. So, what is a non homogenous equation you have L of y general case it is not the what I am going to give you is about the general case not necessarily with constant coefficients. I want to understand the solution structure of this case L y is equal to y double prime this is L y by definition plus p t y prime plus q t y is equal to some r t. You already seen that using the existence uniqueness theory an initial value problem has a unique solution under the that is a proof which we already stated in the beginning. When p q and r are continuous the initial value problem has a unique solution to the homogeneous equation, but the idea is to understand the solution structure. Naturally you cannot expect the linear space now, because if you have two solutions to the space y 1 and y 2 y 1 plus y 2 cannot be a solution to that one you have the linear structure here. So, if you have y 1 and y 2 are solution if you add you get a L of y 1 plus L of y 2 is 2 r t not r t. So, it does not have the linear structure here, but you will see a some affine structure. So, let us let S tilde be the set of also I do not call it is a space be the set of all solutions. When I say that S tilde is the set of all solutions we are not looking into the initial value problem initial value problem has a unique solution. So, here general solution we are looking at it be the set of all solutions to the non homogeneous system. So, let me call it 2 now set of all solutions 2. So, this let us see that suppose a particular solution first you want to understand any solution is called a particular solution. Any solution that is why it is a any particular solution any solution let me particular any solution particular any function or any function function y particular equal to y particular t which satisfies 3 namely L of y particular is equal to r t is called a particular solution. So, L y particular solution satisfies this solution and this is tilde is a set of all is called the particular solution. There is no 3 here. So, it is no 3 it is 2 any solution to 2 is called that is no 3 here we are not in the risk 3 yet it is called a particular solution. So, now look at this is an interesting fact suppose y let me to distinguish a homogeneous equation suppose y equal to y homogeneous is a solution to the homogeneous equation solution to the homogeneous equation which we already know homogeneous equation what does that mean L of y home is equal to 0 or equally y home belongs to s let us recall this notation where s is the set of all solutions to a homogeneous equation and we know that s has two independent solutions only and the dimension of s is equal to 2. Then immediately this will give you immediately what about y is equal to y home plus y particular in other words what I am trying to do that I am taking a one solution from your homogeneous equation and one any particular solution then what would be L of y L of y is equal to L of y home plus y particular. Here is where you use the fact that L is a linear operator since L is a linear operator this is L of y home plus L of y particular, but what is L of y home L of y home is equal to 0 L of y particular is equal to r of t that immediately tells you this is nothing but 0 plus r t that is equal to r therefore, y is a that implies y belongs to s till. So, if you start with a solution any particular solution of your choice you take any s the solution from s add you get a new solution to your s. So, this immediately tells you that s tilde if you take s and if you take any particular solution this is always contained in s tilde. So, what the interesting proposition is that it can immediately prove a proposition in with a one line proof actually s tilde. So, that is what I said is in a affine form y particular solution you see proof is one line, but let me give you a proof. So, you take any s in s tilde. So, you already proved one inclusion you have to prove the other inclusion take any y belongs to s tilde and y particular is given to you that is what I am saying. So, if you can find one particular solution and then take any solution. So, that gives you the complete solution structure. So, any solution to your non homogenous equation can be written as the general solution can be written in this form and that is what the essentially this proposition tells you that implies now consider l of y minus y particular again you consider this that implies l of y minus l of y particular, but what is l of y y is in s tilde. So, l of y is nothing, but the r, but l of y particular is a particular solution and that is also equal to r and hence is equal to 0 that implies y minus y particular belongs to s tilde s it is a solution. So, y minus y particular is a solution to your homogenous equation. Hence, that implies there x is y home in s such that y is equal to y home plus y particular. So, what this tells you is your corollary which is. So, let me state it another corollary there is nothing new and rewriting this same proposition, but because this is the way probably introduced in the universities and other books the. So, this is theorem the proposition which I have stated is basically rewriting let y 1 y 2 be two independent solutions to be two independent solutions to your homogenous equation solutions of the homogenous equation. L y equal to 0 and y particular is a particular solution to the non homogenous equation non homogenous equation then non homogenous equation L y equal to r then the general solution then the general solution of L y equal to r is given by y is equal to y equal to alpha y 1 plus beta y 2 plus y particular where alpha beta are constant. So, that gives you the complete structure of the solutions to your homogenous non homogenous equation. So, the problem here is that if you can find two independent solutions to your homogenous equation and one particular solution whatever form it is you can find the we can write down your general solution and if you want to solve your initial value problem find the general solution use the two conditions of the initial values substitute and determine the thing and that is why the method is again as you see finding two particular two independent solutions is the difficult task. And what is the reduction order tells you that if you can find one solutions to your homogenous equation then using the method of reduction order you can find the second solution. And what we are going to quickly probably describe and a method to find solution find a particular solution even the two independent if you have two independent solutions to your homogenous system that can be used using the method of variation of parameter to determine a particular solution. So, at the end of it if you look at it in theoretically you may have difficulty in integrating out the functions, but end of it the fundamental difficulty is finding one solution to your homogenous equation if that is there you are more or less near the general solution to your non homogenous equation as well. So, let me I have few more minutes in this class. So, let me just introduce the method of variation of parameters, so we will do that in this class. So, again look at how do you the idea of method of variation of parameters is to determine a particular solution using the two known if at all solutions to your homogenous equation that is what we want to do it. So, assume so we have basically assume y 1 y 2 are two independent solutions the idea we are going to use is the same as we have used in the reduction order. In the reduction order if you know one solution y a second solution cannot be a constant multiple of that equation. So, you varied that constant as a function of t and fortunately the reduced to the first order equation. Now, we have two solutions y 1 and y 2 we use these two solutions if you take a linear combination of y 1 and y 2 in the form c 1 y 1 plus c 2 y 2 that is going to be only a solution to your homogenous equation. So, if you want the solution to if you are looking for a solution to your non homogenous equation using the a linear combination of y 1 and y 2 with constants on the coefficients is not possible. So, the idea is to vary both c 1 and c 2 in the coefficient what I am trying to say is that then for any c 1 c 2 any c 1 c 2 constants c 1 y 1 plus c 2 y 2 then for any constant is also is only a solution can only be a solution can only be a solution to L y equal to 0. You are not by putting any linear combination of y 1 and y 2 you are not going to get any solution to the non homogenous equation you will get solutions to only homogenous equation. So, the idea vary c 1 c 2 as functions of functions of function of t. So, look for a solution look for a solution y equal to c 1 t into y 1 t plus c 2 t y 2 t of the non homogenous equation look for a solution y equal to c 1 t into y 1 t plus c 2 t y 2 t of the non homogenous equation of the non homogenous equation homogenous equation. So, what are the things we have? So, we have. So, let us consolidate the facts. So, we have L y 1 equal to 0 L y 2 equal to 0 that is a solution to me and we need use this we need to find c 1 c 2. So, that L y is equal to r. So, that is important thing that is what we are trying to say is it possible to find. So, we need to find c 1 and c 2 two unknown functions satisfying this equation L y equal to r. So, indeed we need two conditions and that is what you have to. So, I again what I want you to do is that the exercise is to do the computations exercise do the computation do the computation. What are the computations you have to do it? With this you find y prime y double prime and compute L of y substituted in the operator L of y. So, I will write down the expression here you will get 2 times 2 you have to see where the computation comes here. Let us see 2 into c 1 prime y 1 prime plus c 2 prime y 2 prime do the computation plus you get c 1 double prime y 1 plus c 2 double prime y 2. You have a factor here and you will have p t into c 1 prime y 1 plus c 2 prime y 2 and the q 2 will get absent because you have y 1 and y 2. Use the fact that L of y 1 you have to use the fact that L of y 1 is equal to 0 L of y 2 equal to 0. Now, our idea is to get a two equations for c 1 and c 2 solvable. So, the first step is to take. So, we are trying to make and sorry this is you want L of y equal to 0 and we want this to be r t that is more important. So, you want that equation to be r t because you want L of y to be the solution to your non homogenous equation. So, we are trying to make a simpler thing first you make this quantity 0. So, these are all imposed whether you can. So, you will have c 1 prime y 1 plus c 2 prime y 2 plus c 2 prime y 1 and y 2 are given to you. So, you have one equation. So, you have an equation 1 then we call this equation 1. Now, look at these expressions I take there are two terms here one term here if you do a computation there if I calculate my d by d t of here c 1 prime y 1 plus c 2 prime y 1 plus c 2 prime y 1 you see this will give you if I compute here c 1 double prime y 1 one more term is there I will write that one from here I will get c 2 prime y 2 and then from here I get c 1 prime y 1 prime plus c 2 prime y 2 plus c 2 prime y 2 plus c 2 prime. So, if I use this thing. So, this is a similar term. So, if I take among the two term if I combine this term with one term here that is nothing but d by d t of this one but this is already chosen by 0 here. So, this is nothing but 0. So, only term if I choose this is the advantage if I choose here I get all this term plus one more term will get cancelled. So, only remaining term will be one more term. So, what I will get is that in that case I will get the c 1 prime y 1 prime plus c 2 prime y 2 prime is equal to r t and the first equation already I get a c 1 prime y 1 plus c 2 prime y 2 is equal to 0. So, you have two equation in two unknowns and this is solvable when this is solvable if this determinant of y 1 y 2 y 1 prime y 2 prime is not equal to 0. But what is this determinant this is nothing but your Ronskian if you recall it this is the Ronskian you see maybe I will what I will do is that I will start from here in the next class and I will do the as a moral analysis done, but I do not want to hurry up, but essentially tells you that you have a Ronskian which will be non-zero because 5 1 m n 2 are independent solution and hence you can solve it for c 1 n c 2 c 1 prime n c 2 prime.