 So, in the last lecture, I could not complete the method of variation of parameters. So, let me quickly recall what I have done in the last few minutes of the lecture, previous lecture and then let me complete it. So, what we are doing is that method of variation of parameters, method of variation of parameters. So, what are the things we have given? We have given two solutions y 1 and y 2, which are solutions to homogeneous equations. Therefore, L y 1 is equal to 0, L y 2 equal to 0 and then our idea is to look for a solution y of the form C t y 1 plus C 1 t y 1 plus C 2 t y 2. So, that y is a solution to the non-homogeneous equation. So, y satisfies L y equal to r. What we have requested you to do a computation and after doing the computation, we have imposed a condition because we have to determine C 1 and C 2 as functions of t. So, we require two conditions. Looking at the differential computation of L y, first to do the computation of L y by computing y prime, y double prime and substituting in L y which should be equal to L of y equal to r. We have seen a term. So, we imposed a condition C 1 prime y 1 plus C 2 prime y 2 is equal to 0. Once you impose this condition, then L y equal to r will reduce to the condition C 1 prime y 1 prime plus C 2 prime y 2 prime equal to r t. So, you see so we have two equations in two unknowns. Here C 1 prime and C 2 prime are the unknowns y 1, y 2, y 1 prime and y 2 prime are given to you. To determine so that therefore, there exists unique C 1 prime C 2 prime if the corresponding determinant of this matrix y 1, y 2, y 1 prime, y 2 prime is not equal to 0. But what is this? This determinant is nothing but the Ronskian of y 1, y 1, y 2 and this nonzero since y 1, y 2 independent solutions y 1, y 2 is independent. So, we are through in the sense that this is nonzero. So, there exists unique y 1 and y 2. How do you in fact you can write down the solutions quickly? In fact, you can solve it. You can find this to a linear system with two unknowns. So, let me write down your C 1 prime. So, you will get your C 1 prime is equal to let me just do it, but do the computation. You will get r y 2 by w y 1, y 2 and C 2 prime if there are any mistakes do the computation and we will see there are any mistake r y 1 by I think the computation is correct, but you can compute it and check for any errors. So, you get C 1 prime, you get C 2 prime, integrate to get C 1 and C 2, integrate to get C 1 and C 2. So, this is a reasonably a very general method and what it shows that to determine a solution to your homogeneous system again we need to know two independent solutions, but the when the equation with constant coefficients with a certain right hand side we can write down the equation in a neat way. This is another method the method of undetermined coefficients. We introduce that also method of undetermined coefficient and determine coefficients. This is for some special class because we want to complete the non homogeneous term with constant coefficient when it is a homogeneous equation with constant coefficient we already know how to write down the equations completely depending on the roots of the characteristic equation. So, consider one special cases two special cases. So, you will take L y is equal to this with a constant coefficient, but certain calculations is also true without the constant coefficient, but for the simplicity we will assume p q are constants p q constants some of the parts we will work even with that, but you can we can write over the a b c, but a is non zero which can be divided you can write it in this form. So, we are interested in this is the L y definition. So, a special right hand side L y equal to e power a t. So, we are interested in solving one to solve yeah this is why we want to solve this this is the equation we want to solve L of y is equal to e power e. Again this suggest by the same arguments which I have given in the class here there is a y here which is missing see y y prime are all proportional y prime is proportional to y then y double prime is also proportional to y and that is why he we propose the solution of the form e power r g, but here on the right hand side e power a t is available. So, you want to get cancel the best way one of the best motivation to look for a solution of that form. So, look for a solution of solution of the form y is equal to some constant we do not know a e power a t. So, if you compute y prime. So, you will have y prime is equal to a a e power a t y double prime is equal to a a square e power a t. So, if you substitute there e power a t will cancel this will give you immediately a a square plus p here p is constant a p a plus q a q into equal to e power a t that right hand e power a t that will get answer. So, that is a into a square plus p a plus q is equal to 1. So, our idea is to find an a. So, that y equal to a e power a t equal to 0. So, we need a to be divided. So, this implies a you can take it to be 1 over a square plus p a plus q and this is possible if that is non-zero. So, that puts a condition if you want to find a solution that immediately not equal to 0. So, this immediately suggest that we have a solution of this form if a is not a root of this equation. This is a thing which we have actually done which we will see again is the kind of resonance type same. So, if you have a right hand side e power a t, then such an e power a t should not be. So, if you are proposing a solution to this equation of the form e power a t to the non-homogeneous equation, only that e power a t should not be a solution to the homogeneous equation that is a minimum thing it is telling. So, exactly telling that e power a t should not be a solution to the homogeneous equation is the condition that a square plus p a plus q is not equal to 0. So, that gives a solution. So, y t equal to e power a t by e power a t by a square plus p a plus q is a solution is a particular solution is a particular solution. We are looking for particular solutions if a square plus p a plus q not equal to 0. So, again so what happens if a square plus p a plus q equal to 0, then look for a solution. Now, you have an idea then look for a solution of the form some constant a into t e power a t. So, you are looking for it you know that why it is a solution some c t into e power a t form and that is t form. So, now proceed it. So, you are looking for the solution of the form this is why our word looking for a. So, compute again do a simple computation you compute y prime in this case y double prime then that will imply do the computation I will skip the computation what you will get is that a into 2 a plus p equal to 1 you will arrive at this formula. So, that will give you a is equal to 1 over 2 a plus p of course, then in that case 2 a plus p should be not 0 if 2 a plus p. So, you see first you got a solution therefore, a 1 by 2 a plus p into t e power a t is a particular solution in this particular case particular solution if this has to be 0 it is in that with that condition you are getting it p a equal to q equal to 0 and you also need this condition. But you need a known 0 condition in that case. So, now what happens if both are 0. So, one more case left out suppose let us go this suppose these 2 are happens that is a square plus p a plus q is equal to 0 2 a plus p not equal to 0 that is also equal to 0. So, what will happen what is this situation first of all this is the more interesting you see 2 a plus p is nothing, but the derivative of this with respect to a that means is nothing, but a is a double root a is a double root that is the situation. So, here when a is a root that means the input function the in the engineering terminology such a kind of right sense sets are always called input function or a forcing function. So, if e power a t if the forcing function has this e of the form e power a t then a is not a root of the auxiliary equation then you have an immediate solution of that form. On the other hand if a is a root of the auxiliary equation or a characteristic equation, but if a is not a double root if a is not a double root if a is not a double root then you have again solution of this form is a y equal to e power a t a particular solution of this equation. If a is a double root that means these two conditions are trying. So, try for a solution of the form next level try for a solution of the form y t is equal to a t square e power a t all this will be independent and then you can see do the computation again a symbol computation you can do and you can see that a is nothing, but the half. So, a complete independent solution can be obtained therefore, you can get your particular solution of this case y particular of t is equal to e power a t by a square plus p a plus q if a square plus p a plus q not equal to 0 you will have this solution e power a t t e power a t by 2 a plus p if a square plus p a plus q not equal to 0 you will have this solution e power a t t e power a t by 2 a plus p if a square plus p a plus q is equal to 0, but 2 a plus p not equal to 0 and you have this situation half of t square e power a t if both are 0 it completes the complete analysis you have. So, you have your particular solution you also know how to compute your complementary functions what are called the solutions to your homogeneous system the same procedure can be adopted. So, I will leave it for you as an exercise to leave it if you want to compute of the form equation of the form other types of signals input functions or forcing functions are available say for example, suppose you have sin omega t or cos omega t one of them you can do that and you can actually determine the solution of the form. So, what you look for is that a solution first you determine a solution of this form if omega is not a root again that kind of same homogeneous system if it is not a solution you can proceed with the same form and you can if that is not a solution next time you try omega sin omega t. So, the exercise is to find the particular solution you find a particular solution. So, this completely analyzes the situation when you are solution when you have a equation second order equation with constant coefficients even with particular signals on the right hand side you see like e power omega t sin omega t and other kinds of things which you can do it completely. So, what I will do is that I will in this particular lecture I would like to recall once again which we have done it in the first lecture first module namely the spring mass system because now you know we have already studied the spring mass system, but now you know how to write down the solution. I will not give the details here, but to complete a thing and you want you want you to understand this equations is the better way you recall your spring mass system. So, this is a quick recovery spring mass dashboard system because now you we will understand exactly what we want to do that one. So, for more details you go back to that problem. So, what is the spring mass system m y double prime plus c y prime plus k is equal to some external force f and we have analyzed this in various situation what is m m is the mass this is c is the damping and k is the spring constant. What I would like the students to do when they read this thing try to understand this equation even in the case of exercises understand in the case of RLC case of RLC circuit because it is a same equation which we have studied completely, but now since you know the solutions we can write it here. So, we have analyzed the various cases. So, let me quickly recall all the cases what is the first case we have studied when c equal to 0 f equal to 0 this is called no damping and no free oscillations is the case of free undamped system free undamped. There is no damping system the system moves either it is a trust or if you have the system starts moving it keeps on moving the phenomena of Newton's law. So, in that case what we have done is that c is equal to 0 c is equal to 0 f is equal to 0 you can divide by m. So, you will have y prime square is equal to omega naught square into yeah there is a y here omega naught square y equal to 0 what is omega naught omega naught is called the square root of k over here this is called the natural frequency of the system this is what we have done it. And this immediately if you write down the auxiliary equation now you see the auxiliary equation will be m square i u star r square plus omega naught square which has complex roots. You see omega naught. So, you get the complex roots and your solution will be a cos omega naught t plus b sin omega naught t. So, an exercise which probably at that time I have not given if I do not remember probably it given write it in the form this is better way to understand the system write it in the form try to do this form y is equal to y t is equal to r cos omega naught t minus delta. So, you have this equations where r is equal to square root of a square plus b square and delta is equal to tan inverse of minus b b. Note that omega naught is given by k by m. So, it is not a constant the constants are so for a mathematical this is a general solution because it consists of two arbitrary constants r and delta here the arbitrary constants are given by n delta. So, this if you look at it r gives you the amplitude cosine varies from between minus 1 and 1. So, this will vary between it will be oscillates fully between r and r with a phase difference delta. So, if you plot this curve you will get the solution it will be plotting there will be a phase difference. So, this will be your minus r this will be your r with a phase difference. So, it will oscillate and the oscillation never stops as expected the case to which we have considered in this that is the first case. So, there is no damping. So, there is no external force thing. So, now we are taking the case c naught equal to 0. So, you have provided some damping to the system like you are spring and putting it in an oil or something like that. So, you are providing a damping and damping is understanding damping in the case of mechanical system as well as in the electrical system which is nothing but the resistance there is important. So, you have f equal to 0 this is damped free oscillations damped, but free oscillations. So, you see in this case it is a homogeneous equation and homogeneous equation you know how to write down here because m c k are non negative. In fact, m and c in this case we are assuming m is always positive you are assuming c positive in this case. So, everything is of course, k is also positive. So, you are assuming everything is positive in this case. So, in this case you can write down your solution which we have already obtained depending on the root of that. So, you will have a general solution of the form e power r 1 t plus b e power r 2 t when is that the roots are distinctive if I write down that one it will be c square minus 4 m k positive and it will be a plus b t into e power r t if c square minus 4 m k equal to 0 and it will be e power minus c t by m 2 m may be a cos some mu t mu t can be exactly write down plus b psi mu t if c square minus 4 m k negative and where what is mu mu is nothing but 1 by 2 m. Square root of 4 m k minus this is what in the last class we have seen it already you see you can write down and where r 1 is in this case what is in this case. So, in this case r 1 r 2 given by minus c plus or minus square root of c square minus 4 m k by 2 a here you are r the root should be minus c by 2 a you see and look at it some interesting observation c square minus 4 m k is the positive case in that case is a positive quantity whether you put minus sign or plus sign it in absolute value it will be thing. So, in this case both r 1 and r 2 are negative this we explain well let me repeat it because now it will be better clear it is negative since r 1 and r 2 are negative this will go to 0 this will go to 0 exponentially. The second case here this term will go to 0 but then there is a t term but then t into e power r t where does it will go the exercise is that. This will also go to 0. So, here is a simple exercise most of you will have a tendency to say that exponential goes faster than polynomial which is a fact but what is the meaning of exponential goes nothing. So, you have to prove this just mere words of telling that exponential goes faster than polynomial is not enough you have to show that t e power r t goes to 0 as t tends to infinity where r is a negative quantity. So, you prove this otherwise that is not true. So, proving this fact is the meaning of that exponential goes faster than the polynomial not here the third cases. So, there are in the first two cases this case and this case there are no oscillations and this term goes faster that means this has to be m and k are fixed quantities all this sign changes because of c and c is the one which we are controlling in the damping. So, when you want to make c square minus 4 m k positive it means that you have to add more damping to the system and if you reduce the damping you will reach a situation here. But even if you reduce the damping it still goes to 0 this term is still goes to 0 but slightly slower way but still exponentially. But if you reduce the damping still further you will reach a situation here but some damping is required that is because if c is positive still there is a negative quantity sitting here u c c is positive. So, this is minus of that. So, still negative quantity sitting here. So, this exponential term still there but the difference when the damping is reduced further and further you will have a term of this form which will have an oscillation this will behaves like an oscillation. It will have an oscillations behaving but then this will kill. So, the oscillations will kill thing and we can write that third case. Third situation you can write it as third situation you can still write the third situation as y t is equal to r e power minus c t by 2 m same way do it as an exercise cos of mu t minus delta u c. So, that this is your now amplitude the amplitude degrees. So, oscillates is when there is c equal to 0 this term is not there. So, it is no oscillations are not reducing but since c is positive the oscillations are still there but it is amplitude reduces. So, if you plot this graph here you can plot two curves here what the curve is also you can plot it this curve is nothing but minus r e power c t by 2 m this curve is nothing but r e power minus c t by 2 m this is at the point minus r this at r and this starts oscillating it touches oscillates like that it decays like that it goes to 0. So, that is the situation when there is a dynamic. Now, the final situations which let me know to do it completely but let me do it completely to do the next case suppose c naught equal to 0 and f naught equal to 0 but I do not want to take arbitrary for a f it is an input function. So, let me try with f naught is equal to cos omega naught t with f naught naught equal to 0. So, you have an external frequency coming here now but then there is a damping is provided. So, this is called forced damped vibrations forced damped vibrations. So, what you can do is that. So, it is a case of non-homogeneous situation and non-homogeneous situation with c naught equal to 0 you know for the homogeneous part you know how to write your homogeneous solution which I already described in the just previous part and plus you can write down a solution particular solution you can write down because it is of the form cos omega t you can do and what we have observed is that as t tends to infinity this goes to 0. So, as t tends to infinity what will have the y will behave like y particular solution you see. So, because y homogeneous part this part goes to 0 as t tends to infinity. So, the behavior of your solution will behave like y and this is not think it is a according to that you can exactly write down your equation write down your solution. So, what I suggest you to now solve this equation because it is a cos omega t using method of undetermined coefficient you can find your y particular solution this we y home we already determined. So, you determine y particular to find the solutions and see that there will be oscillations and d k everything according to that situation. The one interesting case is that we already seen that when there is no damping when there is no damping and no external force acting on it there itself you have vibrations which is not d k. Now, we will have the most probably either unwanted or desirable according to the situation in mechanical system it is undesirable because if there is no damping at the same time an external force coming and acting on the system. So, you will have a situation where c is equal to 0 and f is equal to f naught cos omega t or sin omega t with f naught not equal to 0. When c equal to 0 you have your solutions and this gives this again reduces to two cases because when you have omega t here when c equal to 0 there is a natural frequency omega naught. So, if you looking for this is a natural frequency if you just the previous introduction when you have a solution here cos omega t and if this omega happen to be the solution of this here in the homogeneous equation then the solution will be differ. So, if omega the case one in two cases the natural frequency omega naught is different from omega then there is no problem problem you can write down the solution and nothing interesting then what we have seen nothing interesting nothing new let me write it you will write down the solution you will have some oscillation. So, if you have even if there is no damping it will be you have a homogeneous system which is oscillating which we already seen that a homogeneous solution then you will have a component of a particular solution and there absolutely no problem. The second case when you have an equation omega is equal to omega naught, but then cos omega t cannot be a multiple cos omega t sin omega t cannot be the solution out to your homogeneous system. So, the solution in this case will look like little more interesting you will have a solution due to the homogeneous part cos omega naught t plus c sin omega naught t and then the particular solution will have it another fact the particular solution will be of the form let me write down f naught by 2 m this is ok these are small constant what the fact is that you have to try a solution if omega is a solution to that cos omega t is a solution to your homogeneous system you have to look for a particular solution of the form t cos omega t or t sin omega t and that is what it gives you you will have a sin omega naught t. Now, look at this fact this will oscillate without reducing your vibration amplitude, but here you see a factor this is not only oscillating as t increases this amplitude of this factor the amplitude of this portion increases. So, if you have plot your graph this one you will have something initially there may be some vibration as t increases the vibration this is what is called the undesirable oscillation and as explained in my first module unwanted vibrations in the case of mechanical systems and the tac comma bridge collapse etcetera which we have discussed on the other hand the same system when you are trying for a RLC circuit even if a small current comes and if this equation represent the second order linear equation with constant coefficient represent the equation for I and then what is happens is that if you have an external frequency coming which coincides with your natural frequency that gets its amplitude gets becomes larger and larger and you can get the signal and that is exactly what you do by in the tuning. So, many many examples in this situation can be done worked out. So, just solving equations using the methods or theories not enough one need to interpret the solutions according to the problem type of problem one is study this marvelous completes which I want to tell in this module. But before completing this particular module on second and order linear equations I want to introduce one of the important concept not only for this initial value problem it will also come in boundary value problem and this is also appears extensively in partial differential equations which this is called the a concept of greens function. So, I am not going to do it too much of greens function here my idea is just to motivate you why greens function the greens function comes naturally in this setup. So, greens function basically used to represent solutions need to get representation formula, but why do we expect such a greens function and what is a greens function in the setup of the second order equations. So, let me recall your operator once more may be a little later I will call it. So, let us start with a differential operator d by d x. So, this is a differential operator which we are studying now you are comfortable with that one and you have the finding derivative is a forward problem. So, you already seen that forward problem that means you have your d u d y equal to f given f given f equal to f equal to f equal to f equal to f equal to f given u find d u can be computed that is a forward problem no doubt in that. So, you have a d. So, you can think that d by d x is a mapping probably as I said this some more a motivation one can make it precise define. So, you can think it as a for example, you can say that is a functions from c 1 of some interval t naught to t to going to c of t naught to t you see. So, you take a differentiable continuously differentiable function differentiate. So, if you take u here your d u by d x equal to f belongs to here you see on the other hand if you do this is the differential calculus on the other hand when you go to an integral calculus we are trying to invert it we are trying to loosely speaking loosely we are trying to we are inverting we are inverting d by d x the operator d by d x that is what you are doing it. So, these are different interpretations we are trying to invert means given f find u that is what you are trying to do it find u. So, that d u by d x is equal to f you see and that is what we eventually calculated your calculated u t in terms of in I have used it t since I am throughout using t let me use t only. So, let me use d by d t d by d t you have d by d t you have d by d t and this is formally represented by f of t d t this is a formal representation. So, you have in some sense d going to an inversion of integral. So, in other sense if you have your fundamental theorem essentially tells you that integral of d you are trying to get some identity you see that is what you are basically trying to do that what an integral calculus problem. So, thing is that whether we can do some representations. So, you are trying to represent your solutions u is of these four. So, now you have a general operator l I have a general operator l I can think that at least an operator say from t to c 2 to t naught to t going to say c of t naught to t where l of u. So, you have y here taking going to l of y by definition what is l of y is equal to y double prime p q r given to p and q r given to u y prime plus q d y this belongs to c naught to t. So, you have an operator l mapping from the thing here. So, what is the meaning of solving it? So, what is the ODE now we are trying to do it so far ODE given r continuous given r solve for y right solve for y satisfying l y. So, this is the forward problem basically given thing is forward that is not a problem. So, ODE solving is that you are trying to in the integral calculus setup in a more general ODE you are trying to invert your ODE you are trying to solve for l of y is equal to r. So, I can introduce a mapping. So, I can introduce a mapping as from c of t naught to t you do not have to worry about the spaces right now, but that is not the issue here t naught to t where given r you are finding the solution y which is nothing, but s r I call this is equal to s r. So, if you put initial condition you know that there is a solution given a put the in particular if you put conditions y 0 is equal to 0 y prime of 0 equal to 0 this s will be linear if you put other conditions you would not get linearity, but that is ok. So, you put that conditions you have a basically linearity you are trying to invert this kind of equations. So, our question is that analogous to this one given you you are getting a representation. So, looking for a representation. So, looking for a representation looking for a representation of the form you want to represent y y as a function of t that is nothing, but s r as acting as a function of y. So, this is an operator you want to get a representation of this operator s in an integral form can you have an integral form g of t of psi d psi r d psi. So, can I get a representation in the integral form? It is a slightly more general you cannot have an easy way integral form integral representation. In particular not in particular sorry if you realize that one of the classical theory is finding solutions to integral operator where there is a certain kernel is given. So, given a kernel this we can call it a kernel in general we call it here a greens function. So, when you given a two variable function and you can introduce the integral operators which are linear integral operators. In fact, the development of a linear theory actually began from the integral operations test theory. So, integral operator. So, we are looking for such a function is possible. So, you have a is it possible to find a two variable function g which is a function of t and psi defined for where t psi greater than or equal to say t naught and can we find such a g. So, that you have a solution representation for the and that g should be independent of of course, the forcing term. So, let us see let us try to compute formally which also can be justified suppose assume such a g axis how the assume such a g axis axis. Do not worry about the computational aspects of it which one can justify and prove the precise result. As I said the motivation of greens function is more important we are not going to spend time. So, you compute your y prime t when you compute your y prime of t. So, this we will define say probably from t naught to 2 to make it precise t and you should know how to take derivative this is a simple formula which probably we may do it in the our preliminaries. When you do the computation you can take your t naught to t you now let me write this is with respect to t you are differentiating with respect to t t of psi r psi d psi and then there will be a boundary term. How do you find the boundary term you just put psi equal to t and differentiate. So, that is nothing but g t t that is a boundary term it is easy to calculate you put the limits and differentiate the limits that is a general formula when such a you can integrate from alpha t to beta d integral alpha d to beta t and 2 variable function how to take derivative. Now, compute your y double prime of t when you compute your y double prime of t. So, you have t naught to t I will g t t of t psi r psi d psi plus you have already computed g t put evaluation at t r t that is coming this is fixed. So, that do not count now here there is a difference you should not be differentiating with respect to the first variable you have to differentiate this as a now when you are differentiating you have to differentiate this as a single variable function and you have to differentiate. So, there is a difference between when I write here d t of g t t r t. So, do not write g t at t t that means when you write g t at t t means that you are differentiating only with respect to the first variable and substituting for psi equal to r t here is not like that you are substituting and then differentiate. So, you do some a bit of calculations. So, you have calculated y prime of t y double prime of t substituting the equation. So, you will have L of y where L of y of t L of y of t you can actually do this justifications t because assuming g we are not introduced g yet. So, you will have L of g the corresponding L of g of course, with respect to the first variable the second variable psi is nothing but a integration parameter, but the differentiation and the differential operator is with respect to the first variable not the other one r psi d psi plus you will have few terms here. For example, you will have g t t r t plus you will have a term p t into g t t r t plus you will have a d d by d t of g t t r t. So, you will have a term p t into g t t r t so you have that kind of term you can do that one. So, when you do this one so this motivates as to define we want this to be r the best way to define r is that for each parameter you have to introduce the parameter properly you define that to be 0 for each of that parameter appropriately. So, it introduce that parameter that gives you you want to be r t. So, you want this as a to be 0 for every t you want this g t t to be 1 you want this g t of this is equal to 1. So, that you get your r t and this you demand 0 once that demand 0 d by d t of that also will be 0. So, this force to introduce the greens function introduce the greens function greens function g t psi I introduce g t psi I introduce the this in a very nice clever way 1 have to introduce I introduce g t psi for psi greater than or equal to t. So, I will introduce g t psi for fix psi. So, you have to introduce you fix psi fix psi as a parameter. So, I am introducing fix psi greater than or equal to t naught as a parameter I want to treat it as a parameter. So, what I will do is that. So, you have t naught here I have some psi here on this portion I will introduce the problem. So, I will introduce the problem l g this is the differential operator is with respect to t parameter and psi coming because I am introducing that the problem in for psi to t or whatever it is. So, you see so, I need conditions at psi. So, what are the conditions I will put at psi I will put g t at psi psi equal to 0 I want g psi equal to 0. So, you see r is not coming into picture here. So, you introduce this problem now this problem by uniqueness you have that solution unique solution which is smooth enough. So, that everything can be justified whatever I have done. So, you have introducing this problem on the left side you have to introduce for entire interval you define g t psi equal to 0 because it is defined only for psi greater than t greater than or equal to psi. So, the problem is t is here. So, you want this one. So, g t psi is equal to 0 for t in t naught to psi. So, you can complete it. So, you have a green function you are defining on that interval you are defining that one. Then you actually prove now everything. So, you have your solution operator y. So, that will give you your solution operator y is equal to s r that is equal to integral law. So, it is how this quite often this is introduced here as an arbitrary way, but this is naturally we are demanding a representation g t psi s r acting at t y acting at t r psi d psi. So, you have a nice representation of this thing. You can actually do little more using the variation of parameters etcetera. In fact, one can do a computation little more. Suppose y 1 y 2 be two independent solutions to independent solutions of homogeneous system. Then you know see look at here even though it is a homogeneous system you have a non homogeneous parameter. So, even the solutions of this can be computed using the variation of parameters using exactly what we have done to compute the non homogeneous system. Even for this equation you can use the variation of parameters using the two independent solutions. So, these are all I will leave it as a simple exercise. You can do that one y 1 and y 2 it is a homogeneous solution. Then you can actually compute g of t psi to be nothing, but y 1 of psi y 2 of t careful with psi and t minus y 1 of t and y 2 of psi. So, you have a nice representation of that divided by. So, you have to divide w of you calculate the Ronsky and of 2 and evaluated at psi that is I think you have to do that. So, you have a representation of solutions. So, if you write down now your solution. So, if you can write down for your solution with l y equal to 0 y 8 0 is l y is equal to R with y 8 0 is equal to please not that the solution satisfies the homogeneous conditions. So, you will get y 0 is equal to 0 y prime of 0 is equal to 0. So, you can write down the solution your phi evaluated at t in a nice form. So, you are integrating only with respect to psi here. So, you can write down and you have an you are integrating with respect to psi. So, you can this will come out y 2 will come out. So, you will have phi t into y 2 of t minus y 1 will come out the remaining integral will be psi t into y 1 of t and what is your phi t? phi t is nothing but l y t is equal to g t psi r psi. So, this factor you have to take it. So, is equal to integral t naught to 2 t you get y 1 of psi r psi d psi by w evaluated at psi y 1 y 2 and your psi t will be integral of t naught to t evaluated at y 2 of psi r psi you have w of psi. So, you have a good representation. So, you have a good representation of your t. So, the greens function the main idea of this part is to introduce the greens function. So, you can introduce this and but what I say that this looks simple, but understanding the greens function fundamental solutions are fundamental in the other situations. So, you can see this probably when we deal with the boundary value problems we will come to the greens function little more, but so there is a beautiful representation using greens function. With this we complete the this particular module on linear second order and linear first order equations. In another module we study the nth order linear equations and nth order linear system. There our main focus will be and to understand the stability of equilibrium points. Hence, we will be using the linear algebra extensively. Thank you. Bye.