Mod-2 Lec-1 Introduction to Differential Equation

welcome to lecture series on differential equations for undergraduate students today's topic is introduction the first question arises what do we mean by differential equations before answering this question let us try to remember one thing which we have already done the exponential function yes e to the power x we know that this function has a property that if we differentiate it with respect to x we get the same function that is d y over d x is e to the power x thus we got a relationship that d y over d x is equal to y this relationship is called differential equation so now we are ready to answer what do we is a differential equation a differential equation is a equation involving an unknown function and its derivatives thus d y over d x is equal to k y d 2 y over d x square plus 4 y is equal to 0 1 plus d y over d x square whole to the power 3 by 2 is equal to c times d 2 y over dx square x times dy over dx whole square minus y times dy over dx plus x is equal to 0 x times del u over del x plus y times del u over del y is equal to 2 u del 2 y over del t 2 is equal to c square times del 2 y over del x 2 are all examples of differential equations in these examples we have seen that sum involves ordinary derivatives while some examples involved partial derivatives thus we categorized our differential equations into rna differential equations and partial differential equations an ordinary differential equation is an equation that contains one or several derivatives of an unknown function of one variable only thus in our example the first example d y over d x is equal to k y the second example d 2 i over d x 2 plus 4 y is equal to 0 1 plus d y over d x whole square complete power 3 by 2 is equal to c times d 2 y over d x 2 x times d y over d x square minus y times d y over d x plus x is equal to 0 are all examples of ordinary differential equations here we do have the unknown function y and its derivatives and the function y was of only one variable and we had the ordinary derivative with respect to that variable only now a partial differential equation is an equation involving an unknown function of two or more variables and its derivatives with respect to independent variables thus in our examples x times del u over del x plus y times del u over del y is equal to 2 u del 2 y over del t 2 is equal to c square times delta y over del x 2 are examples of partial differential equations here in the first example u is a function of two variables x and y and the equation involved is partial derivative with respect to x and y while in the second equation y is the unknown function and this function is of two variables t and x and it involve the second order partial derivative with respect to these independent variables thus we have seen that all these equations are involving a function of one or more variables the derivatives ordinary are partial again the derivatives are first order are higher again the derivatives may be having the exponents thus we are now ready to give some basic terminology of the differential equations the first one is order the order of a differential equation is the order of the highest derivative in the of the unknown function involved in the equation thus in our example let us see first example d y over d x is equal to k y this involves the first order derivative only thus the order of this equation is 1 y list the equation d 2 y over d x 2 plus 4 y is equal to 0 this involves the second order derivative and that is the highest derivative involved in this equation thus order of this equation is 2 similarly in this equation 1 plus d y over d x square whole to the power 3 by 2 is equal to c times d 2 y over d x 2 in this equation it involved the first order derivative as well as the second order derivative so the highest derivative involved is the second order derivative thus this equation has order 2. now we see in the last example that x d y over d x square minus y times d y over d x plus x is equal to 0 we see the derivatives involved are only of the first order this order of this equation is 1 only all so this equation is also being called first order equation the second term is degree the degree of a differential equation is the exponent of the highest derivative occurring in it after the equation has been rationalized with respect to derivatives let us understand this more clearly with the help of examples so in our first example we do have that the derivative involved is only that highest derivative is first order and its exponent is 1 thus the degree of this equation is 1 in the second example the derivative which is involved that is second order derivative is the highest derivative and its exponent is 1 thus this equation also has degree 1. now see in this example we do have the derivative involved is a first order only but we see that d y over d x has a exponent 2 so the highest exponent is 2 thus degree of this equation will be 2 now let now let us see one more example we had in our examples that is 1 plus d y over d x square whole to the power 3 by 2 is equal to c times d 2 y over d x 2 in this equation we do have a power as 3 by 2 that is it is involving a square root now this is not called a rational equation because here the power of d y over dx is having 1 over 2 something so we will rationalize it how will rationalize it we will square this equation on both the sides and thus we would get 1 plus dy over dx square whole to the power 3 is equal to c square times 2 y over d x 2 is equal to d x 2 here the highest derivative involved is the second order that is d 2 y over d x 2 and its exponent is 2 thus degree of this equation has 2. now we can categorize the equations in different manners so one more categorization of differential equations is coming as linear differential equation a differential equation is called linear if the function and its derivatives are appearing in the equation with degree one only and separately if this is not happening then it is called non-linear let us understand this more clearly with the help of examples so first let us see the example of linear differential equation see this example x cube times d 3 y over d x 3 minus 3 x square times d 2 y over d x 2 plus 6 x times d y over d x minus 6 y is equal to 0. here we see that the derivatives involved are first order second order and third harder and all these derivatives and the function y are in degree one only none of them have any exponent moreover none of the places i do have that the function and derivatives are appearing simultaneously in any component thus this equation is linear in the unknown function y and its derivative this is called a linear differential equation let's see one more example d 2y over d x 2 minus y is equal to 0 here we do see is that is the derivative is d 2 y over d x 2 and y both are occurring in degree 1 and separately thus this is also an example of linear differential equation let us see some example of non-linear equations see this first example d 3 y over d x 3 plus 2 times d 2 y over d x 2 multiplied with d y over d x plus x square times d y over d x whole to the power 3 is equal to 0. now we see here that derivatives are coming with exponent as the first derivative d y over dx has degree 3 while as in the second component d 2 y over d x 2 that is the second order derivative and the first order derivative are occurring simultaneously thus this is a non-linear equation the second example d2y over dx2 minus m times 1 minus y square times dy over dx plus y is equal to 0 here we see that derivatives are occurring in the first degree only but the function itself is occurring in the second degree moreover in one component we do have that the function and its derivative are occurring simultaneously thus this is also an example of non-linear differential equations let us have something more about linear differential equations a linear differential equation of order n is a differential equation which is written in the form a n x times d n y over d x n plus a n minus 1 x times d n minus 1 y over d x n minus 1 and so on a 1 x times d y over d x plus a naught x times y is equal to f x we see here that all the derivatives are occurring in first degree only and their coefficients are the function of x only none of them involve the y are its derivative this is a linear equation now here the highest occurring derivative is d n y over d x and that is init order thus order of this equation will be n only if its coefficient a in x is not a zero function thus this equation will be called a linear equation of order in if a n x is not a zero function why if a n x is a zero function the first term will not be there and then the highest derivative would be only d n minus one over d x n minus one that is n minus one order so this is a differential equation of order n many times this kind of equations are also being written as a n x times y n plus a n minus 1 times y n minus 1 and so on a 1 x times y 1 plus a not x y is equal to f x what the change we have done here if you see is that we have replaced this yn with dny over dx in that is the innerth derivative we have used another notation y in similarly for n minus 1 of derivative we have used the notation y n minus 1 and so on for second derivative the y 2 and for first derivative y 1 these notations are being used so that it is convenient in writing so this is also one way of writing the differential equations and these notations are being used for convenience now we have seen that these differential equations are functions and their derivatives and their relationship thus a function will satisfy these equations so any function which is satisfying these equation will call its solution so now we come to another term solution a function is said to be a solution of a differential equation if the function and its derivatives when substituted in the differential equation satisfy the equation identically let us try to see it with the help of example let us take this equation x times d y over dx is equal to 2y we can check that y is equal to x square this is a solution how can we check let's see that is for y is equal to x square what will be d y over dx that will be 2 times x now this value of the function that is x square and its derivative 2x we can put into the given equation and thus what we will get x times dou x is equal to 2x square and right hand side is also 2x square thus we get that this equation is being satisfied identically thus what we got that x square is a solution of the differential equation x times d y over d x is equal to 2 y so we have got a solution now let us see what as soon as we talk about solution the question comes existence and uniqueness what do we mean by existence does a differential equation has a solution the answer to this question is called existence then it comes uniqueness does a differential equation has more than one solution if yes can we find out a particular solution that is if yes how can we find out a solution which satisfies a particular condition so now we have seen that a differential equation may or may not have a solution if it is having a solution it may have more than one solutions so now we are going to categorize the solutions also the first categorization is general solution the description of all solutions to differential equation is called the general solution let us elaborate it little bit more we have just seen this example that is x times d y over dx is equal to 2 y now let us see little bit more if i take this function c times x square we can check that this is also a solution of this equation for all values of c we can check it see y is equal to c x square the d y over d x will be 2 times c x now if i incorporate these things into the equation what i would get is that x times 2 c times x is equal to 2 c x square which is nothing but the right hand side thus c x square is also satisfying this equation for all values of c what it says that is we are getting not only one solution but many solutions that is for each value of c i will get one solution so we this c x square will be called the general solution of this differential equation now the second question is uniqueness before answering this let us try to define something more over here initial value problem a problem in which we are looking for the unknown function of a differential equation where the equation of unknown function and its derivatives at some points are known is called the initial value problem this is also referred as ivp in short say for example the equation d y x times d y over d x is equal to 2 y if i incorporate one extra condition that the function value is 5 at x is equal to 1 this is called an initial value problem a solution to the initial value problem is known as particular solution so now let us see this with the help of example just now we had seen that y is equal to c x square is the general solution of x times d y over d x is equal to 2 y now in carbohydrate this condition that is y at 1 is 5 that means in c x square at x is equal to 1 will put what we will get is that c is equal to 5 hence we got that y is equal to 5 x square is a particular solution this solution we have got when we have used this initial condition that is the function value is 5 at x is equal to 1 let us understand it little bit better with the help of graph we see here is that is this is the graph of all the y is equal to c x square here these graphs i have made for different values of c that is all these functions are solution to the differential equation x times d y over d x is equal to y for different values of c we are getting it now in our condition what the initial condition was given that at x is equal to 1 the value of the function should be 5 we see is that is this is that is at 1 if we see here we are getting the value of 5. so this is the graph which we will or this is the function which is coming as a particular solution we see is that 1 no other function has value 5 over here so thus to arrive to a unique solution we require a particular condition now we have started this lecture with this function that is exponential function and we had seen that this is satisfying the equation d y over d x is equal to y now let us take another differential equation d y over d x is equal to 2 y this is closely related to the previous one the only difference is of a constant 2 we have used at the y but now this e to the power x this does not satisfies this equation anymore now how to find out the answer to this equation that is which function will satisfy this equation ok let us take this assumption that the function should be of the same form that is exponential one we have to do little bit experiment with the same function so let us assume that we take a function c times e to the power k x where both this c and k are constants if i differentiate it we get d y over d x is equal to k c times e to the power kx that is k times y thus we have got that function y is equal to c times e to the power kx satisfies the equation d y over d x is equal to k y so we had x started with d y over d x is equal to 2 y so now what i will do i will replace this k by 2 so we got that c times e to the power 2 x is solution to the differential equation d y over d x is equal to 2 y now what we have actually got let's summarize the result we have actually got that c times e to the power k x is a solution to the differential equation d y over d x is equal to k y this is true for every value of c so that's what we have got actually we have got a general solution to the differential equation d y over d x is equal to k y where this constant k is determined by the equation as in our example what we have chosen we have chosen k is equal to 2 this k is also referred as rate constant but this c is not determined by the differential equation that says we have actually find out a whole family of possible solutions and one for every possible value of this constant is a solution let us see the graph of this function this is the family of exponential graph so here i have made this you see is that is here we have different exponential curve all these are e to the power 2x functions but with different values of c so all these functions satisfy the equation that d y over d x is equal to 2 y how do we choose any particular solution that says we require something more for that one what that more let us see is that we say is that from where this function is is starting that is what is the initial value so suppose i say that is my initial value that is y at 0 that is initially is 1 now if i get what is the solution that c times e to the power k x that will be actually 1 that's what we get that c is equal to 1 if i replace c is equal to 1 i will get this function we see that this function we are initially at 0 we are starting at 1 thus we are having this as a function which is satisfying this particular condition so if we change our initial condition say the function is starting at half we will get that a different function that is we will get here y is equal to half times e to the power 2x this curve now if i change my initial condition to minus half that is the function is starting at minus r we see that our function has solution has been changed so thus what we have got that we summarize the result that we have got the initial condition is picking up a particular solution the differential equation gives me a whole lot of general solutions so now i give you some question to think about find the solution to differential equation d y over d x is equal to minus 5 y use the initial condition that at 0 the function is starting with 10. see that is how the graph of this function will look like and what is the difference it has with our example just now what i have done okay moreover you can experiment little bit more see that if you change the initial value to 0 that is the function is starting to 0 now find it out what you are getting does it drastically change your solution think about it and then see so here are some more questions for you to think about find the general solution to the differential equation d y over d x is equal to 2 x think about what should be the solution of this differential equation second question is d y over d x is equal to 3 what will be solution of this differential equation moreover use the initial condition that y 0 is equal to 2 in both these questions and find out a particular solution you can experiment with other initial conditions also and see what you are getting this would give you a little bit insight that is what we mean by differential equation and its solution so we have learned what are differential equations and what are solutions of differential equations now the question arises why do we study differential equations are what we achieve by knowing the solution of differential equations in other words what is the application of these differential equations where in real life we will apply this differential equation so that we do study these differential equations that says that is what is application of differential equation the answer is modeling what is modeling modeling is that actually transforming the physical phenomena into mathematical formulation by changing physical phenomena into mathematical formulation we do know more about that phenomena moreover we may predict the future behavior of that phenomena thus differential equations are very important topic in mathematics science and engineering in all those situations where changes occur and predictions are desirable the most of the systems which come under study they observe certain natural laws or observations these physical changes when we transform into the language of mathematics may turn out to be differential equations thus we try to see which kind of functions or which kind of phenomena will translate into differential equation a typical real word application of differential equations involve those phenomenas which changes with the time and the change is proportional to the rate of the change are the acceleration of the change that is whenever this relationship we can use as the differential equations now whenever a physical law involves a rate of change of a function such as velocity or acceleration etc we do get differential equations thus in physics and engineering application we do get many differential equations an example here i have taken here there is a stationary object which is initially at initial position a force is being applied on it and this stationary object starts moving so from where it starts moving that point i am taking initial position after time t1 it reaches to a position x and it is still continues to move and at time t2 it reaches to position y we have observed here that this object is not covering equal distance in equal time or in other words the equal distance has not been covered in equal time that says that the movement of the object that is the velocity of the object was changing and why it is happening since the force has been applied once only and object is moving on certain surface that surface is also giving some force so this velocity is changing now this is an example where we can say that we can use the differential equation see for simplicity if i use that the change which is occurring is at constant rate that is the distance it is covering with respect to time that is changing with constant rate so if i denote this distance by s then ds by dt is a constant say a and this thus we have got that is the change in the distance is at constance rate this is a differential equation and by solving this we can predict that at time t3 where this object will be this is what very simple application here we do have so now in this application what we have done we have taken one assumption that the change is at constant rate thus we have got that everyday problem what scientist is struggling is how do i translate a physical phenomena into a set of executions which can describe it in real world actually it is not possible to find out a system of equations or a set of equations which describe this phenomena completely and adequately so what we do is we find out a set of equations which can describe this system approximately and when we say is approximately of course we are making certain assumptions of certain observations and thus we come up with some set of equations we try it we see that how it is working we compare it with real world results and then again we can improve upon our set of equations this is what is the modeling we are having and in all those situations where changes are occurring and we want to find out how these changes are happening all these are coming modeling through differential equations now let us try to see this with a simple example the first example is about the population growth we are writing it as unlimited because we had made an assumption that there is no condition or no control on the growth the example is the number of bacteria in a liquid culture is observed to grow at a rate proportion to the number of cells present at the beginning of the experiment there are 10 000 cells and after three hours there are five lakh cells now how many will be there after one day of growth if this unlimited growth continues what is the doubling time of bacteria now let us start with this example what we have been given that is a liquid culture is having certain number of cells and it's growing with time now how do i translate it you see here first we will assume that is from the starting of the experiment the time elapsed that will denote by t and the time we will be counting here in the hours second thing we would be assuming is the unknown function y that is the number of cells at time t why we are calling it a function because it is changing with time t then what we have been given we have been given that the starting of the experiment the number of cells are 10 000 that is y at 0 is 10 000 here i have used one another notation y naught let's see that is this is also we are sometime referring that is the value of the function at that point now what more we have been given in this one that is the rate at which this growth is taking place or that is the number of cells are changing that is called the growth rate so what will the growth rate growth rate will be d y over dt this is given that it is changing according to the size of the population at that time t that means it is given that d y over d x is equal to k y where k is the constant of proportion now we do remember this equation we have seen this equation in our first part of this lecture we do know that this equation has a solution as y t is equal to c times e to the power kt now we are being given initial value that the starting of the experiment we do have y at 10 000 cells now if i incorporate this initial value we have seen in our first examples that is this should be c so c would be 10 000 thus what we have got that this function would be 10 000 times e to the power kt now the question is what will be this k for that we have to see our experiment a little bit more we are being given that after 3 hours it has been observed that the number of cells are 5 lakh so now translate this into the mathematical formulation we have been given so rt is equal to 3 my function would be 10 000 times e to the power 3 k it is given that number of cells at that time are 5 000 so if i equate these two things what i do get e to the power 3 k is equal to 50 now we can solve this to find out the value of k how we will solve it we'll just take the natural logarithmic on both the sides of this one so we do get log of e to the power 3 k is equal to log of 50 which is giving us that 3 k is equal to log 50 thus k is equal to log 50 by 3 and now we will incorporate the value of natural log of 50 which is approximately 3.91 thus we are getting the value of k as 3.91 by 3 is equal to 1.3 so what we have got that is the value of k is 1.3 now this is the rate constant and what rate we would have because we have taken our t as the time in hour so it would be that it is changing 1.3 per hour that is rate of change is 1.3 per hour so now what is our solution we have got that at the time t the number of cells will be 10 000 times e to the power 1.3 t that is this is we have got this function as a solution or this is what we are giving as the number of cells at the time t now let us see this by the graph we see that is this is the graph of the function e to the power 1.3 t and since initially we have started at the value 10 000 we see here in the graph that it is 10 000 is starting so this is the solution of this one now we are ready to answer the question that is what will be the growth after one day that is after 24 hours that is we will incorporate t is equal to 24 in our solution what it will be 10 000 times e to the power 1.3 and in place of t i will put 24 which we will then solve will reach to 10 000 times e to the power 31.2 which will be giving me 3.5 into 10 to the power 17. that is these are the number of cells at the end of one day at the time of 24 hours now let us try to solve the second part of the problem that is what is the doubling time of the bacteria so first we should know what is the doubling time doubling time is actually the amount of time one takes for a given population to double in size so let us explain it let us say that is tau is the doubling time then what we are being said is that is the number of cells at time tau that is y tau should be twice the number of cells at the initial position that is y naught so now incorporate these information in our solution because we do know what is the y t y tau is 10 000 times e to the power k tau and initially the value of the y naught that is the number of cells at the beginning or ten thousand so twice of this would be twenty thousand thus what we got that e to the power k tau is equal to 2 now we do know that in this solution that k is 1.3 so we incorporate this k tau is equal to log 2 after taking logarithmic and incorporating the value of k as 1.3 we do get that tau is equal to approximately 0.69 divided by 1.3 that is approximately 0.533 what is this this is tau that is the time and time how we are measuring in hours so this is doubling time is approximately 0.533 hours which if i change into the minutes i do get approximately 32 minutes so we have here one problem where we are interested to know the growth of bacteria in a liquid culture we had observed that experiment for certain time we had initially started our observations when there were 10 000 cells and after three hours we have observed that there are five lakh cells and one more assumption we have made that the growth is proportional to number of cells present in the culture at that time with the help of these three things we have modeled our problem we said since the rate is changing according to the size of population we got the equation d y over d t is equal to k y and we have solved that equation with the help of other observations which we have made the first observation that initially we have started at 10 000 cells then the fact that after 3 hours we have observed it as 5 lakh from there we have got the rate constant that is k and then we could predict it that is after 24 hours how many bacteria will be in that culture moreover we can also answer the question that is in how much time the population will get doubled and these answers we have made similarly let us see one more example of the similar nature but little bit different that is radioactive decay radioactive decay is a property which characterizes the substances whose atoms undergo spontaneous changes these are being stored either in radioactive isotopes or in step that are unstable forms are in stable form on those materials which decays these decays are normally at a constant rate which when they burst are being counted on a geezer counter thus this burst r being that is atoms are being burst at a constant rate and this isotape that is the radioactive material changes to the stable material that says is that this constant rate would be proportional to the number of atoms present at the isotape at that time so this decay is that is the change in the mass of radioactive material is proportional to the mass present at that time and this would be decreasing we can understand this in another words as if example if there is half of the material present at the isotip then the number of bursts in the atoms will be half of the times whatever it is previously thus this change is in the decreasing order and this is called decay now this problem also we can study as we have used this growth phenomena what is the difference between the growth phenomena and this decay phenomena is the thing is that is there the things were changing to they are growing up and here the population is decreasing or decaying one so we can use the same kind of thing what we have been given that is let us see we have been defining the variable here first that is t is the time and let us say mt is the variable that is the mass of the ratio radioactive isotope at time t it is given that this mass is changing at a constant rate which is proportional to the mass present at that time that is the rate of change of the mass would be dm by dt this is proportional to the mass given at a time thus we can say that dm by dt is equal to minus km now see here i have used minus sign well whatever be the mass present at the isotape that will always be a positive number that will be always a positive and it is decreasing k i have used as a positive constant so i have to use the sign minus because as the time is changing the mass is decreasing so the rate of change should be negative thus we have used here minus k and this solution of this differential equation now we have already seen this kind of differential equation the difference is here that is rather than having plus k we are having here minus k of course we have changed the variable from y to m so the solution would be that is in the solution itself also we would change it to the minus k and the solution would be that m t is equal to m naught times e to the power minus k t here m naught i have used as the constant r that is the the constant in the general solution instead of c why m naught i have used because there must be some initial value known for this experiment also that is from where we are starting observing this isotip to decay that is that m naught is initial mass present on the isotape now we are ready to solve the problem let's see that is what is this problem before going to the problem let us see this function little bit more since this function is negative exponential it is having see we have this function which is decreasing of course this function should be decreasing because what we are having is that solution must be a mass which is a function which is decreasing and we see as initially we are starting at different values we do have different solutions now try to solve one problem let us see what the problem we are going to solve here carbon 14 is a radioactive isotape of carbon that has a high fla half life of 5600 years it is used extensively in dating organic materials that are very very old sometimes thousands of years of so the question is what fraction of original amount of carbon 14 in a sample would be present after 10 000 of the years now let us see is that how we are going to model this problem carbon 14 is a radioactive isotope so it has a property that it will decay and the decay would be at a constant rate and that rate would be proportion to the amount of the radioactive material that is carbon 14 present at that time thus the equation we will have is that dm by dt is minus km we have already seen that the solution of this equation is m naught times e to the power minus kt now for this problem we have to find out what is m naught and what is k r to answer this we have to know is that is after 10 000 year what will be the fraction of amount which will be left that is we are not necessarily want to know what is m naught whatever be m naught after 10 000 years what fraction of that amount will be there so we are more interested in finding first what is k how we are going to find it out we will see in our example what other conditions are given to us we have been given this information that half life now what is the half-life half-life of a substance or a decaying material is the amount of time it takes for 50 percent of the original amount of the substance to decay we have been given here that half life of this is 5600 so let here tau be the half life that is the tau will be the time it will take for material to come as the half of that percentage and let us say that m naught is the amount of the substance present initially then what it will give now because we do know what is that solution that is empty is equal to m naught times e to the power minus k t we have been given that at tau is equal to 5000 so at tau i would get m tau as half times of m naught now put tau as 5 600 in our solution and we do get that m naught times e power minus k tau as half m naught r e to the power minus k tau as half thus what we are getting is that 2 is equal to e to the power k tau and thus tau is log 2 upon k so we are just solving this equation without actually incorporating the value of tau we are given that tau is equal to 5600 so we will incorporate this and we get k is equal to log tau log 2 upon tau which is 0.69 divided by 5 600 which gives approximately 1.2 into 10 to the power minus 4 per year that is the time we have incorporated here in the terms of year because it is it takes a long time so we have got that k is equal to this thus what we have got the solution we have got the solution as n t is equal to m naught times e to the power minus 1.2 times 10 to the power minus 4 t now we are ready to answer our question question is that after 10 000 years the fraction of the original amount will be m at ten thousand divided by m naught because initially the amount is m naught so now use these informations in our solution m at ten thousand will be m naught times e to the power minus 1.2 into 10 to the power minus 4 and at the place of t i will use 10 000 which gives me that it is e to the power minus 1.2 now we can calculate this value which is coming up approximately as 0.3 what does it says after 10 000 years the 30 percent of the original sample will be left on the carbon 14 isotope that is that is again we have seen here in this one that if we are talking about a decay which cannot be stopped by any outside source so we are having the decay is happening and that decays again at a constant rate and that constant rate is proportional to the population size or the material present at that time that again we have used the same kind of differential equation and we have solved this problem also so in both these examples we have seen that we have used the exponential function and its differential equations that is why why i had used this example just because i have started my lecture with exponential function so this is not the only application of differential equation actually we can model many kind of physical phenomena all those phenomenas where changes are taking place and we are interested to know what will be the future behavior of that phenomena so wherever this changing changes are occurring we do have the differential equations knowing the differential equations knowing that how to solve those differential equations we can know the function that is we can know that phenomena what this phenomena is the function says is that if i say this is a function it says is i do know completely about this thing so if any phenomena i am saying is a function i do completely know the phenomena i do know the characteristic of that phenomena i can predict the future behavior of that phenomena or i may actually calculate many things about many characteristics of that particular phenomena that is why this differential equations are very very important here i would conclude today's lecture you