Good morning. With this lecture, we start our module on the analytical methods of solution of ordinary differential equations. Analytical methods are more powerful in the sense that they can give you the solution of differential equations in the form of expressions without having to resort to numerical values. And that means that groups of solutions you can represent together and the solution process is also much less tedious. However, there is one good aspect and one limitation to the analytical method. The limitation is that analytical methods are applicable only on certain special classes of differential equations. Beyond that the huge possibilities of differential equations do not lend themselves to successful analytical treatment. However, the good part of it is that most of the differential equations arising in nature in the physics or in engineering can be to a good extent reduced or simplified to those classes of differential equations which can be analytically solved. So, as we take up the analytical methods of solution of differential equations, we approach the problem slowly. First we consider first order ordinary differential equations and then we take up the problem of second order ODEs. And beyond that after second order, we will generalize whatever aspects of the second order methods can be generalized to higher order. So, first we take up the first order ordinary differential equations. Even before going into the solution method, let us see how typically a differential equation is formed and what it in general represents. For example, let us take this equation which is an equation in x and y that means in x y plane this will represent a curve. However, it will represent a curve with a given value of c and a given value of k. Now, if we are interested in supplying in assigning all sorts of values for these two parameters c and k then we will get different curves. And a differential equation between x and y is capable of representing all these curves together independent of c and k. And from that representation with the help of a differential equation, we will get rid of c and k these parameters and the relationship will be then given by x y and its derivative. So, that will be in some sense an intrinsic representation of all curves of this type. So, for that what we do we differentiate this and get the first derivative which is this. And if you notice between this y and this d y by d x, you can immediately see that the parameter c can be immediately removed. So, you have got two equations y equal to this and d y by d x equal to this. So, between two equations you can immediately eliminate one unknown say c. So, if you do that then we will have. So, for that what you will do you will divide d y by d x with y and then in the division on the right side c will get eliminated. So, you will find d y by d y as d y by d x by y as this divided by this. So, c gets cancelled and you have got k by s, c gets cancelled and most of the powers of x also get cancelled and you get k by s. So, from here you find that one of the parameters c has been eliminated, but the other parameter k remains. So, as many parameters you want to remove that many times you need to differentiate. So, before differentiating further what we can do we can take y on that side and have it in this form. There was another possibility of differentiating it directly and then between these c equations we could eliminate two unknowns which is c and k which are c and k. Now, as you differentiate this you will find that you get this. Now, again you will find that d y by d x has k into something and d 2 y by d x square has k into something else. Now, if you divide these two c has got eliminated already on dividing these two you will find that k also gets eliminated. So, from here you will note that d y by d x d 2 y by d x square turns out to be divided by d y by d x turns out to be this divided by this and that will mean this. Now, you cross multiply and find x y into this will be d y by d x into this and you have got this differential equation. So, this gives you a relationship between x y and its own derivatives d y by d x and d 2 y by d x square. Eliminated are all the variables parameters c and k. Now, this is a compact intrinsic description of the curve of the relationship between y and x. The other items the other quantities that enter into this equation are derivatives of the same function y. So, that means that this differential equation when you want to solve then basically you will get back this constant c and k. Sometimes it happens that while solving you might find some other solutions also which are not incorporated here. Now, you have noticed that every differentiation is accompanied with a loss of information. So, that loss of information is supplied at the time of solution of the differential equation with the help of initial or boundary condition. In the numerical solution of differential equations we have seen that with n initial conditions or n boundary conditions and nth order equation differential equation can be solved completely in principle and that will be available in terms of numbers numerical values. However, if you do not give those initial or boundary conditions if you give only the differential equation that will mean that as many possible differential initial conditions are there that many different forms of solutions you can find and that is why a solution of a differential equation without any conditions specified will be infinite and in that there will be constants of integration involved which will be called as arbitrary constants they can be given any value and that will be a valid solution. So, we talk of different solutions of the same differential equation that is differential equation satisfying different solutions satisfying different set of initial or boundary value boundary value. Now, in the classification of differential equations there are two important terms order and degree order of the differential equation is the order of the highest order derivative highest derivative that is appearing in the differential equation. For example, in this case the order of this differential equation is 2 because this is the highest order derivative that is involved and if the differential equation can be organized in the form of a polynomial equation in y and its derivative as this equation certainly is then the degree in which the highest order derivative appears in that that is called the degree of the differential equation. Now, in a differential equation there could be square root here for example, now square root of the right side appearing here also would be a differential equation in that case the degree of this would be 2 because to free that square root sign to free that radical sign you will need to square on both sides and this term here the second derivative would appear in degree 2 in that case the degree would be 2 in this case this is a second order equation of degree 1. However, when we try to classify differential equation in the form in the classes of homogeneous and non homogeneous bodies there we count the degree as the total combined degree of y d y by d x d 2 y by d x square and so on that is the unknown function and its derivative. So, in that sense here you will find that this term has got degree 1 and degree 1 total degree 2 similarly this term has total degree 2 this term also has got total degree 2 the way x is appearing is not counted in that for example, x could have appeared even as exponential sign cosine logarithmic function and so on that way x is appearing in the equation is not counted in this count of the degree. Now, when we try to classify the equation the differential equations as homogeneous or non homogeneous for that we count the total degree of every term. So, in this case total degree of each term turns out to be 2 and therefore, we call this a homogeneous equation. If there are some terms which are of degree 2 and some other terms which are of degree 3 or 1 then that would be called a non homogeneous equation. In particular a differential equation having the terms with only degree 1 and 0 is called the linear oddies. So, that is a particular case which will be of great interest to us. Now, when you try to solve the differential equation there are three kinds of solutions that you talk about one is the general solution. So, general solution of this differential equation for example, would be this that is in which there will be two arbitrary constants why two because the order of this differential equation is two. That means, its solution process in one way or the other will involve two integrations and in that there will be two constants of integrations involved and they will be the arbitrary constants. So, the general solution is in terms of two arbitrary constants in this case. So, the number of arbitrary constants will be same as the order of the d e. When we supply when we assign particular values to those arbitrary constants then we get a particular solution which we were getting during the numerical solution. There are some situations where there is a solution of a differential equation but it cannot be obtained by giving particular values to the arbitrary constants. Nevertheless it turns out to be a solution in the sense that the function and its derivatives together satisfy this. That is a single solution one solution without involving any arbitrary constant which satisfies this and which cannot be found from the general solution by assigning special value. That kind of a solution is called a singular solution which is possible only in the case of some non-linear equations non-linear differential equations and that is not possible in linear equations as we will see further. Now let us go into the solution processes of first order ordinary differential equations. The simplest possibility is when the differential equation is given in such a manner that you can split it in this manner and the d x taken on this side and psi y taken on this side results into this situation where the variables are completely separated. The left side has only y and its differential and the right side has only x and its differential. In that case independently we can integrate the two and we provide for the situation that the two integrals are allowed to differ by a constant. So, then just it is a straight forward quadrature or integration. So, integration of the left side is equal to integration of the right side plus some constant. So, this is the arbitrary constant which will be involved in the solution of the differential equation. Now the moment you have got this expression after integration which will be an expression in x y and this arbitrary constant c free of the derivatives you can say that you have solved this differential equation. So, this is the simplest possible situation that is where the variables are separable. Sometimes the variables are not already separable the way the equation is given, but they can be separated with some suitable substitution. For example, in this case where the derivative y prime is given as a complicated function of alpha x plus beta y plus gamma. Then you can say that this entire expression if we substitute as v then what happens we get d y by d x is equal to g of v. But then d y by d x two variables x and y are involved now here v gets involved. So, what we do we differentiate this and work out the relationship between v and x. So, as we differentiate this we find d v by d x is alpha plus beta into d y by d x. In the place of that d y by d x we put this g v. So, we get d v by d x is equal to alpha plus beta g v. The moment we get this we find that as we take this entire right side and put it in the denominator on the left side and get d x here then we get d x equal to d v by this. So, with d x no element of v is involved and with d v this entire thing will depend on only v and constants no x. So, then immediately through a quadrature we get the solution. Now, we say that we have solved the differential equation even though we have not found v as an explicit expression of x, but we have found x as an explicit expression of v. It is as good as finding the other not only that if you could find an implicit relationship between x v and c then also we would say that the differential equation has been solved. Now, after finding this relationship between x v and the constant c then one can put v equal to this and that will give us the relationship between x and y which we were looking for that is the solution of the differential equation. Next complicated case is this where the slope function d y by d x is available in the form of a rational function some polynomial of x y divided by some other polynomial by x y of x y that is a rational function of x y. So, in such a situation there are several cases that lend themselves to easy solution. One is when f 1 and f 2 are homogeneous functions of the n s d g say this is a function and this is another function. So, if this is if this is a if you divide the numerator as well as the denominator by x to the power something then if we can represent the two functions after division in this manner that is this is a function purely of y by x that means whenever y and x appear in this expression they appear together in the form of a ratio y by x and the same thing happens for this case also. Then the solution is very easy because in the place of y by x we can substitute u and as we substitute that we find that the d y by d x will turn out to be if you represent u as y by x then y will be u x and then the term d y by d x in the differential equation has to be replaced with d u by d x. So, you differentiate it and find out an expression for d y by d x that will turn out to be u into 1 plus d u by d x into x. So, this expression for d y by d x we can insert here and that gives us this expression for d y by d x in terms of x u and d u by d x and on this side we have simply phi 1 by phi 2 which are both functions of u only. And now separating this separating variables from this equation is easy we take u on this side this entire side is consisting of u only then and here we have only x. So, d x and x we take on the other side and the entire expression depending on u we take on this side. So, we get d x by x and here we will find d u with that phi 2 will join and whatever comes on the other side that will come here. So, that is phi 1 minus u phi 2 that will come in the denominator then we find that this can be separately integrated and this can be separately integrated. Similarly, even if we cannot represent it in the term of in the form of y by x, but if both numerator and denominator turn out to be linear expressions of x and y then also we can handle it almost in this manner.  Now, c 1 and c 2 are spoiling the game then we say we apply a coordinate shift x as capital x plus a and y as capital y plus k and we choose h and k in such a manner that in terms of capital x and y the c 1 and c 2 term vanish constant term vanish. Now, since these are just coordinate shift only shift no scaling etcetera involved. So, therefore, y prime which is d y by d x from here you will get d y as d capital y from here you will get d x as d capital x. So, the derivative remains same. So, in the place of d y by d x we can directly insert this d capital y by d capital x and as we represent x and y as with these two expressions we get this. So, a 1 into x plus h so a 1 x and a 1 h here similarly b 1 into y plus h. So, b 1 y here b 1 y plus k. So, b 1 y here b 1 k here and c 1 remains. Now, we say that what h and k we should choose so that this entire constant term vanishes from both numerator and denominator. So, as we do that we basically need to solve these two this system of two ordinary equations linear equations in two unknowns h and k. Now, if they have unique solution then we choose that h and k and through that substitution this particular case reduces to this situation. On the other hand if this pair of equations for h and k is inconsistent then we can say that in that case a 1 by a 2 should be equal to b 1 by b 2 and in that case as choice of u as a 2 x plus b 2 y will make it separable. On the other hand the third possibility that these two equations having infinite solutions together will not arise because these two being consistent having infinite solutions would require a 1 by a 2 to be equal to b 1 by b 2 and equal to c 1 by c 2. In that case this entire expression and this entire expression will be just multiples of each other. In that case the whole situation would not arise because even before that you would have divided and got a constant. So, that situation will not arise. So, if this pair of equations has unique solution then we choose that solution. On the other hand if it is inconsistent pair then a 2 x plus b 2 y is something which we substitute as u and then it reduces to a separable form. Now we go to some more difficult situation and a particular kind of non-linear differential equations of the first order is called Clare-Rott's equation or it is called the Clare-Rott's form of first order ODEs. That appears when we have y as x y prime plus some complicated function of y prime itself. So, this actually this term is making the differential equation non-linear.  This entire equation both sides which is said to x. On this side the derivative of y is p divided by d x. Here x into p the derivative of this is p plus x d p by d x. The derivative of this f p f of p which is said to x will be derivative with respect to p multiplied with d p by d x. Now when we organize this we find that these two p's cancel each other and we are left with this entire stuff equal to 0 and in that d p by d x is common. So, we organize it in this manner d p by d x is common we get x plus f prime p. Now what is this? This is a product of two terms which is equal to 0. So, that means either this must be 0 or this must be 0 by making this equal to 0 we get one kind of solution by making this equal to 0 we get another kind of solution. So, first let us consider the simple one d p by d x equal to 0. If d p by d x equal to 0 that will mean that p is constant, but p is d y by d x. So, that will mean that y prime is constant m. Then you say that if d y by d x is constant that will mean that y is a linear function of x and that you get as the family of straight lines which is y equal to just integrate this. You just integrate this. So, you find y is equal to m x plus something, but then you already know that from here m is y prime. So, you already know that y is equal to m x plus that something should be f of m. So, this gives you the general solution of this differential equation. However, one particular case rather one singular case that appears here is when this is equal to 0 and if this is equal to 0 then d p by d x does not have to be 0 that means p does not have to be constant and the equation of the curve need not be a straight line. So, when this is equal to 0 that is the singular solution. Then you get x equal to minus f prime p and y is equal to you already know that this is f p and from here you get x into p and x is this. So, you get minus p f prime p. Now, what is this is x equal to some function of p y equal to some other function of p. This function is known because f of y prime that is a known function. So, this is a known function and this is also a known function. So, when x and y both are known functions of a parameter p that essentially gives you a parametric equation of the curve which is the relationship between x and y. So, p appears as a parameter the slope itself appears as parameter. So, and in this there is no arbitrary constant involved and this solution you cannot find from the general equation because this is the equation of a curve and not a straight line this is an equation of a straight line. So, with different values of m here you get different straight lines which are a family of straight lines which is a family of solutions of this differential equation. Now, by giving the value of slope in this as p you do not get different solution, but you get different point on the same solution curve and that solution curve is called the singular solution and it turns out that it is a same slope that is appearing here and here and therefore, through every point that this curve passes a straight line like this passes this curve and this straight line having the same slope at that point that is with that value of m and that value of p here. So, that means all these lines obtained from different values of m will be all tangent to this curve at their respective points of contact. So, you find that a singular solution turns out to be the envelope of family of straight lines which are the general solution which can be found from the general solution of the differential equation. So, this is one very typical case which can arise only in the case of non-linear equations. Now, we slowly make a move towards second order differential equations and at this stage we consider only those second order differential equations which can be easily reduced to first order differential equations. In general, second order differential equations particularly the non-linear ones are much more complicated to solve compared to first order differential equations analytically. So, there are two distinct classes of second order oddies which can very easily reduced to first order oddies. One of them is the one in which the function y the dependent variable y does not appear explicitly. It does appear as it is in the form of its derivative, but y itself does not clearly appear. In that kind of a situation it is very easy to reduce or break down this differential equation into two first order differential equations.  This is a first order differential equation in p. So, we can solve it by the first order method. After solution of that we can say now this expression is equal to d y by d x that will be another first order differential equation which will be then solved. Now at every stage we will get one arbitrary constant. So, finally there will be two arbitrary constants accumulated through the two stages. There is another class of second order oddies in which y appears explicitly, but x itself does not appear y y prime and y double prime. In this case also we use y prime as p, but with a little difference. What we do? We take y double prime as d p by d x, but then x gets involved which was not here. So, we were happy. So, now we need to get rid of this x from here. So, we say d p by d x is d p by d y into d y by d x and d y by d x itself is p. So, we find y double prime as p d p by d y and when we insert this expression for y double prime here then we get this. Now this is a first order differential equation between p and y. As we solve this we get the solution which is p of y and now which we can then solve. So, then d y by d x equal to p of y will give us now p of y goes down d x comes up and we integrate it and it is variable separable. So, we get this as the final solution. So, in these two classes of second order oddies you can apply a suitable substitution to break it down into two first order oddies which can be solved in turn in succession. Now the general theory of first order linear differential equations. Before that we have a little discussion on exact differential equations. As you write a differential equation d y by d x is equal to f of x y. Now if that f of x y can be organized in this manner. Now here no simplification has been made x and y both are involved and then we take this on the d y side. We take this on the d x side and then get everything on the left side. Then we will get m of x y into d x plus n of x y into d y equal to 0. So, from here you will be able to see that d y by d x equal to any function of x y can be always organized in this manner and therefore, that differential equation and this equal to 0 the differential equation that you will get are actually equivalent. So, any first order differential equation can be organized in this manner that is this differential equal to 0. Now, we have already seen from our discussions on multivariate calculus that this is called an exact differential. If there is some function of x y phi if there is some function phi of x y the partial derivatives of which are m and n del phi by del x is m del phi by del y is n. In that case we call this as an exact differential because it is exactly a differential of the function phi because we find that if phi is a function of x y then we try to work out d phi. Then that will turn out to be del phi by del x into d x plus del phi by del y into d y. So, in that case we call it an exact differential. Now, whether there exists some function phi for which this is the x derivative and this is the y derivative that can be easily checked if we differentiate this with respect to y and this with respect to x. That means del m by del y will be the second derivative del 2 phi by del x del y del x and del n by del x will be del 2 phi by del x del y out of continuity of the derivatives these two will be same. So, this is the check del m by del y is equal to del n by del x. So, if we have an exact a differential like this and we want to check whether it is exact or not then we differentiate this with respect to y and this with respect to x. If they turn out to be same then this is an exact differential and in that case the corresponding differential equation is called an exact o d. Which will immediately get solved when we integrate it both sides that is when we write it openly then in place of m we write del phi by del x and in place of n we write del phi by del y. This is an exact differential and that is d phi equal to 0 the solution of that will be phi of x y equal to constant. We immediately get the solution now sometimes it happens that the differential equation as available in hand is not exact, but multiplication with something will make it exact. Now, this is a little working rule for integrating this when you have m and x n separately. So, while integrating what you can do is that first round you integrate this partially with respect to x considering y as constant for the time being and therefore, in the constant of integration you include an arbitrary function of y. Similarly, here when you integrate this part with y with respect to y partially then you consider x as something like a constant and therefore, in the arbitrary constant you include a function of x. And then when you say that these two should mean the same thing that means we try to determine g 1 y and g 2 x such that this entire solution and this entire solution turn out to be same. And that gives you the particular case of g 1 and g 2 which will make it same and that is the solution. So, if these two are not same say if the condition is not satisfied and this is not an exact differential which means that this differential equation is not an exact ODE then sometimes we look for a suitable function with which we can multiply the differential equation which will make it exact. So, that kind of a function f is called an integrating factor. And the succeeding theory of linear differential equations that we will we are going to discuss depends directly on the search for an integrating factor. The general first order ordinary differential equation will look like this even if there is some function of x appearing here we can always divide with that the entire equation to put it in this form. And since it is linear so no term of y can appear here. So, in the first order linear ODE only three distinct terms we can get one is a term with d y by d x from which we have got rid of all the coefficients by dividing with that. Another term can be there which is the y term as coefficient of that any expression of x can appear. And the other term which is making this equation non-homogeneous that can be any function of x including any other constant as a beta etcetera similarly here for that matter. Now, x y and d y by d x are the three important items in which also y and d y by d x. So, this is the d y by d x term this is the y term and this is the term which is p of both this form is called the Leibniz equation. And we try to find out the solution of this for the general first order linear ODE. So, we look for an integrating factor with what to multiply this side so that this side turns out to be exactly the derivative of something. So, we say that suppose that something which we are looking for is f f of x we multiply it with f of x this left side and say f x d y by d x plus f x p x y this we want as the derivative of something. Now, it is clear that derivative of whatever we are going to put that will be y into f because the y can appear only in this manner otherwise this d y by d x term will not be so simply in hand. So, we find that f into y gives us something the derivative of which will have one term which is f into d y by d x the other term will be y into d f by d x.  So, the integrating factor f x must satisfy this differential equation which is also a first order linear ordinary differential equation. The difference between this first order linear differential equation and this first order differential equation is that this is homogeneous. This does not have a term free from f and d f by d x and solving this is extremely simple because this is already in the variable separable form. If we take f below then we find that here we have got d f by f and on this side we have got p x d x. Now, this is dependent only on f this is depending dependent only on x. So, we integrate it the integral of this is l n f and integral of this can be whatever that is as long as that is free from f free from y we do not care how complicated expressions of x are appearing in that. So, this is we keep this in the form of quadrature itself. If l n f is this integral then f is its exponential function e to the power this whole thing. So, we have found the integrating factor after we have found the integrating factor we know that if we multiply this entire equation with this integrating factor then the left side is going to be exact d by d y d by d x of f into y that is this into y. So, we put that. So, the right side left side after multiplication with this is exact derivative of y into this term. So, we say that then we integrate it because on this side we will have q x into this. So, q x into the integrating factor we integrate it and we get y into this is equal to this integral whatever complicated that integral may be plus the constant. Now, this gives us the complete solution of the differential equation. Of course, we can get the expression of y if we multiply both sides with the inverse of this that is e to the power minus integral of p d x with that when we multiply both sides we get the expression of y in terms of x and the arbitrary constant p. Now, notice that when we took this integrating factor f while solution of while solving this in this integral we did not take an arbitrary constant even if we had taken we would get essentially the same solution back. So, in this case whether you take that arbitrary constant or not it will not matter. So, this is the way we solve a first order linear od that is in the hunt for its integrating factor. We find a homogeneous linear od the solution of which is trivial and that provides us the integration integrating factor which when plugged into this by multiplication throughout gives us the left side left hand side of the equation as a as an exact differential coefficient. Now, there are some non-linear equations which can be easily reduced to this labnitz form through certain substitutions. One of them is the Bernoulli's equation in which here apart from q of x a y to the power k term appears which makes it a non-linear equation. However, this y to the power k term can be easily removed by making this substitution. When we substitute z equal to y to the power 1 minus k and then we differentiate this and try to find d z by d x we find that will be 1 minus k into y to the power minus k into d y by d x and then we find that in place of d y by d x we now will try to put expressions involving d z by d x and we will find that d z by d x turns out to be this d z by d x plus this will convert this equation into this form. How? Say we multiply this equation with 1 minus k y to the power minus k. So, then this y to the power minus k multiplied here we will get rid of it 1 minus k will remain. Here also 1 minus k will remain and this y multiplied with 1 minus k and y to the power minus k will get z here. So, 1 minus k p x will remain and z will come. So, here we will get a linear expression of z and on this side the y k y to the power k will be removed through the multiplication of y to the power k and when we multiply when we put y to the power minus k into 1 minus k into d y by d x here this side is just d z by d x. So, now the resulting differential equation that we get in terms of z that is exactly in the form of Leibniz equation. Well as coefficient of z we do not get p x, but we get some constant into p x and in the place of q x we get 1 minus k into q x and this can be subsequently handled in the same way as we handled the previous equation. Another equation is there which is Riccati's equation here the expression on the right side is only slightly complicated compared to a linear equation. If we had removed this equation this term from the equation it would be a Leibniz equation. Now this quadratic term is causing the trouble. However, when we try to solve the differential equation this actually poses more problem than Bernoulli's equation. However, if one solution of this differential equation is known then there is a nice way to get the general solution. In general without knowing even a single solution it is not possible to solve this analytically. However, somewhere from somewhere if you get one solution y 1 then getting the general solution is easy because then you say that let us call the general solution as that known solution plus something which is unknown and then substitute this here. As we substitute that this entire thing as y then on this side we get y 1 prime plus z prime here this is a x simply b x into y 1 plus z c x into y 1 plus z whole square and as we as we simplify this we already know that y 1 satisfies this. So, y 1 prime is already known to be this.  So, b x z x remains from this whole square as we subtract this c y 1 square goes off the other two terms remain one of them is a linear in z that gets put here the other is quadratic in z that is here now you note that this equation is in Bernoulli's form z prime is here a linear expression of z is here with coefficient as x and known expressions of x and here in place of q x y k you have got c x z square that is z to the power 2. So, then using this method we can reduce it to Leibniz form and subsequently solve it. Now this is Riccati's equation there are large number of physical systems and with their offshoot as engineering systems where first order differential equations appear in a big way and they can be solved to predict the behavior of those physical systems. Quite often in the description of curves and orthogonal trajectories first order differential equations are found quite useful. For example, in x y plane a one parameter equation in x and y with c as a parameter represents a family of curves we know that now differential equation of the family of curves can be found by differentiating this with respect to x which will involve d y by d x and then eliminating the parameter c between this equation and the expression which we get by the differentiation and as we eliminate that we get d y by d x as a function of x and y and that gives us the same family of curves in its intrinsic equation. Now if we have got this family of curves in x y plane suppose in this manner and then in many applications we want to know another family of curves which are all orthogonal to it. For example, if this family of curves represent potential curves or contours then the flow lines will be different curves another set of curves another family of curves which will be all orthogonal to these. So this is a an application which is involved in many different types of engineering problems. So in that what we do that if the slope of the curves of this family is given by f 1 then the slopes of the curve of the family which is orthogonal to this whatever they intersect they make right angle they make right angle. So that means the slope of this white curve and the slope of the intersecting blue curve at every intersection point must be right angle that means this fellow's slope and this fellow's slope the two slopes their product should be minus 1. So when we use that fact from analytical geometry then we say that slope of the curves orthogonal to this will be d y by d x equal to minus 1 by f 1 and then this is another differential equation which we can solve and as we solve that we get another family of curves in the form of solution of this this expression this equation. So this equation gives another family of curves and the family of curves given by this phi equal to 0 and the family of curves given by this psi equal to 0 give us two orthogonal families of trajectories of curves that is two families of curves each curve from this family is orthogonal to each curve of the other family. And as we give different values of c we get different curves of one family and as we supply different values of k we get the different solution curves of the second family. So these two together represent orthogonal trajectories now if one of them is the velocity potential potential function of a flow situation then the other one will be the giving the stream line. Similarly, if one of them gives the electric potential contours the second family of curves will give us the directions of the electric fields and so on. In all potential problems and field problems we will find this pair of orthogonal trajectories in 2 d it will be orthogonal trajectories in higher dimension it will be more complicated. So in this lesson we find that these are the important point to make note of the different methods of solving first order differential equations. After this as we discussed earlier we will make a move towards a second order differential equations. In general a second order differential equation would look like this in which x y and its two derivatives will appear. Now in the particular case when we will be talking about linear ODEs second order linear differential equations. So for that the special situation that will arise in that y prime square y prime into y double prime y into y prime such terms will not be present. So there what kind of terms can be present there will be one term with y double prime one term with y prime one term with y and there can be a term without y or its derivative that is this. So typically in many engineering situations or physical situations this side can represent the dynamic model of a plant of a system and this side can represent the input given to that system which drives the system and the response of the system is given in y which we need to solve in order to predict the behavior of that particular physical system. Now in this case which is a linear ordinary differential equation we have terms of degree 1 degree 1 degree 1 and degree 0. Now if this side is not there then we will get a homogeneous equation if r x is 0 then we get the linear homogeneous equation which is this. Now in this a particular case appears when the coefficients of y y prime y and y prime are constant. So then we get this differential equation which is a special case of this and we call it the linear ordinary differential equation with constant coefficient. Similarly here if we get if we have constant coefficient then we get a linear homogeneous ordinary differential equation with constant coefficient which is this. Now this can be considered a special case of this by making the coefficient constant or a special case of this by getting by putting the right hand side as d u. So the linear second order ODE will have its general form here the solution of which we will consider but before that we will consider these special cases. This one this one and this one and finally we will attach this general linear ordinary differential equation of the second degree. So in the next lecture we will start with this simplest case for which the solution will be found very easily. Thank you.


Hey, welcome to my talk. I know who has access to my cloud, do you? I'm Egal Fleckman. I'm an identity security engineer with yourself experience building products for identity and PKI management at Microsoft. And after a few years, I decided to leave and co-found Keto's where we do, we make it easy for companies to become passwordless with easy to use PKI solutions. And there's my Twitter, if you have any questions, I'll be in the Discord and I'll also be on Twitter. So feel free to message me wherever you prefer. And I'll also be in Vegas. So if you wanna swing by and talk to me in person, like let me know. I think here's a picture of me outside. So it looks like I do stuff outside and it's not just coding all day. So let's just jump right into it with a controversial topic. Why security reviews suck? So one of the main things is it's based on what do you think your infrastructure looks like? So it might be you forgot to close a port and in the review you go and say like, now all my ports are closed or you might say this is encrypted and it's not actually encrypted. And it also doesn't change over the years. I've seen teams that have gone over and over again with the same diagram and do not update it to the new microservices they added or they completely changed the platform. And we just kind of like assume that it's still working. And I've also seen some security professionals that care more about the actual diagram than they do from the security. I actually was blocked once until I put a box around my items to show that everything was running in Azure when it was logical that everything was running in Azure. So let's talk about how hackers actually get in and it's a hacker talk. So I had to put a black guy in a black hoodie and let's talk about the first one that is adding a member to a group. So this one is pretty easy. Like once you get access to a group and you add your identity to that group, it's pretty hard to detect because you might have like your engineering group that you add to all your Azure resources. And you're not really checking it that often. You don't change engineers that much. So like you're not gonna be checking the membership. So it's an easy way to get in and when you're checking our back and everything, you're not really gonna check for the group members. Another one is adding an Azure classic administrator. So Azure classic administrator is something that they have been trying to get rid of for like four years now, I think. And now they put them into a different tab so it's harder to add, but it's also harder to detect if someone adds someone and then you don't notice. The third one is adding a service principal as a contributor to a subscription. This was as a pentester, one of my favorite ones to do. You just add the employer to the end of the service principal and no one is gonna touch it because no one wants to break the employment. So once you're in, you're in forever and you will usually have like high contributor owner access. The next one is removing AAD only SQL authentication. That one I haven't really seen, but we're actually forcing everybody in my company to use AAD authentication for SQL and we don't want people to revert back into using SQL or creating their own passwords and everything. So we added that check to make sure that everything is secure. And then another one is adding permissions to Azure AKB policies. So basically you might have good segregation of duties, having different applications, having different access to certain secrets. Some might be able to get them, some others might be able to create them or delete them. But if someone manages to compromise one of those accounts and give themselves access to everything, then when you look at the access policies, you still see the same application. So you don't see anything different, but they might have more access than what they didn't need. And the last one I'm going to talk about is adding IP addresses to the firewall. So most companies have hundreds of IP addresses either from like your microservices that talk to your service or you might have your corporate VPN that has access to it. And adding one extra one either being like a developer being lazy and adding their home address or a bad actor adding their address so they can access the system. It's something that can go unnoticed. So the solution, we created CloudWatcher. CloudWatcher is a PowerShell module that runs on Azure Automation. It's actually in the Azure gallery so you can just go there and download it. I have instructions at the end of the slides. It scans your subscription and if it detects any configuration changes that I just mentioned, it will fail the script and then you can send an alert. We have it that like it emails, sorry calls our own call engineer and let us know. And here you have a run that it shows how it's when there is no changes detected and here's a run that it tells you that there were changes detected. So it'll tell you exactly what it found like resource providers changed and RBAC changes and it will fail the round book. So then you can have like Azure Automation like call you or like send you an email whatever you want it to do. So let's talk about what we actually cover. So RBAC changes, any RBAC change adding removing changing while it would still be adding or removing RBAC we'll catch it. Classic administration changes so you don't have to keep looking for them. We'll look for them. Azure resources providers. So Azure resource provider is something that I didn't really talk about but that's basically how Azure gives for example Key Vault access to your VMs to be able to encrypt their disks. So it's kind of like another way that you can create your custom providers that it will give you certain access to the subscription. So we like to keep that blocked to whatever we are comfortable with in our baseline. Resource creation or deletion. We don't want developers to create kind of like their own VMs or something in the production subscription that it might endanger the rest of the subscription or delete something by mistake that might be like your availability zone or something. So we want to detect them any resource creation or deletion. We also have the change of group membership. That won't, there's an asterisk because we only check first degree. So if you have nested groups we don't go through each group and check all their groups membership. We just check the membership of the first group of any group that is called anywhere in your subscription from RBAC to AD or sorry, AD admins on SQL. Anything that we can find we'll add that group and check it. SQL firewall rule changes. So if they add or remove IPs, SQL server AD. So basically I already talked about it kind of like if you change the AD only or if you change the AD administrator, Azure Key Vault access policies changes and firewalls. So now let's talk about the elephant in the room. I can see you guys all thinking, all right, EGAL is just talking about Azure Security Center. So did I just reinvent the wheel? And the answer is no. You can have a security score of 100% and still be vulnerable to these backdoors by for example, Azure Security Center will tell you if like your management port like port 22 is open to the world, but it won't tell you if it's still closed but an extra IP that is not in your corporate IP address was added. So this one is an addition to Security Center. It does not replace Security Center. We use Security Center and we also use CloudWatcher. And then for the setup, this is a great image of how we actually have it set up. We have multiple subscriptions and they're all talking, they're all cloud watching each other. And here to simplify it, I'm just showing one of the cloud watchers. So you have this production subscription that is the one that you wanna protect in this case. Then you have a cloud watcher in the same tenant that has reader access to this subscription and hopefully this other subscription has different identities managing it. So you don't have the same admin identities managing both subscriptions. And then you have the baseline. We actually have it in another tenant that CloudWatcher only has read access to pull from the storage account and get the baseline JSON. And then we run continuously every few minutes to make sure that everything is still the same and nothing has changed. And if a change is detected in any of the subscriptions will alert our engineers on it. So now let's talk about what's next. What are the next things we wanna build? So one of the things is create a UI to visualize your resources. As I said, I don't think that security reviews right now do like the best job on it. I've never seen a red team go against security review and grab the information from there. They usually just scanned the actual infrastructure. So we wanna create a UI to visualize those resources and maybe add some alerts saying like, hey, like we notice that you haven't enabled SQL AAD. Do you wanna enable that? Or we noticed that you don't have Azure Key Vault firewall rules and just kind of give you better ideas on how to protect your infrastructure, add more resources. So right now we solely focus on SQL and Azure Key Vault. The reason for that is those are the core for our company but we wanna grow it to other Azure services. And the last one is integrated with Pulumi, Terraform and Arm template. So basically we wanna be able to compare your baseline to your infrastructure as code and read the desired configurations to be able to secure it from your desired configuration to your baseline to your actual infrastructure to make sure all three match. And then thank you so much. If you have any questions, please send them to the Discord or send them on Twitter. Here's a GitHub where you can find the read me with all the information about how to install it. And then we have the slides also we posted them online.


Hello and welcome to CMC Markets on Tuesday the 25th of March in the weekly market update. And let's have a quick look at the equity markets. They're a bit of a mixed bag. I think investors are a little bit undecided as to the future direction of equity markets. We're seeing a bit of a range starting to take hold at the moment. The S&P topped out once again around about 1885. It's finding dips pretty much well sought after. If we look at the FTSE 100 and the DAX, they're well below their 2014 highs and yet we have the FTSE MIB trading at multi-year highs. So I think at the moment it's a little bit, there's a little bit of indecision with respect to the next direction or the next move higher or lower in terms of what equity markets are doing. And certainly geopolitical risk is playing into that story. There are concerns about a slowdown in China. There's also concerns about what's going on in Russia, the Crimea and an escalation into Ukraine and possibly even Moldova. You've also got a little bit of unrest in Turkey as well. And I think that's feeding into concerns about emerging market growth going forward. Having said that, we are seeing a bit of an improvement in some of the macroeconomic data that's coming out of Europe. Unfortunately, with earnings currently elevated levels, I think investors are finding it a little bit difficult against a backdrop of a possible dialing back of stimulus by the Federal Reserve to really make a decision one way or the other as to the next move in equity markets. So essentially what we're getting at the moment is an awful lot of chop and as a result that makes trying to predict the next move that much more difficult. So I think for the purposes of this week's market update, I'm going to look at US yields in the context of what Mrs. Yellen said last week and those three words of around six months and try and establish the direction of US rates and potentially the next move in the US dollar. And I'm also going to have a look at gold in that context as well as having a quick look at euro sterling in the context of the improving economic outlook in Europe and last week's UK budget. So I'm going to start with euro sterling and euro sterling is actually managed to remain quite resilient. Despite the fact that this morning, Jens Weidmann, the chairman of the German Bundesbank, suggested that maybe QE probably wasn't as off the table as some people had initially first thought. He didn't rule it in, but on the other hand he didn't explicitly rule it out and he also didn't rule out the possibility of negative deposit rates and that has had the effect of maybe putting a slight cap in the euro. And for the purposes of this particular discussion, let's look at euro sterling and the euro sterling chart because there still remains a significant top around about the 84 level and we can see that from the chart in front of us. The 200 day moving average continues to cap. Also last week we saw a bearish engulfing day on the daily charts. We have pushed back up towards that trend line resistance from the highs in December, but thus far we are finding a little bit of selling interest around that 84 level. And when you actually look at the differing or the divergence in terms of monetary policy or potential divergence in terms of monetary policy with respect to the Bank of England and the European Central Bank, you've got to ask yourself why euro sterling is as high as it is because I think if you look at the two central banks' policy stances, the Bank of England is more likely to tighten policy and the European Central Bank is more likely to ease policy. So on that basis alone, the yield differentials should favour the pound. And the fact is economic data in the UK is continuing to improve. Having said that, inflation data came in much weaker than expected. Well actually it wasn't that much weaker than expected, but it's on a downward path, 1.7% year on year. The likelihood is it could continue to drift lower and that will take some of the pressure off the Bank of England to hike interest rates over the next 12 to 18 months. So let's move on to the path of US rates. Now on the long end of the curve, 5 and 10 year, US Treasury prices have been below the 200 week moving average. The lag at two that has been the two year prices. They actually haven't traded below and closed below the 200 week moving average since 2007. Now if we look at the daily chart which I've got in front of me right now, you can see that for the first time since 2007, we have closed below the 200 week moving average. And that could potentially signal a rise in US yields. Now we're not above the highest levels that we've seen for high four US two year yields. We saw that in the middle of 2013 as can be seen from that spike lower in the Treasury price chart on the daily chart in front of you. But you can see from the weekly chart, the line chart that I'm now just about to put up in front of you, that the direction of travel for US rates could potentially be for lower Treasury prices and high yields going forward. And that will be potentially positive for the US dollar. And that is borne out in the gold price. We've seen a significant decline in the gold price over the past week or so in light of Janet Yellen's comments last week which some have said may well have been misinterpreted by the market. But I certainly think it gives I think it gives colour to a direction of travel for US rates with respect to a slightly stronger dollar over the course of the next six to 12 months. So we broke the downtrend line on the daily chart and we're now heading towards the 50 day and the 200 day moving average. Now those moving averages have crossed over which is potentially positive in terms of a golden cross. However, when we look at the weekly chart it gives us a slightly different story. Because what it does on the weekly chart is we have a bearish engulfing week. And that's potentially a very bearish development. And we could actually see in the short to medium term a move lower towards the 50 and 200 day moving average towards the 50% retracement of the entire up move from the lows that we saw in 2013 around about $11,180 an ounce to the peaks that we saw at the beginning of March around about $1,392. A 50% retracement of that is around about $1,287. So I think there's a good chance we could actually see a lower gold price on the back of a stronger dollar. Obviously that completely discounts any deterioration in the geopolitical backdrop with respect to Russia, Ukraine and Crimea. Okay, so that's pretty much it for this week. Once again, thanks very much for listening. On a side note, just a quick reminder to you all. I will be in Bristol on Thursday for an active trader network. So if you're in the area, please feel free to drop by. There is an invitation along the bottom here. And then on Friday, one of my colleagues, Jasper, will be hosting an active trader network here in the office in London as well. And again, sign up along here for that. Until next week, thanks very much for listening. This is Michael Houston talking to you from CMC Markets.