 Welcome back. In this lecture, we will discuss about the solution of variable coefficient differential equations. We have seen that if the differential equation is constant coefficient whether it is of first order, second order, third order, then the method of solution we have already seen in the previous lectures that you find the characteristic equation solved. After solving or finding the roots, we formulate the solution and we get the general solution and particular solution. And in case if the differential equation is having variable coefficient, then that method fails. For example, if I consider a general differential equation, say a second order. So, consider a second order variable coefficient differential equation, say this is a 0 x y double prime plus a 1 x y prime plus a 2 x y is equal to 0. So, call this equation as 1. Then how to solve it? By the method of characteristic equations and finding the roots will not work. But there are certain situations where special forms of this equation which can be solved just as how we did in a constant coefficient differential equation. So, for example, if we can reduce this variable coefficient differential equation into a constant coefficient differential equation by giving some suitable transformation. For example, if we consider, so consider the Euler Cauchy equation. So, Euler Cauchy equation, so given by x square y double prime plus a x y prime plus b y is equal to 0. So, where a and b are constants. So, this equation though it is a variable coefficient, it is in a nice form. This can be reduced into a constant coefficient differential equation just by applying some transformation. So, consider the transformation, say z is equal to natural logarithm of x or x is equal to e to the power z. So, if you use this transformation, then we are changing the independent variable from x to z. So, if you do this transformation, then what is d y by d x? d y by d x is given by the chain rule, d y by d z into d z by d x, which is equal to by using the transformation 1 by x d z by d x is, z is ln x. So, d z by d x is 1 by x 1 by x d y by d z. So, therefore, differentiating a function with respect to x is equivalent to multiplying 1 by x and taking the derivative of the function with respect to z. That becomes the rule. So, if for this case, then if you find the second derivative, so d square y by d x square is nothing but d by d x of d y by d x. So, we apply the rule stated just above that is differentiating a function, differentiating a function with respect to x is equivalent to 1 by x into differentiating the function with respect to z. So, therefore, we get the second derivative d square y by d x square is equal to d y d x of d y by d x, which is equal to 1 by x times d y d z of d y by d x. So, which is again by definition, 1 by x times d y d z of, what is d y by d x? d y by d x is 1 by x again, 1 by x d y by d z, so which is equal to 1 by x times d y d z of 1 by x is by transformation e to the power minus z, so d y by d z. Now, we can differentiate with respect to z. So, this gives me 1 by x into minus e to the power minus z, so applying the product rule d y by d z plus e to the power minus z times d square y by d z square. So, this is equal to and this e to the power minus z, e to the power minus z is 1 by x and this is also 1 by x, can take 1 by x outside, so this gives me 1 by x square into minus d y by d z plus d square y by d z square. So, therefore, the second derivative d square y by d, d square y by d x square is 1 by x square into minus d y by d z plus d square y by d z square. So, this implies that x square into d square y by d x square is equal to minus d y d z plus d square y by d z square. Now, putting this into the Euler-Corsi equation, so therefore, the Euler-Corsi equation becomes minus d y d z plus d square y d z square plus a times d y by d z plus b y is equal to 0. And therefore, if we simplifying this and taking d square y by d z square first, so this is d square y by d z square plus a minus 1 d y by d z plus b y is equal to 0. So, look at this equation, this equation is a second order constant coefficient differential equation. So, we start up with the Euler-Corsi variable coefficient differential equation, the transformation reduce this into a constant coefficient differential equation of course, is second order and this equation can be solved. Now, we are taking the characteristic root, so it is a constant coefficient differential equation. So, this is a constant coefficient differential equation. So, the method is characteristic roots, find the characteristic roots and these three situations where the roots are real and distinct, the roots are real and equal and roots are complex. So, which we have already seen, say for example, if we take the characteristic roots, the roots are given by lambda 1 lambda 2 is equal to 1 minus a plus or minus square root of a minus 1 the whole square minus 4 b all divided by 2. So, the discriminant of the equation b square minus 4 a c by 2. So, if the roots are real and distinct, so call it then lambda 1 lambda 2 are the real roots. Then the solution is given by, solution is given by Y of, remember the independent variable is now changed to Z, Y of Z is given by c 1 e to the power lambda 1 Z plus c 2 e to the power lambda 2 Z. And again, so if I want to change it back in terms of x to change the independent variable back to x, use the transformation. So, therefore, Y of x is equal to you change Z to x that is c 1 e to the power lambda 1 and Z is ln of x. So, ln of x plus c 2 e to the power lambda 2 ln of x and this is nothing but ln of e to the power lambda 1 ln of x is, so c 1 if you simplify it is x to the power lambda 1 plus c 2 x to the power lambda 2. So, this is a general solution. So, general solution if the roots are real and distinct and similarly, if the roots are real and repeated and the roots are complex can be treated similarly. This is the situation where if the differential equation is having a very special form and in case now we are going to deal with differential equation which is in a general form that is a 0 x y double prime plus a 1 x y prime plus a 2 x y is equal to 0. If this cannot be reduced into a constant coefficient equation and no other methods are available to solve it, then one method of solving is by using the power series solution, the series solution of this differential equation. So, let us see how to solve this by using a power series solution. So, power series solution to the equation a 0 x y double prime plus a 1 x y prime plus a 2 x y is equal to and what we expect is, we expect a solution to this equation called it 1, 1 in the form y of x is equal to summation n goes from 0 to infinity c n x minus x 0 to power n where c 0, c 1, c 2 etcetera are constants. So, we expect a solution to the variable coefficient differential equation 1 in a power series form an infinite power series about a point x 0. Now, the question is, does there exist a power series solution to 1 in the form 2? So, called this form as 2. So, does there exist a power series solution to the above differential equation in the form 2 and if it exists, how to compute the constant c 0, c 1, c 2? So, if s, how to compute c 0, c 1, c 2 etcetera in 2. So, these are the two questions. So, first we address the existence of power series solution to 1. So, under what condition the power series solution for 1 exists and that series is convergent? To answer this question, we need some of the basics ideas, some basic definitions. So, to address the existence problem, there is under what condition one has a power series solution of the form 2. So, we require some conditions and we define and let us again state the form of the equation a 0 x y double prime plus a 1 x y prime plus a 2 x y is equal to 0 and equivalent normal form is equivalent normal form. So, we call this equation 1 and if we take the equivalent normal form y double prime plus p 1 x prime plus p 2 x y is equal to 0, where divide throughout by a 0. So, where p 1 x is equal to a 1 x y a 0 x and p 2 x is p 2 x y a 0 x. So, now, we define what do you mean by a analytic function, a function f is said to be analytic, a real valued function f is said to be analytic at a point x 0, if it is Taylor series expansion about x 0 given by summation n cos from 0 to infinity, the nth derivative of f evaluated at x 0 divided by n factorial into x minus x 0 to the power n. The Taylor series about x 0 exist and converges and converges to f x for all x in some neighborhood of x 0 containing x 0. So, neighborhood in some neighborhood or in this case is some interval containing x 0. So, in this case, we say that a function is analytic. Examples see all polynomial functions are analytic, all polynomial functions are analytic everywhere and functions containing e to the power x sin x cosine x, they are also analytic and another important class of functions analytic is all rational functions. So, rational functions of the form p x by q x, where p x is a polynomial, q x is a polynomial. So, rational functions are also analytic at all point except the denominator, the points at which are denominators not 0, except at the point of denominators equal to 0. So, rational functions is analytic except at those values of x at which the denominator is 0. For example, if we take a rational function 1 by x square minus 3 x plus 2, which is written in the form x minus 1 into x minus 2. So, this is analytic at all points except at x is equal to 1 and x is equal to 2. At these two points x is equal to 1 and x is equal to 2, the rational function is not analytic elsewhere it is analytic. Now, a point is another definition is ordinary point for a differential equation. So, definition the point x 0 is called an ordinary point of the differential equation 1. If x is equal to 1, if differential equation 1 is in the normalized form, call it y double prime plus p 1 x y prime plus p 2 x y is equal to 0. If both functions p 1 x and p 2 x are analytic at x 0, a point x 0 is said to be an ordinary point of the differential equation in the normal form y double prime plus p 1 x y prime plus p 2 x y. If both of the functions p 1 x and p 2 x are analytic at x 0, if that fails, if it is not true then we say the point is a singular point. A singular point if either or both of these functions not analytic at x 0, then so then x 0 is called a singular point of the differential equation. See quickly let us look into an example y double prime plus x y prime plus x square plus 2 y is equal to 0. So, here p 1 x is just x and p 2 x is x square plus 2. So, obviously both are analytic for all points, all points are ordinary points for this differential equation. All points are ordinary points and if we consider another differential equation x minus 1 y double prime plus x y prime plus 1 upon x y is equal to 0. So, here p 1 x is x upon x minus 1 and p 2 x is 1 by x into x minus 1. So, p 1 is analytic for all points except at 1. So, for p 1 x is equal to 1 is a point at which p 1 is not analytic. And here x is equal to 0 and x is equal to 1 for p 2 is not analytic at these two points. So, therefore, the differential equation is analytic for all points except 1 and 0. So, the conclusion is 0 and 1 are the only singular points. So, this tells us x is equal to 0 and x is equal to 1 are singular points of the differential equation. Now, if x 0 is an ordinary point of a differential equation, then we have a sufficient condition to guarantee a power series solution. So, now I state the form of a theorem. Theorem says if x 0 is an ordinary point is an ordinary point of the differential equation, call it y double prime plus p 1 x y prime plus p 2 x y is equal to 0. Then it has two linearly independent non-trivial two linearly independent power series solution of the form y x is equal to summation c n x minus x 0 to the power n, n goes from 0 to infinity and the series converges in some interval, the interval of convergence x minus x 0 less than some r. So, this is an important theorem guarantees the existence of a power series solution to a differential equation. If x 0 is an ordinary point of a differential equation, then about that point we can find, if you have a second order equation, then we can find two linearly independent power series solution. So, existence of two linearly independent power series solution at an ordinary point is guaranteed by this theorem. So, now the method of solution, how to compute the solution? So, method of solution is, by method of solution, what do we want to find? We want to find to find c 0, c 1, c 2, etcetera all these coefficients in the expression y is equal to our series solution is of the form c 0 plus c 1 x minus x 0 plus c 2 x minus x minus x 0 square plus etcetera, which is in compact form we write as c n x minus x 0 to the power n, n goes from 0 to infinity. So, our aim is to find these coefficients, our aim is to find these coefficients. So, since a series converges on x minus x 0 less than r, some number r by the existence theorem, since a series converges for x minus x 0 is less than r about the ordinary point x 0, the series can be differentiated term by term, it may be differentiated term by term on this interval. C d y by d x is derivative of y, which is c 1 plus 2 c 2 x minus x 0 plus 3 c 3 x minus x 0 square plus etcetera, which is summation n goes from 1 to infinity n c n x minus x 0 to the power n minus 1. And similarly, second derivative d square y by d x square, the derivative of the about series, which is 2 c 2 plus 6 c 3 into x minus x 0 plus 12 c 4 into x minus x 0 square plus etcetera, which is written as summation n is equal to 2 to infinity n into n minus 1 into c n x minus x 0 to the power n minus 2. So, now, substituting these values of y d y by d x d square y by d x square in the original equation, now substituting y d y by d x d square y by d x square in the differential equation. And simplifying, we get some constant k 0 plus k 1 into x minus x 0 plus k 2 into x minus x 0 square plus etcetera, is equal to 0, where these constants k 0, k 1, k 2, these are functions of our other constant c 1, c 2, etcetera. So, since the series, since the series, it is valid for all x in the interval, all x in the interval x minus x 0 less than r, that is the interval of convergence. And now, equating right hand side and left hand side, the coefficients of x, x square, x cube etcetera and also the constant terms, we get k 0 is equal to 0, k 1 is equal to 0, k 2 is equal to 0 etcetera. So, we get equations k 0 is equal to 0, k 1 0, k 2 0. If you solve these equations, then we get the values of c 0. So, solve these to obtain values of c 0, c 1, c 2, etcetera. And once we have c 0, c 1, c 2, etcetera, plug into the series form, we get the series solution. The series solution y x is equal to summation c n x minus x 0 to the power n. So, this is the method. So, let us illustrate this by an example. So, let us consider an example. So, what is the problem? The problem is the question is to find the power series from the power series solution of the differential equation, find the power series solution of the differential equation given by y double prime plus x y prime plus x square plus 2 y is equal to 0. So, find the power series solution of the differential equation in powers of… So, the question is to expand or get the power series solution in powers of x. So, that is you are asked to expand, get the solution in powers of x minus x 0, where x 0 is 0. Let us check whether 0 is an ordinary point. So, obviously your p 1 is x, p 2 is x square plus 2. So, all points are ordinary points. So, all points are ordinary points. So, x 0 is equal to 0 is an ordinary point. So, therefore, by the existence theorem, there exists a power series solution. So, two linearly independent power series solution of the form summation c n x x x to the power n. So, we look for a solution y x, y is equal to summation c n x to the power n, n goes from 0 to infinity. So, this is guaranteed by the existence theorem. So, there are two linearly independent power series solutions. So, what we want to do now? Our aim is to find this c 0, c 1, c 2, c n. So, what we do is by differentiating, differentiating we get y prime. y prime is summation n goes from 1 to infinity, n into c n x to the power n minus 1. And similarly, y double prime is summation is going from n is equal to 2 to infinity, n into n minus 1, x to the power n minus 2. Now, plug in these values to the given differential equation. So, substituting in the differential equation, we get summation n into n minus 1 c n x to the power n minus 2, the first term that n goes from 2 to infinity, n goes from 2 to infinity, the first term. And the second term is x into y prime, which is x into summation n goes from 1 to infinity, n into c n x to the power n minus 1. Now, the third term is the sum of two terms, x square plus 2. I take x square into summation into y, summation c n x to the power n, n goes from 0 to infinity plus constant m 2, 2 into summation n goes from 0 to infinity, c n x to the power n, which is equal to 0. Since x is independent of the index, we may rewrite this. So, rewriting, we get summation n goes from 2 to infinity, n into n minus 1, c n x to the power n minus 2, the first term, plus summation n is equal to 1 to infinity, n into c n x to the power n plus summation, if we have just multiplied each term by x, we get x to the power n and n goes from 0 to infinity, the third term. There also, we did a multiplication by x square. So, that results c n into x to the power n plus 2, when multiplied by x square plus 2 times summation n goes from 0 to infinity, c n x n is equal to 0. Now, look at the first and third term, this first term and the third term. The first term, the index is n minus 2, the third term, the index is n plus 2. We want to make all the same uniform index that x n. So, we can do the following to rearrange that. So, consider the first term. So, consider the first term, first term, first summation and replace n minus 2 by a new variable, new dummy index m. Therefore, what we have is m is equal to n minus 2 or n is equal to m plus 2. So, if you use this, then the summation becomes, then this gives the first term in terms of m. m goes from 0 to infinity. Now, n is going from 2 to infinity, that becomes when n is 2, m is 0. So, the summation goes from 0 to infinity, m plus 2 into m plus 1. So, n becomes m plus 2 and n minus 1 is n plus 1 into c, m plus 2 into x to the power m. Now, remember that m is just a dummy variable, m is a dummy variable, dummy index. We can change m to n, change m to n notation. Therefore, it becomes summation n is equal to 0 to infinity, n plus 2 into n plus 1 into c, n plus 2, x to the power n. Similarly, if we do the same thing for the third term, then we will make n plus 2 is equal to m or n is m minus 2. So, this gives in terms of m, this will be m is going from 2 to infinity instead of n is going from 0 to infinity and m is going from 2 to infinity, c m minus 2 and x to the power m. Again, that is m is a dummy index. Therefore, we can plug back the n. This is n goes from 2 to infinity, c n minus 2, x to the power n. Now, the index of all these terms are n with respect to n. So, what we do is the equation now becomes summation n goes from 0 to infinity, n plus 2 into n plus 1, c n plus 2, x to the power n plus summation n goes from 1 to infinity, n c n. Therefore, we do not do any change, x n plus summation, the third term is also changed, n goes from 2 to infinity, c n minus 2, x n plus the last term, 2 times summation n goes from 0 to infinity, c n, x to the power n is equal to 0. Now, we want to make some common range. Since the summation is not uniform, some one is starting from 2 to infinity, another one is 1 to infinity, another one is 0 to infinity. The common summation range is 2 to infinity. So, the other terms we can separate out. For example, the first term, first summation, the case n is equal to 0 and n is equal to 1 can be separated out. Therefore, what we get is this term, which is 2 c 1, 2 times, when n is equal to 0, get 2 c 2, 2 c 2 plus 6 c 3, x plus the common range, n is equal to 2 to infinity, n plus 2 into n plus 1 into c n plus 2 into x to the power n. Similarly, for the second term, that can be split as c 1 x plus the common range, n is equal to 2 to infinity, n c n, x to the power n. And third term, we do not have to change. The last term, we can change this is 2 c 0 plus 2 c 1 x plus summation c n x to the power n, n goes from 2 to infinity. Now, if we rearrange all this, we get this implies that 2 c 2 plus 6 c 3 plus summation, the first term. So, if we combine all these things together, if we add them together, we get 2 c 0 plus 2 c 2 plus 3 c 1 plus 6 c 3 x plus summation, the common summation, n is equal to 2 to infinity, n plus 2 into n plus 1 c n plus 2 plus n plus 2 c n plus c n minus 2 times x to the power n is equal to 0. And this series converges for in the interval x minus x 0 is less than r. So, equating the coefficients of the powers left hand side to 0, because right hand side is already 0, we get 2 c 0 plus 2 c 2 is 0 and 3 c 1 plus 6 c 3 is 0. And this gives c 2 is minus c 0 and c 3 is minus half c 1. And also, we get from the summation term that n plus 2 n plus 2 into n plus 1 c n plus 2 is equal to plus n plus 2 into c n plus c n minus 2 is equal to 0. If we solve it, c n plus 2 is obtained as minus n plus 2 c n plus c n minus 2 divided by n plus 1 into n plus 2 for n greater than 2. Therefore, for each case, n is equal to 2 case, c 4 can be solved, c 4 is minus 4 c 2 plus c 0 by 12. So, this implies that c 4 is 1 by 4 c 0. And similarly, n is equal to 3 case, c 5 is solved to be 3 by 40 c 1. So, therefore, the solution can be written as, the solution can be written as by using these coefficients. So, y of x is equal to c 0 into 1 minus x square plus 1 by 4 x to the power 4 minus etcetera plus c 1 into x minus half x cube plus 3 by 40 x to the power 5 plus etcetera. So, we see that there are two series. So, this is two series solutions, the first one and the second one. They are linearly dependent, two series solutions and their linear combination c 0 of the first one plus c 1 of the second one is a general solution, say general series solution. So, therefore, by doing this method, we can find, if a point is an ordinary point, we can get the series solution and two linearly independent series solution of a differential equation, if the point is an ordinary point. And if the point is singular point, there are methods, Frobenius methods and all that will come in another series. So, with this, I would like to finish. So, we have seen, even if the differential equation is variable coefficient, we can have a solution in the in power series form, series solution. Bye.