CSIR NET MATHEMATICS DECEMBER | Ordinary & Partial Differential Equations | Solutions of Section B

Hello friends, in this video we are going to discuss the problem of ordinary differential equation from CSI and NIT December 2017. Example number 41 from booklet A which is of marks 3. Consider the differential equation x minus 1 y double dash plus x y dash plus 1 by x into y equal to 0. Then x is equal to 1 is the only singular point. Second option x is equal to 0 is the only singular point. Third option, both x is equal to 0 and x is equal to 1 are singular points. And the fourth option, neither x is equal to 0 nor x is equal to 1 are singular points. This problem is about singular point of differential equation. So, first we recall the concept of singular points. So, general form of differential equation, pxy double dash plus qxy dash plus rxy equal to 0. Now, convert into the standard form by dividing the coefficient of y double dash, which is px. Then we have y double dash plus qx divided by pxy dash plus rx divided by pxy equal to 0. Then any point x is equal to x0 is said to be ordinary point if the rational polynomial qx divided by px and rx divided by px are analytic at x is equal to x0. That means both these rational polynomials are analytic at x is equal to x0 then it is called ordinary point otherwise it is called If at least one of them is non-analytic then the point is called singular now for the given differential equation x minus 1 y double dash plus x y dash plus 1 divided by x into y equal to now convert into a standard form by dividing the coefficient of y double dash which is x minus 1 we have y double dash plus x divided by x minus 1 y dash plus 1 divided by x into x minus 1 y equal to 0. The first rational polynomial qx divided by px is equal to x divided by x minus 1. This polynomial is not analytic at x is equal to 1 because px will be 0 when x is equal to. And the another polynomial rx divided by px which is 1 divided by x into x minus 1 is not analytic when x is equal to 0 and x is equal to 1. If you put x is equal to 0 then polynomial will be infinity. Example number 42, from booklet A, which is of marks 3. Let D denote the unit disk given by xy such that x2 plus y2 is equal to 1. And let DC be its complement in the plane. The partial differential equation x square minus 1 del square u by del x square plus 2y del square u by del x del y minus del square u by del y square equal to 0 is first option parabolic for all xy belongs to DC, hyperbolic for all xy belongs to D, third option hyperbolic for all xy belongs to DC and parabolic for all xy belongs to D. So, this problem is about the classification of partial differential equation. So, first we are going to understand the classification of partial differential equation. The general form of partial differential equation is R del square root by del x square plus S del square root by del x del y plus T del square root by del y square f of x, y, z, p, q equal to 0. Equation is said to be elliptic. At a point X, Y and D. S square minus 4RT less than 0 where S is a coefficient of del square root by del x del y R is a coefficient of del square root by del x square and T is a coefficient of del square root by del y square. Then equation is parabolic at a point x y in D if S square minus 4RT equal to 0. Third option hyperbolic at a point x y in D if S square minus 4RT greater than 0. Now, for the given partial differential equation, if we compare with the standard form, then r is equal to x square minus 1, s is equal to 2y and t equal to minus 1. Now, we are going to calculate the value of s square minus 4rt. So, s is equal to 2y, so square of 2y minus 4 times r is equal to x square minus 1 and t is equal to minus 1. So, after the simplification we have 4 times y square plus x square minus 1. Now, this polynomial 4y square plus x square minus 1 greater than 0 if all x y belongs to dc. Where dc is complement of x square plus y square less than equal to 1 which is x square plus y square greater than 1. Therefore, x square plus y square minus 1 greater than 0. S square minus 4RT greater than 0 for all x, y belongs to DC. Therefore, the equation is hyperbolic at all the point of DC and which is the option number 3. Therefore, third answer is correct. Example number 43 from booklet A marks 3. The set of real numbers lambda for which the boundary value problem d square y by dx square plus lambda y equal to 0 with the boundary value. y of 0 is equal to 0 and y of pi equal to 0 as non-trivial solution is first option interval minus infinity to 0 second option set square root n such that n is a positive integer third option n square such that n is positive integer and the fourth option are set of real numbers. So first we are going to find the solution of ordinary differential equation d square y by dx square plus lambda y equal to 0. Corresponding auxiliary equation m square plus lambda equal to 0. The roots of the auxiliary equation depend on the value of lambda. Case 1, lambda less than 0 implies lambda is equal to minus n square where n is any real number. We can rewrite auxiliary equation in the form m square is equal to minus lambda. Therefore, m square is equal to n square as lambda equal to minus n square. Therefore, m has to value m equal to plus or minus n. And hence, corresponding solution of given differential equation is c1 e power nx plus c2 e power minus 1. Now the boundary value first boundary value y of 0 is equal to 0 that means y equal to 0 when x is equal to 0. So y equal to 0 c1 e power 0 1 plus c2 e power 0 is also 1 therefore c1 plus c2 equal to 0 hence we have c2 is equal to minus 1. Second boundary condition y of pi equal to 0 that means y equal to 0 when x is equal to pi therefore 0 is is equal to C1 e power. n pi plus c2 e power minus n pi but c2 is equal to minus c1 so 0 is equal to c1 e power n pi minus c1 e power minus n pi. You can rewrite c1 into e power n pi minus e power minus n pi. As a product of two number equal to 0 therefore c1 equal to 0 or e power n pi minus e power minus n pi equal to e power n pi minus e power minus n pi equal to 0. So, we can rewrite the equation. e power 2n pi equal to 0 but it is possible only when n equal to 0. If n equal to 0 and lambda equal to minus n square therefore lambda equal to 0 but we already assumed that lambda is negative number therefore c1 must be 0 because the other number is not 0 for lambda less than 0. If c1 equal to 0 therefore c2 equal to minus c1 so c2 is also 0 and c1 c2 0 so y equal to 0 which is a trivial solution but we are interested in the non-trivial solution. So for lambda less than 0 we cannot get any non-trivial solution. Case 2 lambda equal to 0 therefore m square is equal to 0 therefore m is equal to 0 0 repeated solution therefore y equal to c1 plus c2 x is a corresponding The corresponding solution of given differential equation. Boundary value y of 0 is equal to 0. 0 is equal to c1 plus c2 into 0. So c1 equal to 0. y of pi equal to 0. Therefore x is equal to pi. So 0 is equal to c1 plus c2 pi. But c1 is already 0. Therefore c2 pi equal to 0 and c2 is equal to 0. Therefore c1 and c2 both are 0. So y equal to 0. Therefore we will have Trivial solution when lambda equal to 0. So for lambda equal to 0 we cannot get any non-trivial solution. Case 3. Lambda greater than 0. So lambda equal to n square for any real number n. So m square is equal to minus lambda. So m square is equal to minus n square. Therefore m is equal to plus or minus ni. We have complex solution. And corresponding solution or differential equation is c1 cos nx plus c2 sin nx. Now boundary values. y equal to 0. When x is equal to 0, so substitute x is equal to 0, then we have 0 is equal to c1, cos 0 is equal to 1, and sin 0 is equal to 0, therefore c1 is equal to 0. y of pi is equal to 0, that means substitute x is equal to pi, and y is equal to 0, therefore we have 0 is equal to c1 cos n pi plus c2 sin n pi. But c1 is already 0, therefore we have only one term, 0 is equal to c2 sin n pi. We are interested in the non-trivial solution. 0 is equal to sin n pi. Therefore it is possible for all integer value of n. That means n is an integer. Therefore lambda is equal to n square is an integer. Hence which is the third option where n square says that n is a positive integer. n is positive or negative integer. We will have the same set. Therefore third option. n square says that n is a positive integer is a correct answer. Example number 44, booklet A, marks 3. Let uxt be the solution of initial value problem utt minus uxx is equal to 0 with the initial value ux0 is equal to x cube and utx0 is equal to sin x. Then value of u at pi pi is first option 4 pi cube, second option pi cube, third option 0 and the fourth option 4. Here we are going to discuss the solution method of partial differential equation. Differential operator method for the simplicity we can write down the differential equation in the form of uxx minus utt equal to 0. Differential operator d which represent the partial derivative of function with respect to x and d dash represent the partial derivative of function with respect to t. Then the partial differential equation can be written in the form of operator d square minus d-square u Now, assume d is equal to m and d dash is equal to 1. Then, we have auxiliary equation m square minus 1 equal to 0. Now, the roots of auxiliary equation are 1 and minus 1. Therefore, the solution uxt is equal to function phi1 x plus m1 t plus function phi2 x plus m2 t. Where M1, M2 are the roots of the auxiliary equation which is 1 and minus 1. Therefore u x t is equal to phi 1 x plus t plus phi 2 x minus t where m1 equal to 1 and m2 is equal to minus 1 is a solution of given differential equation. Now initial values first initial value u of x 0 is equal to x cube. So So evaluate u when t is equal to 0. Then we have u is equal to phi1 x Because t is equal to 0. ux0 is equal to x cube. Therefore, we have x cube is equal to phi1x plus phi2x. First equation. Now, second initial condition u2x0 is equal to sinx. Differentiate u with respect to t. Then we have phi1 dash x plus t minus phi2 dash x minus t. Evaluate at t equal to 0. phi1-x minus phi2-x. UTx0 is equal to sinx. Therefore, we have sinx is equal to phi1-x minus phi2-x. Integrate both sides with respect to x. We have minus cosx is equal to phi1x minus phi2x. Therefore, we have two equations. x cube is equal to phi1x plus phi2x. And minus cos x is equal to phi 1 x minus phi 2 x. Using these two equations we are able to find the value of phi 1 and phi 2. Phi 1 equal to x cube minus cos x divided by 2. And phi 2 x is equal to x cube plus cos x divided by 2. Now phi 1 is a function of x plus t and phi 2 is a function of x minus t. So in the function phi 1 just x replaced by x plus t. Then phi 1 of x plus t is equal to x plus t cube minus. Cos x plus t divided by 2 and in phi 2 x replaced by x minus t therefore we have phi 2 x minus t equal to x minus t cube plus cos x minus t divided by 2. Therefore u x t is equal to x plus t cube minus cos x plus t divided by 2 plus x minus t cube plus cos x minus t divided by 2. This is The solution of initial value problem. We want to evaluate u at pi pi that means x is equal to pi and t equal to phi in the solution. We have u pi pi equal to pi plus pi cube minus cos pi plus pi divided by 2 plus pi minus pi cube plus cos pi minus pi divided by 2. Cos pi plus pi cos 2 pi has value 1 and cos pi minus pi that means cos 0 has also value 1. Then we have 8 pi cube minus 1 pi minus pi 0 plus 1. Minus 1 plus 1 cancel 8 pi cube divided by 2 is equal to 4 pi cube which is the first option and it is the correct answer. Thanks for watching. For more videos like share subscribe press the bell icon of next gen mess.