CSIR NET MATHEMATICS | Calculus of Variations | Minimize the Functional J(x)(हिंदी में)

Hello friends, this Inside the video Calculus of Variation's We will describe the problem. From CSI NET June 2019 Example No. 44 Booklet C Marks 3 Let x⋆ be the curve which minimizes the functional jx is equal to the integral from 0 to 1 x squared t plus x dot squared t dt Satisfying x of 0 is equal to 0 x of 1 equals 1 then the value of x times 1 by 2 is We have 4 options available. First option root e divide by 1 plus e second option 2 root e divided by 1 plus e third option root e divided by 1 plus 2e And the fourth option: 2 root e divided by 1 plus 2e These are functional GXs. It is called the Lagrangian function. Then we will have the Longaraj function Langaraj of TX x dot is equal to x squared t plus x dot square We need to find x star t minimizing the functional Then we will have to use the Langraja equation for it. Langraja equation d/dt (x minus d) d/dt (x) = 0 partial derivative of l Its value with respect to x will appear 2x and the partial derivative of l with respect to the value of x dash 2x dash equals 0 derivative of 2x dash with respect to t 2x double dot equal to 0 So, ours will come today, second order linear differential equation x double dot minus x equals 0 auxiliary equation m squared -1 equals 0 The factors will be m minus 1. m plus 1 equals 0 m1 equal to 1 and m2 equal to minus 1 So this ordinary differential The solution to the equation will arrive. x of t equals c1 e power m1 t plus c2 e power M2T m1 equals 1, m2 equals minus 1 Then the solution will come for c1 e to the power t plus c2 times e to the power minus t Here c1 and c2 are arbitrary constant is and its value We are given to find x of 0 is equal to 0. x of 1 is equal to We will use the first condition. x of 0 is equal to 0 So 0 is equal to c1. e power 0 plus c2 e power 0, then c1 plus c2 equal to 0, the first equation obtained For the second equation, we will use x of 1 equals 1 So 1 equals c1 times e to the power 1 plus c2 e to the power minus 1 So we have two equations. We will solve these two equations. Then we will obtain the values of c1 and c2. and that value is c1 equal to 1 over e minus e to the power of minus 1 and c squared is equal to -1 divided by e minus e power minus, the value of both of these We will get it if we keep it in x. x_t is equal to 1 divided by e minus e to the power of minus 1 e to the power of t minus 1 over e minus e to the power of minus 1 times e to the power of minus t its final form will come 1 divided by e minus e^(-1) × e^t minus e to the power of minus t We still need to find the value x times 1 by 2 So we will take t equal to 1 by 2. So we will get 1 divided by E minus E to the power of minus 1 times E to the power of 1 divided by 2 minus E to the power of minus 1 divided by 2 e raised to the power of -1 equals 1 over E times E to the power of -1/2 equals 1 divided by E to the power of 1/2 Then, after keeping it, we will get E divided by square minus 1. E minus 1 divided by E to the power of 1 over 2 The answer will be E divided by the square root of E. E power 1 by 2 in the denominator And there will be two factors of e squared minus 1. e minus 1, e plus 1 e minus 1 cancels with e minus 1 Then we will have it. root e divided by e plus 1 Our first answer is So the correct answer to this example is first option: root e divided by 1 plus e Thank you for watching for more videos Like, share, subscribe Raise the bell icon of Next Gen Mess