CSIR NET MATHEMATICS | How to solve the problem of Differential Equation using Linear Algebra?

Hello friends, in this video we are going to discuss the problem of ordinary differential equation which require the concept of linear algebra from the CSIR-NET exam June 2018. Example, consider the ordinary differential equation y double dash plus p x y dash plus q x y equal to 0, where p and q are smooth functions, let y1 and y2 be any two solution of ODE, let W(x) be the corresponding Wronskian. Then, each of the following is always true. First option, if y1 and y2 are linearly dependent, then x1, x2 such that Wronskian of x1 is equal to 0 and Wronskian of x2 not equal to 0. Second option, if y1 and y2 are linearly independent, then Wronskian of x is equal to 0 for all x. Third option, if y1 and y2 are linearly dependent, then Wronskian of x not equal to 0 for all x. And if y1 and y2 are linearly independent, then VX is not equal to 0 for all x. Now, to solve this problem, we should know the condition of linear independence of function. The condition is, two functions y1 and y2 are linearly dependent if the W(x) is equal to determinant y1 y2, y1 dash y2 dash is equal to 0 for all x. In other words, it should be identical 0. We begin with Ordinary differential equation y double dash minus y equal to 0 will satisfy the condition where px is equal to 0 and qx is equal to minus 1. So, p and q are the smooth functions. Now, we are taking two solutions of this differential equation y1 equal to e power x and y2 equal to 2 into e power x. Now, simply y2 is a constant multiple of y1. Therefore, y1 and y2 are linearly dependent. Now, we are going to calculate the Wronskian. So, Wronskian of x is equal to e power x plus 2 into e power x, e power x and the differentiation of 2 into e power x is also 2 into e power x, then 2 into e power x minus 2 into e power x is equal to 0. So, VX is equal to 0 for all x, for the y1 equal to e power x and y2 equal to 2 into e power x. Therefore, the third option y1 and y2 are linearly dependent, then Wronskian of x not equal to 0 for all x, which is wrong. Now, similarly, the first option is y1 and y2 are linearly dependent, then x1 and x2 such that Wronskian of x1 equal to 0 and Wronskian of x2 not equal to 0, which is also not possible because Wronskian of x is equal to 0 for all x. The first option is also wrong. Now, we consider the another two solution of y dash, y double dash minus y equal to 0. The first solution is y1 equal to e power x and y2 equal to e power minus x. And as we know, both are the independent functions. Wronskian y1, y2, y1 dash, y2 dash is equal to e power x, e power minus x, e power x and differentiation of e power minus x is minus into e power minus x. Therefore, multiplication e power x into minus e power minus x has value minus 1 and minus times e power x into e power minus x is also minus 1. Wronskian has value minus 2 for all x. This second option, if y1 and y2 are linearly independent, then Wronskian of x is equal to 0 for all x, which is also wrong. Now, therefore, the remaining fourth option should be correct, which is y1 and y2 are linearly independent, then Wronskian of x not equal to 0 for all x. Thanks for watching. For more videos, like, subscribe, press the bell icon of next channel.