 Namaste, Myself, Mr. Virajdar Bala Saheb, Assistant Professor, Department of Humanities and Sciences, Walchand Institute of Technology, Solapur. In this video lecture, we will discuss application of First Order and First Degree Ordinary Differential Equation, Part 2. At the end of this session, students will be able to find orthogonal trajectories of the family of Cartesian curve. Already, we discussed in video one regarding working rule to find orthogonal trajectory of Cartesian curve. In this video session, we will illustrate more examples on orthogonal trajectory of family of Cartesian curve. Let us pause the video for a while and write answer to the given question. Question is, form a differential equation of the family of curve x into y equal to c, where c is a parameter. Come back, I hope you have written answer to this question. Here, I will going to explain the solution of this question. Given curve is x into y equal to c denoted by equation 1. Now, we have to form it is a differential equation. For that, we have to differentiate this curve with respect to x. So, that the derivative of left hand side by using product rule as x keeping as it is and derivative of y with respect to x is d y by d x plus y as it is and derivative of x with respect to x is 1 is equal to derivative of constant c is 0. This is equation 2. Now, this is a differential equation of given curve x into y equal to c. Now, let us consider the example. Example 1, find the orthogonal trajectories of the family of curve y square equal to c in bracket 1 plus x square, where c is parameter. Solution, let us denote the given curve y square equal to c in bracket 1 plus x square by equation 1. Now, differentiate both this side with respect to x. We get derivative of y square is 2 y into d y by d x is equal to c as it is and in bracket derivative of 1 is 0 plus derivative of x square is 2 x. Now, cancel 2 from both this side. We get y into d y by d x is equal to c into x denote this equation by 2. Now, we have to eliminate parameter c between equation 1 and 2. For that from equation number 1, we can find value of c as y square upon 1 plus x square. Now, substitute this value of c in right side of equation 2. We get y into d y by d x is equal to x into y square upon 1 plus x square. Now, 1 y gets cancelled from both this side which gives d y by d x equal to x into y upon 1 plus x square denote this equation by 3. This is the differential equation of given cartesian curve. In order to find orthogonal trajectory, we have to replace d y by d x by minus d x upon d y in this equation 3. We get minus d x upon d y equal to x into y upon 1 plus x square denote this equation by 4. This is the differential equation of orthogonal trajectory. Now, we have to solve this equation number 4 using any one of the suitable method of solution of first order and first degree differential equation. Another method of separating the variable is applicable so that we can separate the variable as 1 plus x square upon x into d x is equal to minus y into d y. In the left hand side we can simplify as 1 upon x plus x square upon x gives x and bracket close into d x is equal to minus y into d y. This is called as variable separable form. Now, on integrating both this side integration of 1 by x plus x into d x is equal to integration of minus y into d y which implies in the left hand side we can separate the integration for each term we get integration of 1 by x into d x plus integration of x into d x equal to minus integration of y into d y. Therefore, integration of 1 by x with respect to x is log x plus integration of x with respect to x is x square by 2 is equal to minus sin as it is and integration of y with respect to 2 y is y square by 2 plus k constant of integration. Therefore, x square by 2 plus log x and when we bring minus y square by 2 to left side it becomes plus y square by 2 is equal to constant k. Now, multiply by 2 on both this side we get x square plus y square plus 2 into log x equal to 2 into constant k. This is the family of a curve which is orthogonal to given family of a curve. Let us consider another example, example 2. Find the orthogonal trajectories of the family of curve y equal to x plus a into e raise to x. Now, solution let us denote the given curve y equal to x plus a into e raise to x by equation 1. Again differentiate both this side with respect to x we get d y by d x equal to derivative of x is 1 plus a constant as it is and derivative of e raise to x is e raise to x denote this equation by 2. Now, we have to eliminate parameter a between 1 and 2 for that from equation 1 we can write value of a into e raise to x as y minus x and substitute this value in the right hand side of equation 2. We get d y by d x is equal to 1 as it is plus value of a into e raise to x is y minus x. This equation is denoted by 3. This is the differential equation of given cartesian curve. Now, to find orthogonal trajectory in equation number 3 we have to replace d y by d x by minus d x upon d y. We get minus d x upon d y is equal to 1 plus y minus x. Now multiply minus sign on both this side which gives d x upon d y is equal to x minus y minus 1 denote this equation by 4. This is the differential equation of orthogonal trajectory. Now we have to solve this equation number 4. Here the method of separation of variable is not applicable, but equation 4 we can write as d x upon d y minus x equal to minus in bracket y plus 1. This is the linear differential equation of first order and first degree with x as a dependent variable. That is of the type d x upon d y plus p dash x is equal to q dash where p dash and q dash are functions of y. But after comparison here the value of p dash is minus 1 and value of q dash is minus of y plus 1. Now in order to find the solution of this differential equation first of all we find it is a integrating factor. By formula integrating factor equal to e to power integration of p dash with respect to 2 y. But here p dash is minus 1 so that it is equal to e raise to integration of minus 1 into d y which is equal to e raise to minus y. Therefore integrating factor is equal to e raise to minus y. Now general solution of this differential equation is x into integrating factor is equal to integration of integrating factor into value of q dash into d y plus constant k. Now substitute the value of integrating factor and q dash in this equation we get x into e raise to minus y is equal to integration of e raise to minus y in bracket minus of y plus 1 bracket close into d y plus k. Therefore x into e raise to minus y is equal to now minus sign you can take outside the integration and into integration of e raise to minus y in bracket y plus 1 into d y plus constant k. Now we have to integrate right hand side integration using integration by parts. Here y plus 1 created as a first function denoted by u and e raise to minus y is a second function denoted by v. Therefore x into e raise to minus y is equal to minus in bracket keeping y plus 1 as it is and into integration of e raise to minus y is minus e raise to minus y minus integration derivative of y plus 1 with respect to y is 1 and into integration of e raise to minus y is minus e raise to minus y into d y bracket close plus constant k. Therefore x into e raise to minus y is equal to minus in bracket minus of y plus 1 into e raise to minus y. Then minus sign you can take outside becomes plus into integration of e raise to minus y is again minus e raise to minus y bracket close plus constant. Therefore x into e raise to minus y equal to when we multiply this minus sign inside we get y plus 1 into e raise to minus y then minus minus plus e raise to minus y plus constant k. Therefore x into e raise to minus y equal to in the right hand side in the first two term when we take e raise to minus y is common outside in bracket we get y plus 1 plus 1 that gives y plus 2 and into e raise to minus y plus constant k. This is the required equation of family apacre which is orthogonal to given family apacre. To prepare this video lecture I use this book as a references. Thank you.