 Namaste, Myself, Mr. Birajdar Bala Sahib, Assistant Professor, Department of Humanities and Sciences, Walchand Institute of Technology, Solapur. In this session, we will discuss applications of first order and first degree ordinary differential equations, part one. Learning outcome, at the end of this session, students will be able to find orthogonal trajectories of the family of Cartesian curve. Let us start with introduction. In this session, we will discuss the application of first order and first degree ordinary differential equations. As we know that the mathematical equations of applications like electrical circuits, RL circuit, RC circuit and mechanical applications like Newton's law of cooling, motion of particle, etcetera are the first order and first degree ordinary differential equations. So that, method of solving first order and first degree differential equations plays an important role to solve such a application problems with their initial conditions. Here mainly three types of applications, orthogonal trajectory, mechanical engineering applications like Newton's law of cooling and motion of particle, electrical circuit problems. Let us pause the video for a while and write the answer to the question. Question is, list the method of solving the first order and first degree ordinary differential equation. Come back, I hope you completed the solution. Here I will going to give the answer. As we know that the solution of first order and first degree ordinary differential equation is obtained according as classification of that equation as variable separable form, homogeneous differential equation, non-homogeneous differential equation, exact differential equation, reducible to exact form, linear differential equation and reducible to linear form. These different forms we already discussed in previous videos. Now, one of these form is required to learn examples on applications. Now in this session, we will discuss orthogonal trajectories. Now definition, two families of curves are called orthogonal trajectories of each other. If every member of one family cuts each member of another family at right angles. For example, the family of straight line y equal to m x and a family of circle x square plus y square equal to a square are orthogonal to each other. When we trace these two curves in x y plane, we come to know that how they are orthogonal to each other. Let us see the diagram. This is our x axis and it is a y axis and these are the family of a circle x square plus y square equal to a square and the red color lines are the family of a straight line y equal to m x and this is the equation of a circle we know that. When we observe this figure, we come to know that every member of the straight line cuts each member of the circle at right angle. So, that these two curves are orthogonal to each other. In this session, we are finding orthogonal trajectory for Cartesian curves. Now working rule, let us denote the given equation of curve f of x comma y comma c is equal to 0 as equation 1, where c is a parameter. In step 1, differentiate given curve with respective to x and eliminate parameter c, we get a differential equation f of x comma y comma dy by dx equal to 0 denote this equation by 2. This is the differential equation of given Cartesian curve. Now in step 2, replace dy by dx by minus dx upon dy in above equation 2, we get f of x comma y comma minus dx upon dy equal to 0. This is the equation number 3. This differential equation is for orthogonal trajectories. Now in step number 3, we have to solve the differential equation 3 obtained in step number 2. We get a family of a curve which is orthogonal to given family of a curve called as orthogonal trajectories. Now illustration example 1, find the orthogonal trajectories of the family of a curve a into y square equal to x cube, where a is parameter solution. Let us denote the given equation e into y square equal to x cube by equation 1. Now differentiate this equation with respective to x, we get in left side a as it is and derivative of y square with respective to x is 2 y into dy by dx is equal to derivative of x cube is 3 x square. From this we get dy by dx equal to 3 into x square upon 2 a y denote this equation by 2. Now we have to eliminate parameter a between equation 192 for that from equation 1 we can find value of a as x cube upon y square and put it in equation 2. We get dy by dx equal to 3 x square upon 2 y as it is and value of a is x cube upon y square which is equal to now here y y cancel in the denominator and x square 1 x square gets cancel from numerator and denominator. Finally, we get dy by dx equal to 3 y upon 2 x denote this equation by 3. This is the differential equation of given family of a curve. Now in order to find orthogonal trajectory we have to replace dy by dx by minus dx upon dy in equation 3. We get minus dx upon dy equal to 3 y upon 2 x denote this equation by 4. This is the differential equation of orthogonal trajectory. Now we have to solve this equation using one of the suitable method of first order and first degree ordinary differential equation. Here variable separable method is applicable we can separate the variable x and dx one side and y and dy another side. We get minus 2 x dx equal to 3 y dy. Now integrating both side integration of minus 2 x into dx equal to integration of 3 y dy. Therefore, now in left side minus 2 as it is and integration of x is x square by 2 is equal to in right side 3 as it is and integration of y is y square by 2 plus constant of integration k. Therefore, in left side 2 2 gets cancel minus x square is equal to 3 by 2 into y square plus k. Now multiply by 2 on both the side we get minus 2 x square equal to 3 into y square plus 2 into k. This is the required family of a curve which is orthogonal to given family of a curve. Let us consider another example number 2. Find the orthogonal trajectory of the family of curve x square plus 2 into y square equal to c where c is the parameter solution denote the given equation x square plus 2 into y square equal to c as equation 1. Again differentiate both the side with respect to 2 x we get 2 x plus 2 into derivative of y square with respect to x is 2 into y into dy by dx. So, that 2 into 2 is 4 y into dy by dx is equal to derivative of right hand side constant is 0. Now dividing 2 on both the side we get x plus 2 y into dy by dx equal to 0. denote this is the equation 2 and this is the differential equation of given curve. Now replace dy by dx by minus dx upon dy in equation 2 we get x plus 2 y in bracket minus dx upon dy equal to 0. This implies x minus 2 y into dx upon dy equal to 0 denote it as equation 3. This is the differential equation of orthogonal trajectory. Now we have to solve this equation number 3 by using variable separable method for that we write equation 3 as x equal to 2 y dx upon dy which implies x into dy equal to 2 y into dx. Now by cross division of x and y we get 1 upon y into dy equal to 2 into 1 by x into dx. This is the variable separable for. Now integrating both side integration of 1 upon y dy equal to 2 into integration of 1 by x dx. Therefore, integration of 1 upon y is log y is equal to 2 into integration of 1 by x is log x plus log k is a constant of integration. Therefore, log y equal to 2 log x you can write log of x square by property m into log a equal to log a raise to m and plus log k. Therefore, log y equal to log of k into x square by property log of a plus log of b equal to log of a b. Now cancel logarithm from both the side we get y equal to k into x square it is the required family of a curve which is orthogonal to given curve. To prepare this video lecture I refer these two as a references. Thank you.