 Hi and welcome to the session. Let us discuss the following question. Question says, for the following differential equation, find a particular solution satisfying the given condition. Given differential equation is cos dy upon dx is equal to a where a belongs to real numbers and given condition is y is equal to 1 when x is equal to 0. Let us now start with the solution. Now we are given cos dy upon dx is equal to a where a belongs to real numbers and given condition is y is equal to 1 when x is equal to 0. Now this differential equation further implies that dy upon dx is equal to cos inverse a. Now separating the variables in this equation we get dy is equal to cos inverse a dx. How integrating both the sides of this equation we get integral of dy is equal to integral of cos inverse a dx. Now this further implies integral of dy is equal to cos inverse a multiplied by integral of dx. Now using this formula of integration we get integral of dy is equal to y and here we get cos inverse a multiplied by x plus c where c is the constant of integration. Now we can write this equation as y is equal to x cos inverse a plus c. Let us name this equation as 1. Now the given condition in the question is y is equal to 1 when x is equal to 0. So in this equation we will put y is equal to 1 and x is equal to 0. Now substituting 1 for y and 0 for x in this equation we get 1 is equal to 0 plus c. Now this further implies c is equal to 1. Now substituting this value of c in equation 1 we get y is equal to x cos inverse a plus 1. Now subtracting 1 from both the sides of this equation we get y minus 1 is equal to x cos inverse a. Now dividing both sides by x we get y minus 1 upon x is equal to cos inverse a. Now this further implies cos y minus 1 upon x is equal to a. So the required solution for the given differential equation is cos y minus 1 upon x is equal to a. This completes the session. Hope you understood the solution. Take care. Have a nice day.