 Hi and welcome to the session. Let us discuss the following question. Question says, find a particular solution of the differential equation dy upon dx plus y cot x is equal to 4x cos x. Given that y is equal to 0 when x is equal to pi upon 2, let us now start with the solution. Now given differential equation is dy upon dx plus y cot x is equal to 4x cos x. Now clearly we can see this is a linear differential equation in this form where p is equal to cot x and q is equal to 4x cos x. Now we can write this is a linear differential equation of the type dy upon dx plus dy is equal to q where p is equal to cot x and q is equal to 4x cos x. Therefore integrating factor is equal to e raised to the power integral of p dx that is e raised to the power integral of cot x dx. Now using this formula of integration we can find this integral. So we get integrating factor is equal to e raised to the power log of sin x we know integral of cot x dx is equal to log sin x. Now this is further equal to sin x only using this law of logarithms we get this term is equal to sin x. Now we know solution of the equation is y multiplied by integrating factor is equal to integral of q multiplied by integrating factor dx plus c. Now substituting 4x cos x for q and sin x for integrating factor in this expression we get y sin x is equal to integral of 4x cos x multiplied by sin x dx plus c. Now we will evaluate this integral. Now we know cos x and sin x are reciprocal of each other. So we can write this integral as integral of 4x dx. So we get y sin x is equal to integral of 4x dx plus c. Now this further implies y sin x is equal to 4 multiplied by integral of x dx plus c. Now using this formula of integration we can find this integral and we get y sin x is equal to 4 multiplied by x square upon 2 plus c. We know this integral is equal to x square upon 2. Now in this term we will cancel common factor 2 from numerator and denominator both and we get y sin x is equal to 2x square plus c. Now let us name this equation as equation 1. Now we are given that for the given differential equation y is equal to 0 when x is equal to pi upon 2. So we will substitute y is equal to 0 and x is equal to pi upon 2 in the equation 1. Now equation 1 becomes 0 is equal to 2 multiplied by square of pi upon 2 plus c. Now this further implies 0 is equal to 2 multiplied by pi square upon 4 plus c. Now we will cancel common factor 2 from numerator and denominator both. Now subtracting pi square upon 2 from both the sides of this equation we get minus pi square upon 2 is equal to c or we can simply write c is equal to minus pi square upon 2. Now substituting this value of c in equation 1 we get y sin x is equal to 2x square minus pi square upon 2 where sin x is not equal to 0. So this is the required particular solution of the given differential equation. So this is our required answer. This completes the session. Hope you understood the solution. Take care. Have a nice day.