 Hi friends, I am Pooja and today we will discuss the following question. For the following differential equation, find a particular solution satisfying the given condition. dy upon dx minus 3y cot x is equal to sin 2x and the given condition is y is equal to 2 when x is equal to pi by 2. Let us now begin with the solution. So the given differential equation is dy by dx minus 3y into cot x is equal to sin 2x. Let us mark this as equation 1 and the given condition is y is equal to 2 when x is equal to pi by 2. Now this is a linear differential equation of the form dy by dx plus py is equal to q therefore the integrating factor of this equation is given by e raise to the power integral p dx. Now by comparing the equation dy by dx plus py is equal to q with this equation 1 we can clearly see that p is equal to minus 3 cot x. So we get integrating factor is equal to e raise to the power integral minus 3 cot x dx and this is equal to e raise to the power minus 3. Now integral of cot x is log mod sin x. So we get e raise to the power minus 3 into log mod sin x and this is equal to e raise to the power minus log sin cube x and this is further equal to e raise to the power log 1 upon sin cube x and we can write this as this is equal to e raise to the power log cos x cube x and this is further equal to cos cube x because e raise to the power log cos x cube x can be written as cos x cube x therefore we get the integrating factor as cos x cube x. Now multiplying equation 1 by cos x cube x we get cos x cube x into dy by dx minus 3 y cot x into cos x cube x is equal to sin cube x into cos x cube x and this implies cos x cube x into dy by dx minus 3 y cot x into cos x cube x is equal to 2 sin x into cos x upon sin cube x. Now this implies we can write the left hand side as d upon dx of y into cos x cube x and this is equal to now the right hand side becomes 2 into cot x into cos x. Now integrating both the sides of this equation with respect to x we get integral d upon dx of y into cos x cube x dx is equal to integral 2 cot x cos x dx and this implies now integrating left hand side we get y into cos x cube x is equal to integrating 2 cot x cos x we get minus 2 cos x plus c where c is a constant and we mark this as equation 2. Now to find the value of c we will use the given condition that when x is equal to pi by 2 y is equal to 2. Now putting the value of x and y in equation 2 we get there 4 2 cos x cube pi by 2 is equal to minus 2 into cos x pi by 2 plus c and this implies 2 into now cos x pi by 2 is equal to 1 so we have 1 cube is equal to minus 2 into again cos x pi by 2 is equal to 1 so we have minus 2 into 1 plus c and this implies 2 is equal to minus 2 plus c which further implies c is equal to 4. Now substituting this value of c in equation 2 we get y into cos x cube x is equal to minus 2 into cos x plus 4. Now dividing both the sides by cos x cube x we get this implies y is equal to minus 2 into sin square x plus 4 into sin cube x. Hence the required solution is y is equal to 4 sin cube x minus 2 sin square x. This is our answer. Hope you have understood the solution Bye and take care.