 Hi and how are you all today? The question says solve the following differential equation or here this is the differential equation which is given to us. It is dy by dx plus y cot x is equal to x square cot x plus 2x. Now this given equation is a linear differential equation of the form dy by dx plus py equal to q. Now here p is equal to cot x and q is equal to x square cot x plus 2x right. So we have the p and the q let us find out the integrating factor first. Integrating factor is equal to e raised to the power integral p dx that is e integral cot x dx which is equal to e log sin x that is further equal to sin x right. So this is the integrating factor. Now we have the integrating factor hence the solution is given by y into integrating factor equal to integral q into integrating factor into dx plus c. So we have y sin x equal to integral x square cot x plus 2x into sin x dx plus c is further equal to y sin x equal to let us multiply these two have integral x square cot x sin x dx plus integral 2x sin x dx plus we have y sin x is equal to x square into cot x can be written as cos x upon sin x into sin x dx plus 2 integral x sin x dx plus c. On simplifying it further we have y sin x equal to integral x square cot x dx plus 2 integral x sin x dx plus c. Now here we have integral x square cot x dx plus 2 let us integrate it by using bypass. So we have first function first function into integral of second that will be x square upon 2 minus integral derivative of first function that is cos x into integral of again second function this is the second function that I am taking and this is the first function x square upon 2 plus c. So now we have y sin x equal to integral x square cos x dx plus x square sin x minus integral x square cos x dx plus c these two will get cancelled out we have y sin x equal to x square sin x plus c and this is the required answer to the given question. So hope you understood it well and enjoyed it do remember your bypass theorem and have a nice day.