 Hi and welcome to the session I am Shashi and I am going to help you with the following question. Question is if y is equal to 3 cos log x plus 4 sin log x show that x square y2 plus xy1 plus y is equal to 0. Let us start the solution now we are given y is equal to 3 cos log x plus 4 sin log x. Now differentiating both sides with respect to x we get dy upon dx is equal to 3 multiplied by minus sin log x multiplied by 1 upon x plus 4 multiplied by cos log x. Now this is further equal to minus 3 upon x sin log x plus 4 upon x multiplying both sides by x we get x dy upon dx is equal to minus 3 sin log x plus 4 again differentiating this equation with respect to x we get equation as 1. So we can write again differentiating both sides of equation 1 with respect to x we get d square y upon dx square plus dy by dx multiplied by 1 is equal to minus 3 cos log x multiplied by 1 upon x plus 4 multiplied by minus sin log x multiplied by 1 upon x or we can write this implies x d square y upon dx square plus dy upon dx is equal to 1 upon x multiplied by minus 1 that is minus 1 upon x multiplied by 3 cos log x plus 4 sin log x. Now multiplying both sides by x we get x square d square by upon dx square plus x dy upon dx is equal to minus 3 cos log x plus 4 sin log x clearly we can see 3 cos log x plus 4 sin log x is equal to y. So we will substitute for 3 cos log x plus 4 sin log x y so we can write x square d square y upon dx square plus x dy upon dx is equal to minus y so this implies x square d square y upon dx square plus x multiplied by dy upon dx plus y is equal to 0. Now this implies x square y2 plus x y1 plus y is equal to 0. So we were required to prove this only so we can write x square y2 plus x y1 plus y is equal to 0 hence proved this completes the solution hope you enjoyed the session good bye.