 Hi and welcome to the session. I am Pika here. Let's discuss a question which says show that the given differential equation is homogeneous and solve it. As d y by d x minus y plus x sin of y over x is equal to 0. So let's start the solution. The given differential equation is x d y by d x minus y plus x sin of y over x is equal to 0. Let us get this equation as number 1. Or this can be written as x d y by d x is equal to y minus x sin of y over x. Or this can be written as d y by d x is equal to y over x minus x over x sin of y over x. Or this can be written as d y by d x is equal to y over x minus sin of y over x. Let us get this equation as number 2. Now right hand side of the above equation or equation 2 is of the form g of y over x. And so it is a homogeneous function of degree 0. Therefore the given differential equation that is the equation 1 is a homogeneous differential equation. We will solve equation 2 by substituting y is equal to v x. So for y is equal to v x. Therefore d y by d x is equal to v plus x sin of d v over d x. Now we will substitute the value of y and d y by d x in equation 2. So from equation 2 we have v plus x d v over d x is equal to v minus sin v d v over d x is equal to v minus sin v minus v or x d v over d x is equal to minus sin v. Now on separating the variables we have d v over sin v is equal to minus d x over x. By integrating both sides we have integral of d v over sin v is equal to minus integral of d x over x. Or we can write the left hand side of this equation as integral of cosecant v d v and this is equal to minus integral of d x over x. Now the integral of cosecant v d v is log of mod cosecant v minus cot v and the integral of d x over x is log of mod x plus log c where log c is a constant of integration. Or we can write this equation as log of mod cosecant v minus cot v plus log of mod x is equal to log c. Now by replacing y by v x or v by y over x we get log of mod cosecant y over x minus cot y over x. Plus log of mod x is equal to log c or this can be written as log of mod cosecant y over x minus cot y over x into mod x is equal to log c. Now this can be written as log x into 1 over sin y over x minus cos y over x over sin y over x and this is equal to log c. And this can be written as x into 1 minus cos y over x over sin y over x is equal to c. And this can be written as x into 1 minus cos y over x is equal to c sin y over x. Hence the general solution of the given differential equation is x into 1 minus cos y over x is equal to c sin y over x. So this is the answer for the above question. I hope the solution is clear to you and you have enjoyed the session. Bye and take care.