 Hello and welcome to the session. In this session we discussed the following question which says, find the particular solution satisfying the given condition for the following differential equation, dy by dx minus y upon x plus cosec of y upon x is equal to 0, y is equal to 0 when x equal to 1 is the condition given. Let's move on to the solution now. The given differential equation is dy by dx minus y upon x plus cosec of y upon x equal to 0. Let this be equation 1. Now this is a homogeneous differential equation. Now we can rewrite the equation 1 as dy by dx is equal to y upon x minus cosec of y upon x. Now we take this as equation 2. Next we put y equal to vx. So this means we have dy by dx is equal to v plus x into dv by dx. Then substituting the values of y and dy by dx that is this and this in equation 2 we get v plus x into dv by dx is equal to vx upon x that is in place of y we put vx minus cosec of vx upon x. Now this further gives us v plus x into dv by dx is equal to x cancels with x. So we have v here minus cosec of v since x cancels with x. Thus we have v plus x into dv by dx is equal to v minus cosec v. Now this v cancels with this v and so we are left with x into dv by dx is equal to minus cosec v. Or you can say we get dv upon cosec v is equal to minus dx upon x or you can say we have sin v dv is equal to minus dx upon x. Now integrating both sides we get integral sin v dv is equal to minus integral dx upon x. So this gives us minus cosec v is equal to minus log of modulus x plus c further we get cos v is equal to log of modulus x minus c. Now in place of v we put y upon x so we get cos of y upon x is equal to log modulus x minus c. Now the given condition in the question is y is equal to 0 when x is equal to 1. So let this be equation 3. Now we are given x equal to 1, y equal to 0. So in this equation we put x equal to 1 and y equal to 0. So we get cos 0 is equal to log 1 minus c. Now cos 0 is 1, log 1 is 0. So this gives us c is equal to minus 1. So now we get the value of c as minus 1. Now from equation 3 we have cos of y upon x is equal to log of modulus x plus 1. So this is the required particular solution of the given differential equation. So this completes the session. Hope you have understood the solution of this question.