 Hello and welcome to the session I am Deepika here. Let's discuss the question which says show that the given differential equation is homogeneous and solid. y dx plus x log of y over x into dy minus 2x dy is equal to 0. Let's start the solution. The given differential equation is y dx plus x log of y over x dy minus 2x dy is equal to 0. Let us give this equation as number 1. Now equation 1 can be written as x log of y over x minus 2 into dy is equal to minus y tx or dy by tx is equal to minus y over x log of y over x minus 2. Again this can be written as dy by tx is equal to minus y over x upon log of y over x minus 2. Let us give this equation as number 2. Now the right hand side of 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. To solve this equation we will put y is equal to dx or v is equal to y over x. Therefore dy by dx is equal to v plus x into dv over dx. Now on substituting the value of y and dy by dx in equation 2 v half that is from equation 2 v half v plus x dv over dx is equal to minus v over log v minus 2 or this can be written as x dv over dx is equal to minus v over log v minus 2 minus v or x dv over dx is equal to minus v minus v log v plus 2 v over log v minus 2 or dv over dx is equal to v minus log v over log v minus 2. On separating the variables we have log v minus 2 over v into 1 minus log v into dv is equal to dx over x or this can be written as 2 minus log v upon v into log v minus 1 into dv is equal to dx over x. Now integrating both sides we have integral of 2 minus log v over v into log v minus 1 dv is equal to integral of dx over x. Now the integral on the left hand side can be written as integral of minus of log v minus 1 plus 1 over v into log v minus 1 dv is equal to integral of dx over x or this can be written as integral of minus 1 over v dv plus integral of 1 over v into log v minus 1 into dv is equal to integral of dx over x or this can be written as now integral of minus 1 over v dv is minus log mod v plus here substitute log v minus 1 is equal to t then we have 1 over v dv is equal to dt so this integral is of the form dt over t which is log of mod p and that is log of log v minus 1 and this is equal to log mod x plus log c where log c represents the constant of integration or this can be written as log of log v minus 1 over v is equal to log cx this can be written as log v minus 1 over v is equal to cx now on replacing v by y over x this can be written as log y over x minus 1 over y over x is equal to cx or log y over x minus 1 is equal to cx into y over x or log y over x minus 1 is equal to cy the general solution of the given differential equation that is of equation one is cy is equal to log y over x minus 1 so this is our answer I hope the solution is clear to you and you have enjoyed the session bye and have a nice day