 Hello and welcome to the session. I am Deepika here. Let's discuss a question which says, show that the given differential equation is homogeneous and solve it. x minus y into dy minus x plus y into dx is equal to 0. Let's start the solution. Now given differential equation is minus y dy minus x plus y into dx is equal to 0 or we can rewrite this differential equation as x minus y dy is equal to x plus y into dx or dy by dx is equal to x plus y over x minus y. Now we know that a differential equation of the type dy by dx is equal to f of xy can be expressed in the form dy by dx is equal to g of y over x where g of y over x is a homogeneous function of degree 0. Then the given differential equation is a homogeneous differential equation. So we will try to make this differential equation in the form of a function of y over x. Now we can rewrite this differential equation as dy by dx is equal to 1 plus y over x over 1 minus y over x and this is of the form g of y over x. Let us give this as number 1. Now right hand side of the upper equation is of the form g of y over x and so it is a homogeneous function of degree 0 so the given differential equation a homogeneous differential equation. Now we will solve this homogeneous differential equation by putting y is equal to vx on differentiating both sides with respect to x we have 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 1 we have v plus x into dv over dx is equal to 1 plus v over 1 minus v or we can say x into dv over dx is equal to 1 plus v over 1 minus v minus v or x into dv over dx is equal to 1 plus v minus v plus v square over 1 minus v or we have x into dv over dx is equal to 1 plus v square over 1 minus v. Now let us separate the variables so we have 1 minus v over 1 plus v square into dv is equal to dx over x or by integrating both sides we have integral of 1 minus v over 1 plus v square dv is equal to integral of dx over x. Again the left hand side can be written as integral of 1 over 1 plus v square dv minus 1 by 2 into 2v over 1 plus v square dv is equal to integral of dx over x or we have. Now this is tan inverse v because derivative of tan inverse v is 1 over 1 plus v square minus 1 by 2. Now here put 1 plus v square is equal to t so we have 2v into dv is equal to dt. So integral of 2v over 1 plus v square is log of 1 plus v square and this is equal to log of mod x plus c. We can write tan inverse v is equal to 1 by 2 log of 1 plus v square plus log of mod x plus c. Now on replacing v by y over x we get tan inverse y over x is equal to 1 by 2 log of 1 plus y square over x square plus log of mod x plus c. We have tan inverse y over x is equal to 1 over 2 log of x square plus y square over x square plus log of mod x plus c. tan inverse y over x is equal to 1 over 2. Now log m over n is log m minus log m so this is log of x square plus y square minus 1 by 2 log x square plus log mod x plus c. tan inverse y over x is equal to 1 over 2 log of x square plus y square minus 1 over 2 into 2 log x we have tan inverse y over x is equal to 1 over 2 into log of x square plus y square. The required solution is tan inverse y over x is equal to 1 over 2 into log of x square plus y square plus c. So this is our answer for the above question. I hope the solution is clear to you and you have enjoyed the session. Bye and take care.