 Hi 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 1 plus e raised to power x over y into dx plus e raised to power x over y Into 1 minus x over y into dy is equal to 0 Now, let us first understand when the differential equation is Called a homogeneous differential equation now a differential equation which can be expressed in the form dy by dx is equal to f of xy or dx by dy is equal to g of xy where f of xy and g of xy are homogeneous function of degree 0 Then a differential equation is called a homogeneous differential equation. So this is a key idea behind our question We will take the help of this key idea to solve our question. So let's start the solution now the given differential equation is 1 plus e raised to power x over y into dx plus e raised to power x over y Into 1 minus x over y Into dy is equal to 0. Let us give this equation as number 1 Or we can rewrite this differential equation as 1 plus e raised to power x over y Into dx is equal to Minus e raised to power x over y into 1 minus x over y into dy Or again, we can rewrite this equation as dx over dy is equal to e raised to power x over y into x over y minus 1 over 1 plus e raised to power x over y Let us give this equation as number 2 now We see the right hand side of the above differential equation that is of equation 2 is of the form x of x over y 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 homogeneous differential equation we will put x over y is Equal to v that is x is equal to v y now on differentiating both sides with respect to y We get dx over dy is equal to v plus y Into dv over dy Now on substituting the value of x and dx over dy in equation 2 we get v Plus y into dv over dy is equal to e raised to power v y over y that is e raised to power v Into v y over y that is v minus 1 over 1 plus e raised to power v e raised to power v Into v minus 1 over 1 plus e raised to power v or Y Into dv over dy is equal to e raised to power v into v minus 1 over 1 plus e raised to power v minus v or Y into dv over dy is equal to v into e raised to power v minus e raised to power v minus v minus v into e raised to power v over 1 plus e raised to power v or y into dv over dy is equal to minus of e raised to power v plus v over 1 plus e raised to power v Now on separating the variables we have 1 plus e raised to power v over e raised to power v plus v Into dv is equal to minus dy over y and integral of 1 plus e raised to power v over e raised to power v plus v dv is equal to negative of integral dy over y let us give this equation as number three Now to solve the integral on the left hand side, we will substitute e raised to power v plus v is equal to t Then 1 plus e raised to power v dv is equal to dt So this integral is of the form integral of dt over t which is log of mod t that is log of mod e raised to power v plus v and this is equal to minus log of mod y plus log c Here log c represents the constant of integration or log of mod e raised to power v plus v is equal to log c over mod y or on substituting the value of v here we get log mod of e raised to power x over y plus x over y is equal to log c over mod y or We can rewrite this equation as e raised to power x over y plus x over y is equal to c over y or We can write this equation as y into e raised to power x over y plus x is equal to c hence the general solution of the given differential equation that is of differential equation 1 is y Into e raised to power x over y plus x is equal to c So this is our answer for the above question This completes our session. I hope you have enjoyed the session. Bye and take care