 Hello and welcome to the session. I am Shashi and I am going to help you with the following question. Question says, find the general solution of the following differential equation. Given differential equation is y log y dx minus x dy is equal to 0. Let us now start with the solution. Now first of all we will rewrite the given differential equation. Now differential equation is y log y dx minus x dy is equal to 0. How adding this term on both sides of this equation we get y log y dx is equal to x dy. Now separating the variables in this equation we get dy upon y log y is equal to dx upon x. Now integrating both the sides of this equation we get integral of dy upon y log y is equal to integral of dx upon x. Now let us name this expression as 1. Now first of all we will find out this integral. So here we can write integral of dy upon y log y. Now we can find this integral by substitution method. We know derivative of log y is 1 upon y. So we will find this integral by substitution method. Now here we can write put log y is equal to t. Now differentiating both sides with respect to y we get 1 upon y is equal to dt upon dy. Now this further implies dy upon y is equal to dt. Now we can write this integral is equal to integral of dt upon t. Substituting t for log y and dt for dy upon y in this integral we get this integral as integral of dt upon t. Now using this formula of integration we get this integral is equal to log t plus log c1. Here log c1 is the constant of integration. Now substituting log y for t here we get log of log y plus log c1 is equal to integral of dy upon y log y. Now we will evaluate this integral. Now using this formula of integration we get this integral is equal to log x plus log of c2. Here log c2 represents constant of integration. Now let us name this expression as 2 and this expression as 3. Now we will substitute values of these integrals from expression 2 and 3 in expression 1. We know this is the expression 1. Now we get log of log y plus log c1 is equal to log x plus log c2. Now this further implies log of log y is equal to log x plus log c2 minus log c1. Now substituting log c for log c2 minus log c1 we get log of log y is equal to log x plus log c. Now applying this law of logarithm on right hand side of this expression we get log of log y is equal to log of xc. Now this further implies log y is equal to xc. Here we have applied this law of logarithms. Now this further implies y is equal to e raise to the power xc or we can say cx. Now again applying this law of logarithms in this expression we get this expression how the required solution of the given differential equation is y is equal to e raise to the power cx. This is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.