 Hello and welcome to the session. In this session we discuss the following question that says find the general solution of the differential equation x into log x into dy by dx plus y is equal to 2 upon x into log x. Let's proceed with the solution now. The given differential equation is x into log x into dy by dx plus y is equal to 2 upon x into log x. Now, dividing both sides by x into log x we get dy by dx plus y upon x into log x is equal to 2 upon x into log x into 1 upon x into log x. This means we have dy by dx plus y upon x into log x is equal to or this log x passes with this log x and here we are left with 2 upon x square. Now, this differential equation is a real differential equation of the form dy by dx plus py is equal to q and here we have p is equal to 1 upon x into log x and q is equal to 2 upon x square. So, therefore the integrating factor given as is equal to e to the power of integral p dx. That is this is equal to e to the power of integral p which is 1 upon x log x dx. Now, let us find the integral of 1 upon x log x dx for this we assume let log x be equal to p. So, differentiating both sides we get 1 upon x dx is equal to dt. Therefore, we have the given integral as t upon t which is equal to log of modulus t plus the constant of integration or putting the value of t as log x. We have log of log x plus c is the integral 1 upon x log x dx. This is equal to e to the power of log of log x and so this is equal to log x. That is the integrating factor is equal to log x. Now, let us assume this equation to be equation 1. Now, by multiplying both sides of equation 1 by the integrating factor we get log x into dy by dx plus y upon x into log x the whole is equal to log x into 2 upon x square. This further gives us dy by dx into log x plus y upon x is equal to upon x square. To find the differential of y into log x we get y upon x plus log x into dy by dx. So, the left hand side can be written as dy by dx of y into log x and this is equal to 2 into x to the power minus 2 into log x. Integrate both sides with respect to x. So, integrating both sides with respect to y into log x is equal to integral of 2 into x to the power of minus 2 log x dx. So, further we get y into log x is equal to 2 into integral of x to the power of minus 2 into log x dx. So, further we have y into log x is equal to 2 into let us find the integral of this by bioparts. This is our first function. This is our second function. On integrating this bioparts we get first function that is log x into integral of the second function that is integral of x to the power of minus 2 dx minus integral of differential of the first function that is dy dx of log x into integral of the second function which is x to the power of minus 2 dx this whole dx. This whole plus the constant of integration which is c. Now, further we have y into log x is equal to 2 into log x into integral of x to the power of minus 2 is given as x to the power of minus 1 upon minus 1 minus integral of now differential of log x is 1 upon x into integral of x to the power of minus 2 is x to the power of minus 1 upon minus 1 dx and this whole plus c which is the constant of integration. Now, further we have y into log x is equal to 2 into minus log x integral of 1 upon x square dx the whole plus the constant of integration c. Now, next we have y into log x is equal to minus 2 log x into integral of 1 upon x square dx is minus 1 upon minus 1 plus the constant of integration c. So, we now have y into log x is equal to minus 2 log x upon x minus 2 which gives us y into log x is equal to minus 2 upon x into log x plus 1 the whole plus constant of integration c. y into log x equal to minus 2 upon x into log x plus 1 the whole plus the constant of integration c is the general solution given differential equation. This completes the fashion. Hope you have understood the solution of this question.