 Hi and welcome to the session, I am Deepika here. Let's discuss a question which says find the general solution of the differential equation x dy by dx plus 2y is equal to x square log x. Let's start the solution. The given differential equation is x dy by dx plus 2y is equal to x square log x or this can be written as dy by dx plus 2 over x into y is equal to x log x. Let us give this equation as number one. Now equation one is a linear differential equation dy by dx plus py is equal to q where p which is a coefficient of y is equal to 2 over x and q is equal to x log x. Therefore integrating factor is equal to e raised to power integral of 2 over x dx which is equal to e raised to power 2 log x and this is again equal to e raised to power log x square and this is equal to x square. Now to find the solution we will multiply the equation one with the integrating factor because by this way the reference side of this equation will become differential of some function. So on multiplying both sides of equation one with x square vk dx x square plus 2 over x into x square y is equal to x cube log x dy by dx into x square plus 2xy is equal to x cube log x or this can be written as dy dx of yx square is equal to x cube log x. Now integrating both sides with respect to x we have integral of dy dx of yx square dx is equal to integral of x cube log x dx. This can be written as yx square is equal to let us write the integral on the right hand side as i plus c where i is equal to integral of x cube log x dx. Let us give this equation as number two. Let us integrate this by parts therefore i is equal to let us take log x as the first function and x cube as the second function. log x into integral of x cube dx minus integral of dy dx of log x into integral of x cube dx and this is equal to log x into x is to power 4 over 4 minus integral of 1 over x into x is to power 4 over 4 dx and this is equal to x is to power 4 over 4 into log x minus 1 over 4 into x is to power 4 over 4. Let us take x is to power 4 over 16 common from these two terms so we have i is equal to x is to power 4 over 16 into 4 log x minus 1 on substituting the value of i in equation 2. We have yx square is equal to x is to power 4 over 16 into 4 log x minus 1 plus c. This can be written as y is equal to x square over 16 into 4 log x minus 1 plus c into x is to power minus 2. Hence the general solution of the given differential equation is y is equal to x is to power 4 over 16 into 4 log x minus 1 plus c into x is to power minus 2. 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.