 Hi and welcome to the session. Let us discuss the following question, question says solve the differential equation. Given differential equation is e raised to the power minus 2 root x upon root x minus y upon root x multiplied by dx upon dy is equal to 1. Let us now start with the solution. First of all we will rewrite the given equation that is e raised to the power minus 2 root x upon root x minus y upon root x dx upon dy is equal to 1. Now multiplying both the sides of this equation by dy upon dx we get dy upon dx is equal to e raised to the power minus 2 root x upon root x minus y upon root x. Now adding y upon root x on both the sides of this equation we get dy upon dx plus y upon root x is equal to e raised to the power minus 2 root x upon root x. Now clearly we can see this is the linear differential equation or we can say this is the equation in this form. So for solving this equation first of all we will find out integrating factor. We know integrating factor of this equation is equal to e raised to the power integral of p dx. Now comparing these two equations we get p is equal to 1 upon root x. So integrating factor in this equation is equal to e raised to the power integral of 1 upon root x dx. Now let us find out this integral. We know this integral can be written as integral of x raised to the power minus 1 upon 2 dx 1 upon root x is equal to x raised to the power minus 1 upon 2. Now using this formula of integration we get this integral is equal to x raised to the power minus 1 upon 2 plus 1 upon minus 1 upon 2 plus 1 plus c. Now simplifying further we get 2 multiplied by root x plus c. Now we get integrating factor is equal to e raised to the power 2 root x. Clearly we can see this integral is equal to 2 root x. Now let us name this equation as equation 1. Now multiplying both the sides of this equation 1 by e raised to the power 2 root x we get. Now we know solution of this linear differential equation is given by this expression. Now solution of the given differential equation is y multiplied by integrating factor that is e raised to the power 2 root x is equal to integral of we know q in the given differential equation is e raised to the power minus 2 root x upon root x. So here we will write e raised to the power minus 2 root x upon root x multiplied by integrating factor and we know integrating factor is equal to e raised to the power 2 root x. Here we will write dx plus c. Now we will evaluate this integral. Now we can write y multiplied by e raised to the power 2 root x is equal to integral of 1 upon root x dx plus c. We know e raised to the power minus 2 root x multiplied by e raised to the power 2 root x is equal to 1 as these 2 terms are reciprocal of each other and we have written root x as it is in the denominator. Now we will evaluate this integral. Now using this formula of integration we can find this integral we know 1 upon root x is equal to x raised to the power minus 1 upon 2 and integral of x raised to the power minus 1 upon 2 dx is equal to x raised to the power minus 1 upon 2 plus 1 upon minus 1 upon 2 plus 1 plus c where c represents the constant of integration and here we will write left hand side that is y multiplied by e raised to the power 2 root x as it is. Now here we have replaced n by minus 1 upon 2. Now simplifying further we get y multiplied by e raised to the power 2 root x is equal to x raised to the power 1 upon 2 upon 1 upon 2 plus c. Now this further implies y multiplied by e raised to the power 2 root x is equal to 2 root x plus c. We know minus 1 upon 2 plus 1 is equal to 1 upon 2 and x raised to the power 1 upon 2 is equal to root x and x raised to the power 1 upon 2 upon 1 upon 2 is equal to 2 multiplied by root x. So here we can write 2 root x plus c is equal to y multiplied by e raised to the power 2 root x. So this is the required solution of the given differential equation. This completes the session. Hope you understood the solution. Take care and have a nice day.