 Hi and welcome to the session. Let us discuss the following question. Question says, for the differential equation, xy dy upon dx is equal to x plus 2 multiplied by y plus 2. Find the solution curve passing through the point 1 minus 1. First of all, let us understand that if fx is the solution curve, then integral of f dash x dx is equal to fx plus c, where c is determined by using given conditions. This is the key idea to solve the given question. Let us now start with the solution. Now the given differential equation is xy dy upon dx is equal to x plus 2 multiplied by y plus 2. Now separating the variables in this equation, we get y upon y plus 2 multiplied by dy is equal to x plus 2 upon x multiplied by dx. Now integrating both the sides of this equation, we get integral of y upon y plus 2 dy is equal to integral of x plus 2 upon x dx. Now adding and subtracting 2 in the numerator of this integrand, we get integral of y plus 2 minus 2 upon y plus 2 dy is equal to integral of 1 plus 2 upon x dx. Now this implies integral of 1 minus 2 upon y plus 2 dy is equal to integral of 1 plus 2 upon x dx. Now let us name this expression as 1. First of all, let us find out this integral. Let this integral be i1. So we can write i1 is equal to integral of 1 minus 2 upon y plus 2 dy. Now this integral can be further written as integral of dy minus 2 multiplied by integral of dy upon y plus 2. Using this formula of integration, we can find this integral. So here we can write this integral is equal to y. We will write minus 2 as it is and using this formula of integration, we will find this integral. And this integral is equal to log of y plus 2 plus c1, where c1 is the constant of integration. We can find this integral by using substitution method. So here we have shown how we can find out the integral of dy upon y plus 2. Here we have substituted t for y plus 2. Now differentiating both the sides with respect to y, we get dy is equal to dt. Now this integral can be written as integral of dt upon t. Now this is further equal to log t plus c. Now substituting y plus 2 for t here, we get log y plus 2. So we get integral dy upon y plus 2 is equal to log of y plus 2 plus c. Now we will evaluate this integral. Let us assume that i2 is equal to integral of 1 plus 2 upon x dx. Now this integral can be further written as integral of dx plus 2 multiplied by integral of dx upon x. Now using this formula of integration, we get this integral is equal to x. And using this formula of integration, we get this integral is equal to 2 log x plus c2, where c2 is the constant of integration. Now from equation 1, we know i1 is equal to i2. Now i1 is equal to y minus 2 log y plus 2 plus c1 and i2 is given by this expression. So here we can write x plus 2 log x plus c2. Now this implies y minus x is equal to log x square plus log y plus 2 whole square plus c2 minus c1. Using this law of logarithms, we have written 2 log x as log of x square and 2 log y plus 2 can be written as log of y plus 2 whole square. Now substituting c for c2 minus c1, we get y minus x is equal to log x square plus log of y plus 2 whole square plus c. Let us name this equation as equation 2. Now we know given curve passes through point 1 minus 1. So we will substitute 1 for x in this expression and minus 1 for y in this expression and we get minus 1 minus 1 is equal to log 1 square plus log of minus 1 plus 2 whole square plus c. Now this implies minus 2 is equal to log 1 plus log 1 plus c. Now we know log 1 is equal to 0. So we get minus 2 is equal to c or we can simply write c is equal to minus 2. Now substituting c is equal to minus 2 in equation 2, we get y minus x is equal to log of x square plus log of y plus 2 whole square plus minus 2. Now applying this law of logarithms in these two terms we get y minus x is equal to log of x square multiplied by y plus 2 whole square minus 2. Now adding 2 on both the sides of this equation we get y minus x plus 2 is equal to log of x square multiplied by y plus 2 whole square. So the required solution curve is y minus x plus 2 is equal to log of x square multiplied by y plus 2 whole square. This completes the session. Hope you understood the solution. Take care and have a nice day.