 Hi and welcome to the session. Let us discuss the following question. Question says, find the general solution for the following differential equation. Given differential equation is dy upon dx plus y is equal to 1, where y is not equal to 1. Let us now start with the solution. Now we are given with differential equation dy upon dx plus y is equal to 1. Now in this equation also we shall separate the variables x and y. Now clearly we can see if we subtract y from both the sides in this equation we get dy upon dx is equal to 1 minus y. Now multiplying both the sides of this equation by dx we get dy is equal to 1 minus y multiplied by dx. Dividing both the sides of this equation by 1 minus y we get dy upon 1 minus y is equal to dx. Now this can be further written as dy upon minus 1 multiplied by y minus 1 is equal to dx. Taking minus 1 common from this bracket we get y minus 1 inside the bracket. Now after separating the variables we will integrate both the sides of this equation. So we can write integral of dy upon minus 1 multiplied by y minus 1 is equal to integral of dx. Now this further implies negative of integral dy upon y minus 1 is equal to integral of dx. Multiplying and dividing this integral by minus 1 we get this integral. Now first of all let us find out this integral. Now in this integral we will put y minus 1 equal to t. Now differentiating both the sides with respect to y we get dy is equal to dt. Now substituting t for y minus 1 and dt for dy in this integral we get this integral. Now we know integral of dx upon x is equal to log x plus c. So integral of dt upon t is equal to log t plus c. Now we know t is equal to y minus 1. So we will substitute y minus 1 for t here and we get log y minus 1 plus c is equal to integral of dy upon y minus 1. Now multiplying both the sides of this equation by minus 1 we get integral of dy upon y minus 1 is equal to negative of integral of dx. Now we know this integral is equal to log of y minus 1. So here we will write log of y minus 1. Now we will write this is equal to sign as it is. Here we will write this negative sign as it is. Now integral of 1 with respect to x is equal to x only. So here we will write x plus log a. Now this equation can be further written as log of y minus 1 minus log a is equal to minus x. Subtracting log a from both sides of this equation we get this equation. Now applying the laws of logarithms on left hand side we can write left hand side as log of y minus 1 upon a is equal to minus x. Now this further implies y minus 1 upon a is equal to e raised to the power minus x. Here we have applied this rule of logarithm. Now multiplying both the sides of this equation by a we get y minus 1 is equal to a multiplied by e raised to the power minus x. Now adding 1 on both the sides we get y is equal to a multiplied by e raised to the power minus x plus 1. So the required solution of the given differential equation is y is equal to a multiplied by e raised to the power minus x plus 1. This completes the session. Hope you understood the solution. Take care and have a nice day.