 Hi and welcome to the session. Let us discuss the following question. Question size. Find a particular solution of the differential equation. Given differential equation is, x plus 1 multiplied by dy upon dx is equal to 2 multiplied by e raised to the power minus y minus 1 given that y is equal to 0 when x is equal to 0. Let us now start with the solution. Now the given differential equation is, x plus 1 dy upon dx is equal to 2 e raised to the power minus y minus 1. Now dividing both the sides of this equation by x plus 1, we get dy upon dx is equal to 2 e raised to the power minus y minus 1 upon x plus 1. Now taking reciprocal of both the sides, we get dx upon dy is equal to x plus 1 upon 2 e raised to the power minus y minus 1. Now let us consider 2 e raised to the power minus y minus 1. Now we know e raised to the power minus y is equal to 1 upon e raised to the power y. So here we can write 2 upon e raised to the power y minus 1. Now this is further equal to 2 minus e raised to the power y upon e raised to the power y. Now we get 2 e raised to the power minus y minus 1 is equal to 2 minus e raised to the power y upon e raised to the power y. So substituting for this term 2 minus e raised to the power y upon e raised to the power y, we get dx upon dy is equal to e raised to the power y multiplied by x plus 1 upon 2 minus e raised to the power y. Now this further implies dx upon dy is equal to e raised to the power y x upon 2 minus e raised to the power y plus e raised to the power y upon 2 minus e raised to the power y. We know on adding these 2 terms we get this term. So we can write this term as sum of these 2 terms. Now subtracting this term from both the sides of this equation we get dx upon dy minus e raised to the power y x upon 2 minus e raised to the power y is equal to e raised to the power y upon 2 minus e raised to the power y. Now clearly we can see this is a linear differential equation and this equation is in the form dx upon dy plus p1 x is equal to q1. Now comparing these 2 equations we get p1 is equal to minus e raised to the power y upon 2 minus e raised to the power y and q1 is equal to e raised to the power y upon 2 minus e raised to the power y. Now integrating factor is equal to e raised to the power integral of p1 dy. Now substituting corresponding value of p1 here we get integrating factor is equal to e raised to the power integral of minus e raised to the power y upon 2 minus e raised to the power y dy. Now let us find out this integral. Now we will find this integral by using substitution method. So we can write put 2 minus e raised to the power y is equal to t. Differentiating both the sides with respect to y we get minus e raised to the power y dy is equal to dt. Now the given integral is equal to integral of dt upon t. Now we know formula of integration integral of dx upon x is equal to log x plus c. So applying this formula here we get integral of dt upon t is equal to log t plus c where c is the constant of integration. Now substituting 2 minus e raised to the power y for t here we get log of 2 minus e raised to the power y plus c is equal to given integral. Now we get integrating factor is equal to e raised to the power log of 2 minus e raised to the power y. We know this integral is equal to log of 2 minus e raised to the power y. We have already shown this here. Now using this law of logarithms we get integrating factor is equal to 2 minus e raised to the power y. Now we know solution of the given differential equation is given by this expression. Now we will substitute corresponding values of integrating factor and q1 in this expression. Now we get x multiplied by 2 minus e raised to the power y is equal to integral of e raised to the power y upon 2 minus e raised to the power y multiplied by 2 minus e raised to the power y dy plus c. Now here 2 minus e raised to the power y will cancel 2 minus e raised to the power y and we get integral of e raised to the power y dy plus c is equal to x multiplied by 2 minus e raised to the power y. Now using this formula of integration we can find this integral and we get this integral is equal to e raised to the power y plus c and here we will write x multiplied by 2 minus e raised to the power y as it is. Now let us name this equation as equation 1. Now for the given differential equation we are given that y is equal to 0 when x is equal to 0. So we will substitute y is equal to 0 and x is equal to 0 in equation 1. Now equation 1 becomes 0 is equal to e raised to the power 0 plus c. Now we know e raised to the power 0 is equal to 1 only. So here we can write 0 is equal to 1 plus c. Now subtracting 1 from both the sides we get minus 1 is equal to c or we can simply write c is equal to minus 1. Now we will substitute this value of c in equation 1 and we get x multiplied by 2 minus e raised to the power y is equal to e raised to the power y minus 1. Now this further implies 2x minus x multiplied by e raised to the power y is equal to e raised to the power y minus 1. Now adding this term on both the sides we get 2x is equal to e raised to the power y multiplied by 1 plus x minus 1. Now adding 1 on both the sides of this equation we get 2x plus 1 is equal to e raised to the power y multiplied by 1 plus x. Now dividing both the sides by 1 plus x we get 2x plus 1 upon 1 plus x is equal to e raised to the power y or we can simply write e raised to the power y is equal to 2x plus 1 upon x plus 1. Now taking log on both the sides we get y is equal to log of 2x plus 1 upon x plus 1 where x is not equal to minus 1. So this is the required particular solution of the given differential equation. This completes the session. Hope you understood the solution. Take care and keep smiling.