 Hi friends, I am Purva and today we will discuss the following question, find the particular solution of the differential equation 1 plus e raised to the power 2x into dy plus 1 plus y square into e raised to the power x dx is equal to 0. Given that y is equal to 1 when x is equal to 0. Let us now begin with the solution. Now we are given 1 plus e raised to the power 2x into dy plus 1 plus y square into e raised to the power x dx is equal to 0 and we are also given that y is equal to 1 when x is equal to 0. Now separating the variables of the above differential equation we have 1 upon 1 plus y square dy plus e raised to the power x upon 1 plus e raised to the power 2x dx is equal to 0. This implies 1 upon 1 plus y square dy is equal to minus e raised to the power x upon 1 plus e raised to the power 2x dx. Now integrating both the sides we get integral of 1 upon 1 plus y square dy is equal to minus integral of e raised to the power x upon 1 plus e raised to the power 2x dx. Now this implies integral of 1 upon 1 plus y square is tan inverse y and this is equal to minus of. Now in this integral if we put e raised to the power x is equal to t we get e raised to the power x dx is equal to dt. So integral of e raised to the power x upon 1 plus e raised to the power 2x dx is equal to integral of dt upon 1 plus t square. So we get minus integral of dt upon 1 plus t square plus c where c is the constant of integration and here we have e raised to the power x is equal to t. Now this implies tan inverse y is equal to minus tan inverse t because integral of 1 upon 1 plus t square is tan inverse t plus c and we have e raised to the power x is equal to t. So we get this implies tan inverse y is equal to minus of tan inverse e raised to the power x plus c. Now this further implies tan inverse y plus tan inverse e raised to the power x is equal to c we mark this as equation 1. Now it is given that y is equal to 1 when x is equal to 0. So putting x is equal to 0 and y is equal to 1 in equation 1 we get therefore tan inverse 1 plus tan inverse e raised to the power 0 is equal to c and this implies now tan inverse 1 is equal to pi by 4. So we have pi by 4 plus now e raised to the power 0 is equal to 1. So we have tan inverse 1 which is equal to pi by 4 and this is equal to c as e raised to the power 0 is equal to 1 and tan inverse 1 is equal to pi by 4. Now this implies pi by 2 is equal to c or c is equal to pi by 2. Now substituting this value of c in equation 1 we get tan inverse y plus tan inverse e raised to the power x is equal to pi by 2. Hence the required solution is tan inverse y plus tan inverse e raised to the power x is equal to pi by 2. This is our answer. Hope you have understood the solution. Bye and take care.