 Good morning friends. I am Purva. Let's work out the following question. For the following differential equation, find a particular solution satisfying the given condition. 1 plus x squared into dy by dx plus 2xy is equal to 1 upon 1 plus x squared and the given condition is y is equal to 0 when x is equal to 1. Let us now begin with the solution. So, the given differential equation is 1 plus x squared into dy by dx plus 2xy is equal to 1 upon 1 plus x squared and we are given that y is equal to 0 when x is equal to 1. Now, on dividing both the sides of this equation by 1 plus x squared we get dy upon dx plus 2xy upon 1 plus x squared is equal to 1 upon 1 plus x squared whole square. Let us mark this as equation 1. Now, the above equation is the linear differential equation of the form dy by dx plus py is equal to q. Therefore, the integrating factor of this equation is given by e raised to the power integral p dx. Now, on comparing this equation 1 with the equation dy by dx plus py is equal to q, we can clearly see that p is equal to 2x upon 1 plus x squared. So, we get integrating factor is equal to e raised to the power integral 2x upon 1 plus x squared dx and this is equal to e raised to the power. Now, integral 2x upon 1 plus x squared is equal to log 1 plus x squared and this is further equal to 1 plus x squared. e raised to the power log 1 plus x squared is equal to 1 plus x squared. Therefore, we get the integrating factor as 1 plus x squared. Now, multiplying both the sides of equation 1 by 1 plus x squared we get 1 plus x squared into dy by dx plus 2xy is equal to 1 upon 1 plus x squared. And this implies, now we can write 1 plus x squared into dy by dx plus 2xy as d upon dx of y into 1 plus x squared and this is equal to 1 upon 1 plus x squared. Now, integrating both the sides of this equation with respect to x we get integral d upon dx of y into 1 plus x squared dx is equal to integral 1 upon 1 plus x squared dx. And this implies, now integrating left hand side we get y into 1 plus x squared is equal to integrating 1 upon 1 plus x squared we get tan inverse x plus c where c is a constant let us mark this as equation 2. Now, to find the value of c let us use the condition when x is equal to 1 y is equal to 0. So, putting the value of x and y in equation 2 we get therefore 0 is equal to tan inverse 1 plus c and we have this implies 0 is equal to now tan inverse 1 is equal to pi by 4. So, we have pi by 4 plus c and this implies c is equal to minus pi by 4. Now, substituting this value of c in equation 2 we get y into 1 plus x squared is equal to tan inverse x minus pi by 4. Hence, the required solution is y into 1 plus x squared is equal to tan inverse x minus pi by 4. This is our answer. Hope you have understood the solution. Bye and take care.