 Good morning friends. I am Purva and today I will help you with the fallback question the integrating factor of the differential equation 1 minus y square into dx upon dy plus yx is equal to a y Where y is greater than minus 1 and less than 1 is a 1 upon y square minus 1 B 1 upon under root of y square minus 1 C 1 upon 1 minus y square d 1 upon under root of 1 minus y square Now the integrating factor of the linear differential equation dx upon dy plus P1x is equal to q1 is given by e raised to the power integral P1 dy This is the key idea behind a question Let us now begin with the solution Now the given differential equation is 1 minus y square into dx upon dy plus y into x is equal to a y and We can write this as this implies dx upon dy plus y upon 1 minus y square Into x is equal to a into y upon 1 minus y square This differential equation is of the form dx upon dy plus P1x is equal to q1 Therefore the integrating factor of this equation is given by e raised to the power integral P1 dy Now comparing this equation with the equation given in the key idea We can clearly see that P1 is equal to y upon 1 minus y square Therefore we get integrating factor is equal to e raised to the power integral y upon 1 minus y square dy And this is equal to e raised to the power now integral of y upon 1 minus y square is minus 1 upon 2 into log of 1 minus y square And we can write this as this is equal to e raised to the power minus log under root of 1 minus y square and This is further equal to e raised to the power log 1 upon under root of 1 minus y square And this is equal to 1 upon under root of 1 minus y square Because e raised to the power log 1 upon under root of 1 minus y square is equal to 1 upon under root of 1 minus y square And this is same as option D Hence the correct option is D This is our answer. Hope you have understood the solution. Bye and take care