BITSAT Mathematics Question (2021) — Solution

Step-by-step solution

The given differential equation can be written as d x d y + x 1+y^ 2 = ^ -1 y 1+y^ 2 (i) Now, eq. (i) is a linear differential equation of the form d x d y + P _ 1 x= Q _ 1 where P _ 1 = 1 1+y^ 2 and Q _ 1 = ^ -1 y 1+y^ 2 Therefore, I.F =e^ 1 1+y^ 2 d y =e^ ^ -1 y Thus, the solution of the given differential equation is given by x e^ ^ -1 y = ( ^ -1 y 1+y^ 2 ) e^ ^ -1 y dy + C (ii) Let I = ( ^ -1 y 1+y^ 2 ) e^ ^ -1 y dy On substituting ^ -1 y=t, so that ( 1 1+y^ 2 ) dy = dt , we get I= t e^ t d t=t e^ t - 1 e^ t d t=t e^ t -e^ t =e^ t (t-1) or I =e^ ^ -1 y ( ^ -1 y-1 ) On substituting the value of 1 in equation (i1), we get x e^ ^ -1 y = e^ ^ -1 y ( ^ -1 y-1 )+ C or x= ( ^ -1 y-1 )+ C e^ ^ -1 y which is the general solution of the given differntial equation.