Since, the above differential equation is a linear differential equation ∴ I.F. =∫e∫2x.1+x2dx=elog(1+x2)=1+x2 Then, the solution of the differential equation ⇒y(1+x2)=∫
dx
1+x2
+c ⇒y(1+x2)=tan−1x+c ....(1) If x=0 then y=0 (given) ⇒0=0+c ⇒c=0 Then, equation (1) becomes, ⇒y(1+x2)=tan−1x Now put x=1 in above equation, then 2y=