The function `y=f(x)` is the solution of the differential equation `(dy)/(dx)+(x y)/(x^2-1)=(x^4+2x)/(sqrt(1-x^2))` in `(-1,1)` satisfying `f(0)=0.` Then `int_((sqrt(3))/2)^((sqrt(3))/2)f(x)dx` is

The function y=f(x) is the solution of the differential equation (dy)/(dx)+(xy)/(x^(2)-1)=(x^(4)+2x)/(sqrt(1-x^(2)))in(-1,1) satisfying f(0)=0. Then int_((sqrt(3))/(2))^((sqrt(3))/(2))f(x)dx is

The function y=f(x) is the solution of the differential equation (dy)/(dx)+(xy)/(x^(2)-1)=(x^(4)+2x)/(sqrt(1-x^(2))) in (-1, 1) satisfying f(0)=0 . Then underset((-sqrt(3))/(2))overset((sqrt(3))/(2))intf(x)dx is

The function y=f(x) is the solution of the differential equation [dy]/[dx]+[xy]/[x^2-1]=[x^4+2x]/sqrt[1-x^2] in (-1, 1), satisfying f(0)=0. Then int_[-sqrt3/2]^[sqrt3/2] f(x)dx is (A) pi/3 - sqrt3/2 (B) pi/3 - sqrt3/4 (C) pi/6 - sqrt3/4 (D) pi/6 - sqrt3/2

The function y=f(x) is the solution of the differential equation [dy]/[dx]+[xy]/[x^2-1]=[x^4+2x]/sqrt[1-x^2] in (-1, 1), satisfying f(0)=0 . Then int_[-sqrt3/2]^[sqrt3/2] f(x)dx is (A) pi/3 - sqrt3/2 (B) pi/3 - sqrt3/4 (C) pi/6 - sqrt3/4 (D) pi/6 - sqrt3/2

The fx^n y=f(n) is the sol of diff eq^n (dy)/(dx)+(xy)/(x^2-1)=(x^4+2x)/sqrt(1-x^2) in (-1,1) satisfying f(0)=0 then int_(-sqrt3/2)^(sqrt3/2) f(n)ndx is

Find the solution of the differential equation x sqrt(1+y^(2))dx+y sqrt(1+x^(2))dy=0

Find the solution of the differential equation x\ sqrt(1+y^2)dx+y\ sqrt(1+x^2)dy=0.

Find the general solution of the differential equation x sqrt(1-y^(2))dx + y sqrt(1-x^(2)) dy =0.

If y=f(x) satisfies the differential equation (dy)/(dx)+(2x)/(1+x^(2))y=(3x^(2))/(1+x^(2)) where f(1)=1 , then f(2) is equal to