y=int(dx)/((1+x^(2))^(1/2))" and "y=0" when "x=0" ,then value of "y" when "x=1" ,is "

If y=int1/(1+x^2)^(3/2)dx and y=0 when x=0 , then value of y when x=1 is

If y=int1/(1+x^2)^(3/2)dx and y=0 when x=0 , then value of y when x=1 is

If y=int1/(1+x^2)^(3/2)dx and y=0 when x=0 , then value of y when x=1 is

(x^(2)+y^(2))dx-2xydy=0 , given y=0 when x=1

x^2dy+(y-1/x)dx=0 at y(1)=1 then find the value of y(1/2)

(x+1)(dy)/(dx) -1 = 2e^(-y) , y=0, " when " x=1

Solve the differential equation (dy)/(dx) = y + int_(0)^(1)y(x)dx , given that the value of y is 1 , when x = 0 .

Solve the differential equation (dy)/(dx) = y + int_(0)^(1)y(x)dx , given that the value of y is 1, when x = 0.