[" The equation of the curve satisfying the differential equation "y_(2)(x^(2)+1)=2xy_(1)" passing through "],[" the point "(0,1)" and having slope of tangent at "x=0" as "3" is "]

The equation of the curve satisfying the differential equation Y_2 ( x^2 +1) = 4 xy _1 , passing through the point (0, – 4) and having slope of tangent at x = 0 as 4 is:

The equation of the curve satisfying the differential equation y^(2)(x^(2)+1)=2xy passing through the point (0,1) and having slope of tangnet at x=0 as 3, is (Here y=(dy)/(dx) and y_(2)=(d^(2)y)/(dx^(2)))

The equation of the curve satisfying the differential equation y_2 (x^2+1)=2x y_1 passing through the point (0,1) and having slope of tangent at x=0 as 3 (where y_2 and y_1 represent 2nd and 1st order derivative), then

The equation of the curve satisfying the differential equation y^2 (x^2 + 1) = 2xy passing through the point (0,1) and having slope of tangnet at x = 0 as 3, is (Here y=(dy)/(dx)andy_2=(d^2y)/(dx^2))

The equation of the curve satisfying the differential equation x^(2)dy=(2-y)dx and passing through P(1, 4) is

The equation of the curve satisfying the differential equation x^(2)dy=(2-y)dx and passing through P(1, 4) is

The equation of the curve satisfying the differential equation y(x+y^3)dx=x(y^3-x)dy and passing through the point (1,1) is