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 (a)` y=f( x)` is a strictly increasing function (b)` y=f( x )` is a non-monotonic function (c)` y=f( x) )` has a three distinct real roots (d)`y=f( x)` has only one negative root.

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(x+y^3)dx=x(y^3-x)dy and passing through the point (1,1) is

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 (a) ( b ) (c) y=f(( d ) x (e))( f ) (g) is a strictly increasing function (h) ( i ) (j) y=f(( k ) x (l))( m ) (n) is a non-monotonic function (o) ( p ) (q) y=f(( r ) x (s))( t ) (u) has a three distinct real roots (v) ( w ) (x) y=f(( y ) x (z))( a a ) (bb) has only one negative root.

A curve y=f(x) satisfy the differential equation (1+x^(2))(dy)/(dx)+2yx=4x^(2) and passes through the origin. The function y=f(x)

The equation of the curve passing through the point (1,1) and satisfying the differential equation (dy)/(dx) = (x+2y-3)/(y-2x+1) is

If y=f (x) satisfy the differential equation (dy)/(dx) + y/x =x ^(2),f (1)=1, then value of f (3) equals:

y = f(x) which has differential equation y(2xy + e^(x)) dx - e^(x) dy = 0 passing through the point (0,1). Then which of the following is/are true about the function?

Find the equation of a curve passing through (0,1) and having gradient (-(y+y^3))/(1+x+x y^2) at (x , y) .

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

Given the curves y=f(x) passing through the point (0,1) and y=int_(-oo)^(x) f(t) passing through the point (0,(1)/(2)) The tangents drawn to both the curves at the points with equal abscissae intersect on the x-axis. Then the curve y=f(x), is