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)=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.

An equation of the curve satisfying x dy - y dx = sqrt(x^(2)-y^(2))dx and y(1) = 0 is

Verify that the function y'''=0 is a solution of the differential equation y=x^(2)+2x+1.

Let y=f(x) be a function satisfying the differential equation (x d y)/(d x)+2 y=4 x^2 and f(1)=1 . Then f(-3) is equal to

A function y=f(x) satisfies the differential equation (d y)/(d x)+x^2 y=-2 x, f(1)=1 . The value of |f^( prime prime)(1)| is

The graph of the function y = f(x) passing through the point (0, 1) and satisfying the differential equation (dy)/(dx) + y cos x = cos x is such that

Find the equation of the curve passing through (1,0) and which has slope 1+(y)/(x) at (x,y)

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

The curve passing through the point (1,1) satisfies the differential equation (dy)/(dx)+(sqrt((x^2-1)(y^2-1)))/(x y)=0 . If the curve passes through the point (sqrt(2),k), then the value of [k] is (where [.] represents greatest integer function)_____

Solution of the differential equation {1/x-(y^2)/((x-y)^2)}dx+{(x^2)/((x-y)^2)-1/y}dy=0 is