For `y gt 0 and x in R, ydx + y^(2)dy = xdy` where y = f(x). If f(1)=1, then the value of f(-3) is

If (x+f(y))=f(x)+y AA x,y in R and f(0)=1 then find the value of f(7).

Let y=f(x) satisfies (dy)/(dx)=(x+y)/(x) and f(e)=e then the value of f(1) is

IF x cos ( y //x ) ( ydx + xdy)=y sin ( y // x) ( xdy - ydx ) y (1) = 2 pi then the value of 4 ( y (4) )/(pi ) cos (( y (4))/(4)) is :

Let y=f (x) and x/y (dy)/(dx) =(3x ^(2)-y)/(2y-x^(2)),f(1)=1 then the possible value of 1/3 f(3) equals :

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

The solution of the differential equation ydx-(x+2y^(2))dy=0 is x=f(y). If f(-1)=1, then f(1) is equal to

Let f: R rarr R be a differentiable function satisfying f(x+y)=f(x)+f(y)+x^(2)y+xy^(2) for all real numbers x and y. If lim_(xrarr0) (f(x))/(x)=1, then The value of f'(3) is

f(x)= y+(1)/(y) , where y gt 0 . If y increases in value, then f (x)