" 3.(i) "xdy+ydx=xydx,y(1)=

Solve the initial value problem: (xdy-ydx)=xydx,y(1)=1

The solution of the differential equation xdy + ydx= xydx when y(1)=1 is

Solve the initial value problem: x(xdy+ydx)=ydx,y(1)=1

Solve the initial value problem: x(xdy+ydx)=ydx ,y(1)=1.

xdy-ydx = yes

If x^(3)dy+xydx=x^(2)dy+2ydx,y(2)=e then y(-1)=

If x^(3)dy+xydx=x^(2)dy+2ydx,y(2)=e then y(-1)=

y ^ (2) (xdy + ydx) + xdy-ydx = 0

The curve,which satisfies the differential equation (xdy-ydx)/(xdy+ydx)=y^(2)sin(xy) and passes through (0,1), is given by

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 :