" 47."x(xdy-ydx)=ydx,(1)=1

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

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

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

xdy-ydx = yes

xdy-ydx=xy^(2)dx

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 :

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 :

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

xdy-ydx = 2x^(3)y(xdy+ydx) if y(1) = 1, then y(2) +y((1)/(2)) =