`xdx+ydy+[xdy-ydx]/[x^2+y^2]=0`

The solution of the differential equation xdx+ydy =(xdy-ydx)/(x^(2)+y^(2)) is tan(f(x, y)-C)=(y)/(x) (where, C is an arbitrary constant). If f(1, 1)=1 , then f(pi, pi) is equal to

Solve: xdy+ydx= (xdy-ydx)/(x^(2)+y^(2))

Solve xdx+ydy=xdy-ydx.

The solution of (xdx+ydy)/(xdy-ydx)=sqrt((a^(2)-x^(2)-y^(2))/(x^2+y^(2))), is given by

The solution of the differential equation xdy-ydx+3x^(2)y^(2)e^(x^(3))dx=0 is (where, c is an arbitrary constant)

Which of the following is a homogeneous differential equation? (A) (4x + 6y + 5) dy - (3y + 2x + 4) dx = 0 (B) (x y)dx-(x^3+y^3)dy=0 (C) (x^3+2y^2)dx+2x ydy=0 (D) y^2dx+(x^2+x y-y^2)dy=0

e^ydy/dx + x^3cosx^2 = 0

Let y(x) be a solution of xdy+ydx+y^(2)(xdy-ydx)=0 satisfying y(1)=1. Statement -1 : The range of y(x) has exactly two points. Statement-2 : The constant of integration is zero.

Show that the given differential equation xdy-ydx=sqrt(x^(2)+y^(2)) dx is homogenous and solve it.

The solution of the equation xdy-ydx=sqrt(x^2-y^2)dx subject to the condition y(1)=0 is (A) y=xsin(logx) (B) y=x^2 sin(logx) (C) y=x^2(x-1) (D) none of these