The joint equation of lines `y=xa n dy=-x` is `y^2=-x^2,` i.e., `x^2+y^2=0` Statement 2: The joint equation of lines `a x+b y=0` and `c x+dy=0` is `(a x+b y)(c x+dy)=0` , wher `a , b , c , d` are constant.

The solution of the differential equation ydx+ (x +x^2 y) dy =0 is

The solution of the equation (x^2y+x^2)dx+y^2(x-1)dy=0 is given by

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

y dx + (x - y^(2))dy = 0

If y(x) satisfies the differential equation y^(prime)-ytanx=2xs e c x and y(0)=0 , then

If the pairs of lines x^2+2x y+a y^2=0 and a x^2+2x y+y^2=0 have exactly one line in common, then the joint equation of the other two lines is given by (a) 3x^2+8x y-3y^2=0 (b) 3x^2+10 x y+3y^2=0 (c) y^2+2x y-3x^2=0 (d) x^2+2x y-3y^2=0

Solve the differential equation y e^(x)/(y) dx = (x e^(x)/(y) + y^(2))dy ( y ne 0) .

xy=( x+y )^n and dy/dx = y/x then n= 1 b. 2 c. 3 d. 4

The general solution of the differential equation e^(x) dy + (y e^(x) + 2x)dx = 0 is

For the primitive integral equation y dx+y^2dy=xdy ; x in R ,y >0,y(1)=1, then y(-3) is (a) 3 (b) 2 (c) 1 (d) 5