" (iii) "3y^(2)-2y+(1)/(3)=0

For all y ne 0, (1)/(y)+(1)/(2y)+(1)/(3y) =

To remove the first degree terms in the following equations origin should be shifted to the another point then calculate the new origins from list - II {:(" List - I "," List - II "),("(A) "x^(2)-y^(2)+2x+4y=0,"(1) (5,-7) "),("(B) "4x^(2) +9y^(2)-8x+36y + 4 = 0,"(2) (1,-2) "),("(C) "x^(2) + 3y^(2) + 2x + 12y + 1 = 0,"(3) (-1,2) "),("(D) "2(x-5)^(2)+3(y+7)^(2)=10,"(4) (-1,-2) "),(,"(5) (-5,7) "):} The correct matching is

Solution of D.E (dy)/(dx)=(2x+5y)/(2y-5x+3) is,if (y(0)=0) (1) x^(2)-y^(2)+5xy-3y=0 (2) x^(2)+y^(2)+5xy-3y=0 (3) x^(2)-y^(2)+5xy+3y=0 (4) x^(2)-y^(2)-5xy-3y=0

Prove that the equation 2x^(2)+3xy-2y^(2)-x+3y-1=0 represents a pair of perpendicular straight lines.

D.E., having the solution y=c_(1)+c_(2)e^(3x) , is A) y_(2)=3y B) y_(2)=3y_(1) C) y_(3)+3y_(1)=0 D) y_(2)+3y=0

Solve:(1)/(2(2x+3y))+(12)/(7(3x-2y))=(1)/(2)(7)/(2x+3y)+(4)/(3x-2y)=2 where 2x+3y!=0 and 3x-2y!=0

If the points (x_(1),y_(1)),(x_(2),y_(2)), and (x_(3),y_(3)) are collinear show that (y_(2)-y_(3))/(x_(2)x_(3))+(y_(3)-y_(1))/(x_(3)x_(1))+(y_(1)-y_(2))/(x_(1)x_(2))=0

(1)/(y)-(1)/(y-2)=3,y!=0,2

The tangent from the point of intersection of the lines 2x – 3y + 1 = 0 and 3x – 2y –1 = 0 to the circle x^(2) + y^(2) + 2x – 4y = 0 is