1.`x^(3)y^(3)+1`
2.`xy(x^(2)y^(2)+1)`
3. `(xy+1)(x^(2)y^(2)-xy+1)`

(dy)/(dx) = (2xy)/(x^(2)-1-2y)

Using horizontal method : Add x^(2)+y^(2)- 2xy , -2x^(2)-y^(2)-2xy and 3x^(2) +y^(2) +xy

The differential equation of all conics whose centre klies at origin, is given by (a) (3xy_(2)+x^(2)y_(3))(y-xy_(1))=3xy_(2)(y-xy_(1)-x^(2)y_(2)) (b) (3xy_(1)+x^(2)y_(2))(y_(1)-xy_(3))=3xy_(1)(y-xy_(2)-x^(2)y_(3)) ( c ) (3xy_(2)+x^(2)y_(3))(y_(1)-xy)=3xy_(1)(y-xy_(1)-x^(2)y_(2)) (d) None of these

y(xy+1)dx+x(1+xy+x^(2)y^(2))dy=0

Add : x^(3) - x^(2)y + 5xy^(2) + y^(3) , -x^(3) - 9xy^(2) + y^(3), 3x^(2)y + 9xy^(2)

Consider the expression (3)/(2)x^(2)y-(1)/(2)xy^(2)+6x^(2)y^(2) In the term -(1)/(2)xy^(2) , what is the coefficient of x ?

Consider the expression (3)/(2)x^(2)y-(1)/(2)xy^(2)+6x^(2)y^(2) In the term -(1)/(2)xy^(2) , write down the numerical coefficient and the literal coefficient.

Simplify: 2x^(2)+3xy -3y^(2)+x^(2)-xy+y^(2)

Multiply : 3 - (2)/(3)xy + (5)/(7)xy^(2) - (16)/(21)x^(2)y by -21 x^(2)y^(2)

If cottheta+tantheta=x and sectheta-costheta=y then (i) sintheta costheta=-x (ii) sintheta tantheta=-y (iii) (x^2y)^(2/3)-(xy^2)^(2/3)=1 (iv) (x^2y)^(1/3)+(xy^2)^(1/3)=1