`y = (ax + b) e^(-2x)`

Form the differential equations of the following families of curves by elimnating the parameters (arbitrary constants) given against them in the brackets. (i) y = c(x-c)^(2), (c) (ii) xy = a e^(x) + b e^(-x), (a, b) (iii) y = (a+bx)e^(Kx), (a,b) (iv) y = a cos (nx + b), (a,b) (v) = y = a e^(3x) + be^(4x), (a,b) (vi) y = ax^(2) + bx, (a,b) (vii) ax^(2) + by^(2) = 1 (a,b)

If (a^(2) + b^(2)) (x^(2) + y^(2)) = (ax + by)^(2) , prove that : (a)/(x) = (b)/(y) .

The solution of the differential equation (dy)/(dx) = (3e^(2x) + 3e^(4x) )/( e^(x) + e^(-x) ) is a) y= e^(3x) + C b) y=2e^(2x) + C c) y= e^(x) + C d) y= e^(4x) + C

If 4x ^(2) + 12 xy - 8x + 9 y ^(2) - 12 y = (ax + by) (ax+ by -4), then the value of a ^(2) +b ^(2) is

If 4x^(2) + 12xy -8x+9y^(2) -12y = (ax+by) (ax+by-4) , then the value of a^(2) +b^(2) is

{:(x + y = a + b),(ax - by = a^(2) - b^(2)):}

If A={(x,y):y=e^(x),x in R} and B={(x,y):y=e^(-2x),x in R}, then

If y = ax ^(2) + b// x then x ^(2) y _(2) =