Show that every member of the family of functions y = (6 ln(x) + C)/x , x > 0, is a solution of the differential equation x2y
' + xy = 6. (Simplify as much as possible.) y = 6 ln(x) + C x ⇒ y' = 6x · (1/x) − (6 ln(x) +C) x2 LHS = x2y' + xy = x2 · x2 + x · 6 ln(x) + C x = + 6 ln(x) + C = = RHS, so y is a solution of the differential equation.
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Let the two numbers be x and y, then,
xy = -12 . . . (1)
x + y = -10 . . . (2)
From (2), x = -10 - y . . . (3)
Putting (3) into (1), gives
(-10 - y)y = -12
-10y - y^2 = -12
y^2 + 10y - 12 = 0
Therefore, the two numbers are
and 
xy = -12 . . . (1)
x + y = -10 . . . (2)
From (2), x = -10 - y . . . (3)
Putting (3) into (1), gives
(-10 - y)y = -12
-10y - y^2 = -12
y^2 + 10y - 12 = 0
Therefore, the two numbers are