Expert: Ahmed Salami - 2/16/2006

Question
1) Find the equation of the tangent line to the graph: y^2 + ln(xy) = 2
at the point (e, 1)

2) Solve the differential equation: dy/dx= 4/(9+x^2) given the initial condition y(0)=3

Answer
Hi Tina,
Sorry for the time it took.
The slope of the tangent line at any point is the value
of dy/dx at that point. To find dy/dx,
y^2 + ln(xy) = 2
splitting the log expression, we have
y^2 + lnx + lny = 2
differentiating each term with respect to x,
(2y)dy/dx + 1/x + (1/y)dy/dx = 0
(2y)dy/dx + (1/y)dy/dx = -1/x
(2y + 1/y)dy/dx = -1/x
dy/dx = (-1/x)/(2y + 1/y)
     = (-1/x)/[(2y^2 + 1)/y]
     = -y/x(2y^2 + 1)
At the point (e,1)
dy/dx = -1/e[2(1^2) + 1]
     = -1/3e
The equation of the tangent is cartesian method;
slope = (y - y1)/(x - x1)
where (x1,y1) is any point on the line i.e (e,1) in this
case.
Therefore,
-1/3e = (y - 1)/(x - e)
cross multiplying
-1(x - e) = 3e(y - 1)
-x + e = 3ey - 3e
3ey + x - 4e = 0
or explicitly in terms of y,
y = (4e - x)/3e
 = 4/3 - x/3e

To solve dy/dx = 4/(9 + x^2)
i.e dy = 4/(9 + x^2) dx
we integrate both sides
$dy = $4/(9 + x^2) dx   where $ the integral sign
y = $4/(9 + x^2) dx
To find $4/(9 + x^2) dx, we make use of the substitution
x = 3tan#
# = arctan(x/3)
dx/d# = 3(sec#)^2
dx = 3(sec#)^2 d#
9 + x^2 becomes
9 + 9(tan#)^2 = 9[1 + (tan#)^2]
But 1 + (tan#)^2 = (sec#)^2
So,
9 + x^2 = 9(sec#)^2
and
y = $4/(9 + x^2) dx
 = 4$dx/(9 + x^2)
 = 4$3(sec#)^2 d#/ 9(sec#)^2
 = 4$d#/3
 = (4/3)# + c   ( c is a constant)
 = (4/3)arctan(x/3) + c
Using the initial condition
y(0) = 3 i.e at x = 0, y = 3
we have
3 = (4/3)arctan(0/3) + c
3 = (4/3)arctan(0) + c
but arctan(0) has the values 0, 180, 360 ,etc
Taking it as 0,
3 = 0 + c
c = 3
Therefore,
y = (4/3)arctan(x/3) + 3
I hope i have helped. You can always get back to me.
Regards.