Calculus Help!?

Question

Working on old AP Calc BC practice tests, I ran across this, and no one in my class knows how to solve it:

a) In the interval 0
(cot x) (dy/dx) + y = (csc x)

b)Find the solution of hte differential equation in part a that satisfies the condition that y = 0 when x = pi/3



Is it a separable equation? If so, how do I separate the terms??

If not, please help!! Thanks!!

Answer 1

The key here is to rewrite it into the form:
y' + a(x)y = b(x), create an integrating factor, redefine the left side using the Product Rule, then solve. It's somewhat complicated, so I'll step it out:

Step 1: we want to get into the form y' + a(x)y = b(x).
(cot x) (dy/dx) + y = (csc x)
Multiply everything by tan x:
dy/dx + y tan x = csc x tan x

csc x tan x = sin x / sin x cos x = 1/cos x = sec x
dy/dx + y tan x = sec x

Step 2: Now that we have an ordinary differential equation of the form: dy/dx + a(x) y = b(x), we can define an integrating factor (μ) of the form: μ = e^∫a(x) dx.
μ = e^∫tan x dx
μ = e^(ln |sec x|)
μ = |sec x|

Since x is from 0 to π/2, sec x is positive anyway, so we can drop the absolute value.

Step 3: Now that we have μ defined, we multiply the entire equation by μ.

dy/dx + y tan x = sec x
μdy/dx + μ * y tan x = μ sec x
sec x dy/dx + y sec x tan x = sec² x

Step 4: Define the left side using the product rule of derivatives:
sec x dy/dx + y sec x tan x is the derivative of a product:
(fg)' = gf' + fg'
where f = y and g = sec x, f' = dy/dx, and g' = sec x tan x.
Thus, we can rewrite it to be (fg)' or (y sec x)'
(y sec x)' = sec² x

Step 5: Integrate:
y sec x = ∫sec² x

Simplify and solve:
y sec x = sec x sin x + C
y = sin x + C (answer for a.)

Now, for b:
y = 0 when x = π/3
sin π/3 = √3/2
0 = √3/2 + C
C = -√3/2

Final equation: y = sin x - √3/2

Answer 2

Did you mean cactus?

here you go
http://en.wikipedia.org/wiki/Cactus