Expert: Paul Klarreich - 8/27/2009

Question
How do i go about finding the general solution to dy/dx=2cosxe^(2/3sinx)y^2/3 giving solution in implicit form. I am fine with normal equations but trig gives me nightmares.

Answer
Questioner: Don
Country: United Kingdom
Category: Calculus
Private: No
Subject: Calculus
Question: How do i go about finding the general solution to dy/dx=2cosxe^(2/3sinx)y^2/3 giving solution in implicit form. I am fine with normal equations but trig gives me nightmares.
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Hi, Don,

Spend a few minutes each day doing some trig.  Do it right after your blood-pressure medication or whatever, and you will sleep better.  (I also give family advice.)

I think you do a separation of variables:
dy
-- =  2cosx e^(2/3sinx) y^2/3
dx

Cross-multiply or whatever:

dy
------ =  2cosx e^(2/3sinx)dx
y^2/3

Integrate each side separately, as if each were a problem in its own right.  

On the right side, a nice u-substitution of u = 2/3 sin x should do it.  On the left it is just:

y^-2/3 dy

You will get an equation with x's and y's and that should do it.