Differential Equations problem

  • Thread starter L.D.50
  • Start date
  • #1
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Homework Statement



dx/dy = cos(y) - xtan(y)

I need to find the general solution of the problem



Homework Equations



y' + Py = Q
Where P and Q are functions of x

dy/y = - Pdx
ln(y) = -integral(Pdx)+c
y = e^(-int(Pdx)+c)



The Attempt at a Solution



Now I have no idea what to do to be honest. Nothing I try and do to separate the variables to get it in the form y' + Py = Q works. There is always something left over that complicates things even more. Ive tried many trig substitutions, to no avail. Here is one thing I tried.

dx/x = [(1/x)cosy - tany]dy
dx/x = (1/x)cosy dy - tan y dy

But then I have two dy terms. Is this correct? I really have no idea how to proceed from here.

Ive also tried:

dx/dy = cos(y) - xtan(y)
(dx/dy)cosy = cos^2(y) - xsiny
(dx/dy)2cosy = 1 + cos(2y) - 2xsiny

But now once again I have no idea how to proceed. Any help would be greatly appreciated!
 

Answers and Replies

  • #2
63
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The fact that your equation involves dx\dy rather than dy\dx means that just by adding xtany to both sides you get the required form! and you can find the integrating factor in terms of y.

ie use
x' + Px = Q
Where P and Q are functions of y
 

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