Elementary differential equations: Linear equations

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SUMMARY

The discussion focuses on solving the linear differential equation dr/d∅ + r tan∅ = cos∅. The integrating factor μ(∅) is calculated as exp[∫tan∅], resulting in μ(∅) = -cos∅. The user attempts to solve the equation but arrives at a different solution than the textbook, which states r = (∅ + c) cos∅. The discrepancy arises from the integration process and the manipulation of cos²∅ into 1/2 + 1/2 cos(2∅).

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Mdhiggenz
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Homework Statement



dr/d∅+rtan∅=cos∅

μ(∅)=exp[∫tan∅]
μ∅=exp[-ln[cos∅]=-cos∅

-cos∅(dr/d∅)-rtan∅cos∅=-cos^2∅
dr/d∅[-rcos∅]=-cos^2∅

rcos∅=∫cos^2∅

changed cos^2∅ to 1/2+1/2cos2∅
rcos∅=1/2∅ + 1/4 sin2∅+c

the books answer is not even close to mine they have r=(∅+c)cos∅

Where did I go wrong?

Homework Equations





The Attempt at a Solution

 
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Mdhiggenz said:

Homework Statement



dr/d∅+rtan∅=cos∅

μ(∅)=exp[∫tan∅]
μ∅=exp[-ln[cos∅]=-cos∅

##e^{-\ln\cos\phi}=e^{\ln[(cos\phi)^{-1}]} =\frac 1 {\cos\phi}##
 
Thanks LC!
 

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