Review Problemswe get the following system forv01andv02(see (9) on page 195 in the text):v01(θ)cos4θ+v02(θ)sin4θ=0−4v01(θθ+4v02(θθ=tan4θ.Multiplying the Frst equation by sin 4θand the second equation by (1/4) cos 4θ, and addingthe resulting equations together, we getv02(θ)=14sin 4θ⇒v2=−116cos 4θ+c3.±rom the Frst equation in the above system we also obtainv01(θ−v02(θ)tan4θ=−14sin24θcos 4θ=−14(sec 4θ−cos 4θ)⇒v1(θ−14Z(sec 4θ−cos 4θ)dθ=−116ln|sec 4θ+tan4θ|+116sin 4θ+c4.Takingc3=c4=0,weobtainyp(θ±−116ln|sec 4θθ|+116sin 4θ²cos 4θ+±−116cosθ²sin 4θ=−116(cos 4θ)ln|sec 4θθ|,and a general solution to the original equation isy(θc1cos 4θ+c2sin 4θ−116(cos 4θ|sec 4θθ|.25.Since the auxiliary equation, 4
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.