This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Review Problems
We apply the method of variation of parameters to ﬁnd a particular solution to the original equation. We look for a particular solution of the form yp (θ) = v1 (θ) cos θ + v2 (θ) sin θ, where v1 (θ) and v2 (θ) satisfy the system (9), Section 4.6. That is, v1 cos θ + v2 sin θ = 0, −v1 sin θ + v2 cos θ = sec θ . Multiplying the ﬁrst equation by sin θ, the second equation by cos θ, and adding them together yield v2 sin2 θ + v2 cos2 θ = sec θ cos θ ⇒ v2 = 1 ⇒ v2 (θ) = θ. From the ﬁrst equation in the above system we also get v1 = −v2 tan θ = − tan θ ⇒ v1 (θ) = − tan θ dθ = ln  cos θ, where we have taken the zero integration constant. So, yp (θ) = cos θ ln  cos θ + θ sin θ , and y (θ) = c1 cos θ + c2 sin θ + cos θ ln  cos θ + θ sin θ is a general solution to the original equation. Diﬀerentiating we ﬁnd that y (θ) = −c1 sin θ + c2 cos θ − sin θ ln  cos θ + θ cos θ. Substitution of y (θ) and y (θ) into the initial conditions yields y (0) = c1 = 1, y (0) = c2 = 2 ⇒ c1 = 1 , c2 = 2 , and so the answer is y (θ) = cos θ + 2 sin θ + cos θ ln  cos θ + θ sin θ. 255 ...
View
Full
Document
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.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

Click to edit the document details