Solved

In the Previous Problem, Separate Variables Using u(r,θ)=R(r)Θ(θ)u ( r , \theta ) = R ( r ) \Theta ( \theta )

Question 5

Multiple Choice

In the previous problem, separate variables using u(r,θ) =R(r) Θ(θ) u ( r , \theta ) = R ( r ) \Theta ( \theta ) . The resulting problems are


A) r2R+rRλR=0,ΘλΘ=0,Θ(θ) =Θ(θ+2π) r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } - \lambda R = 0 , \Theta ^ { \prime \prime } - \lambda \Theta = 0 , \Theta ( \theta ) = \Theta ( \theta + 2 \pi )
B) r2R+rR+λR=0,Θ+λΘ=0,Θ(θ) =Θ(θ+2π) r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } + \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( \theta ) = \Theta ( \theta + 2 \pi )
C) r2R+rRλR=0,Θ+λΘ=0,Θ(θ) =Θ(θ+2π) r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } - \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( \theta ) = \Theta ( \theta + 2 \pi )
D) r2R+rRλR=0,Θ+λΘ=0,Θ(θ) =Θ(θ+π) r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } - \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( \theta ) = \Theta ( \theta + \pi )
E) r2R+rR+λR=0,Θ+λΘ=0,Θ(θ) =Θ(θ+π) r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } + \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( \theta ) = \Theta ( \theta + \pi )

Correct Answer:

verifed

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Related Questions