Question

the value of b for which u=e^bx cos(5y) is harmonic​

Answer 1

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Answer 2

SOLUTION

SOLUTIONConcept:

SOLUTIONConcept:If u(x, y) is harmonic function, then it must satisfy Laplace’s equation uxx + uyy = 0

SOLUTIONConcept:If u(x, y) is harmonic function, then it must satisfy Laplace’s equation uxx + uyy = 0Calculation:

SOLUTIONConcept:If u(x, y) is harmonic function, then it must satisfy Laplace’s equation uxx + uyy = 0Calculation:Given:

SOLUTIONConcept:If u(x, y) is harmonic function, then it must satisfy Laplace’s equation uxx + uyy = 0Calculation:Given:u(x, y) = ebx cos 5y

SOLUTIONConcept:If u(x, y) is harmonic function, then it must satisfy Laplace’s equation uxx + uyy = 0Calculation:Given:u(x, y) = ebx cos 5y∴ ux = bebx cos 5y

SOLUTIONConcept:If u(x, y) is harmonic function, then it must satisfy Laplace’s equation uxx + uyy = 0Calculation:Given:u(x, y) = ebx cos 5y∴ ux = bebx cos 5y∴ uxx = b2ebx cos 5y

SOLUTIONConcept:If u(x, y) is harmonic function, then it must satisfy Laplace’s equation uxx + uyy = 0Calculation:Given:u(x, y) = ebx cos 5y∴ ux = bebx cos 5y∴ uxx = b2ebx cos 5y∴ uy = -5ebx sin 5y

SOLUTIONConcept:If u(x, y) is harmonic function, then it must satisfy Laplace’s equation uxx + uyy = 0Calculation:Given:u(x, y) = ebx cos 5y∴ ux = bebx cos 5y∴ uxx = b2ebx cos 5y∴ uy = -5ebx sin 5y∴ uyy = -25ebx cos 5y

SOLUTIONConcept:If u(x, y) is harmonic function, then it must satisfy Laplace’s equation uxx + uyy = 0Calculation:Given:u(x, y) = ebx cos 5y∴ ux = bebx cos 5y∴ uxx = b2ebx cos 5y∴ uy = -5ebx sin 5y∴ uyy = -25ebx cos 5y∵ for harmonic function, uxx + uyy = 0

SOLUTIONConcept:If u(x, y) is harmonic function, then it must satisfy Laplace’s equation uxx + uyy = 0Calculation:Given:u(x, y) = ebx cos 5y∴ ux = bebx cos 5y∴ uxx = b2ebx cos 5y∴ uy = -5ebx sin 5y∴ uyy = -25ebx cos 5y∵ for harmonic function, uxx + uyy = 0∴ b2 ebx cos 5y – 25ebx cos 5y = 0

SOLUTIONConcept:If u(x, y) is harmonic function, then it must satisfy Laplace’s equation uxx + uyy = 0Calculation:Given:u(x, y) = ebx cos 5y∴ ux = bebx cos 5y∴ uxx = b2ebx cos 5y∴ uy = -5ebx sin 5y∴ uyy = -25ebx cos 5y∵ for harmonic function, uxx + uyy = 0∴ b2 ebx cos 5y – 25ebx cos 5y = 0∴ ebx cos 5y (b2 - 25) = 0

SOLUTIONConcept:If u(x, y) is harmonic function, then it must satisfy Laplace’s equation uxx + uyy = 0Calculation:Given:u(x, y) = ebx cos 5y∴ ux = bebx cos 5y∴ uxx = b2ebx cos 5y∴ uy = -5ebx sin 5y∴ uyy = -25ebx cos 5y∵ for harmonic function, uxx + uyy = 0∴ b2 ebx cos 5y – 25ebx cos 5y = 0∴ ebx cos 5y (b2 - 25) = 0∴ b2 – 25 = 0

SOLUTIONConcept:If u(x, y) is harmonic function, then it must satisfy Laplace’s equation uxx + uyy = 0Calculation:Given:u(x, y) = ebx cos 5y∴ ux = bebx cos 5y∴ uxx = b2ebx cos 5y∴ uy = -5ebx sin 5y∴ uyy = -25ebx cos 5y∵ for harmonic function, uxx + uyy = 0∴ b2 ebx cos 5y – 25ebx cos 5y = 0∴ ebx cos 5y (b2 - 25) = 0∴ b2 – 25 = 0∴ b = ± 5