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What Is the Solution of the Following Initial Boundary Value u(x,t)=e16π2tsin(πx4) u(x, t)=e^{-16 \pi^{2} t} \sin \left(\frac{\pi x}{4}\right)

Question 33

Multiple Choice

What is the solution of the following initial boundary value problem?
4  What is the solution of the following initial boundary value problem? 4   =   , u(0, t)  = 0, u(2, t)  = 0, u(x, 0)  = sin(4?x)  A)    u(x, t) =e^{-16 \pi^{2} t} \sin \left(\frac{\pi x}{4}\right)    B)    u(x, t) =e^{-64 \pi^{2} t} \sin (4 \pi t)    C)    u(x, t) =\sum_{n=1}^{\infty} e^{-n^{2} \pi^{2} t} \sin \frac{n \pi x}{4}   D)    u(x, t) =\sum_{n=1}^{\infty} e^{-n^{2} \pi^{2} /} \sin (4 n \pi x)  =  What is the solution of the following initial boundary value problem? 4   =   , u(0, t)  = 0, u(2, t)  = 0, u(x, 0)  = sin(4?x)  A)    u(x, t) =e^{-16 \pi^{2} t} \sin \left(\frac{\pi x}{4}\right)    B)    u(x, t) =e^{-64 \pi^{2} t} \sin (4 \pi t)    C)    u(x, t) =\sum_{n=1}^{\infty} e^{-n^{2} \pi^{2} t} \sin \frac{n \pi x}{4}   D)    u(x, t) =\sum_{n=1}^{\infty} e^{-n^{2} \pi^{2} /} \sin (4 n \pi x)  , u(0, t) = 0, u(2, t) = 0, u(x, 0) = sin(4?x)


A) u(x,t) =e16π2tsin(πx4) u(x, t) =e^{-16 \pi^{2} t} \sin \left(\frac{\pi x}{4}\right)
B) u(x,t) =e64π2tsin(4πt) u(x, t) =e^{-64 \pi^{2} t} \sin (4 \pi t)
C) u(x,t) =n=1en2π2tsinnπx4 u(x, t) =\sum_{n=1}^{\infty} e^{-n^{2} \pi^{2} t} \sin \frac{n \pi x}{4}
D) u(x,t) =n=1en2π2/sin(4nπx) u(x, t) =\sum_{n=1}^{\infty} e^{-n^{2} \pi^{2} /} \sin (4 n \pi x)

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