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) $

Question

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) $

Quizplus · Accepted Answer

The Answer of 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 u(x,t)=e−16π2tsin⁡(πx4) u(x, t)=e^{-16 \pi^{2} t} \sin \left(\frac{\pi x}{4}\right) u(x,t)=e−16π2tsin(4πx​)

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