Topic: Engineering Mathematics
Solution of the below PDE
$\frac{\partial u}{\partial t}=4\frac{\partial^{2}u}{\partial x^{2}}$
$u(0,t)=u(\pi,t)=0$,
$u(x,0)=\sin(3x)$
- A. $u(x,t)=e^{-36t}\sin(3x)$
- B. $u(x,t)=-e^{-36t}\sin(3x)$
- C. $u(x,t)=-e^{9t}\sin(3x)$
- D. $u(x,t)=e^{-9t}\sin(3x)$
Answer: A. $u(x,t)=e^{-36t}\sin(3x)$
Explanation: This is a standard 1D heat equation $\frac{\partial u}{\partial t} = \alpha \frac{\partial^2 u}{\partial x^2}$ with $\alpha = 4$. The general solution satisfying the boundary conditions $u(0,t)=u(\pi,t)=0$ is $u(x,t) = \sum B_n e^{-\alpha n^2 t} \sin(nx)$.
Applying the initial condition $u(x,0) = \sin(3x)$, we can see that only the term with $n = 3$ is non-zero, and its coefficient $B_3 = 1$.
Substituting $n=3$ and $\alpha=4$ into the exponential term yields $e^{-4(3)^2 t} = e^{-36t}$.
Thus, the solution is $u(x,t) = e^{-36t}\sin(3x)$.