1. ## Please give me a hand!

this Q

2. Originally Posted by Zachary
this Q

Just substitute the function into the heat equation..

$\displaystyle \frac{\partial u}{\partial t}=-\alpha^2k^2\exp\left(-\alpha^2k^2t\right)\sin\left(kx\right)$

$\displaystyle \frac{\partial^2 u}{\partial x^2}=-k^2\exp\left(-\alpha^2k^2t\right)\sin\left(kt\right)$

It should be evident now that $\displaystyle u\left(x,t\right)=\exp\left(-\alpha^2k^2t\right)\sin\left(kx\right)$ satisfies the heat equation.

Does this help?

3. $\displaystyle u=e^{-\alpha^2k^2t}\sin(kx)$

$\displaystyle u_t=-\alpha^2k^2e^{-\alpha^2k^2t}\sin(kx)$

$\displaystyle u_x=e^{-\alpha^2k^2t}\cos(kx)\cdot k$

$\displaystyle u_{xx}=-e^{-\alpha^2k^2t}\sin(kx)\cdot k^2$

$\displaystyle \alpha^2u_{xx}=\alpha^2(-e^{-\alpha^2k^2t}\sin(kx)\cdot k^2)$

$\displaystyle = -\alpha^2k^2e^{-\alpha^2k^2t}\sin(kx)=u_t\quad \blacksquare$

4. Hi,

And now what ?

5. No wonder i got wrong even though i have tried N-times. Because i thought the equation is like this :

$\displaystyle u=e^{-\alpha^2k^2t\sin(kx)}$