Answer
$ \dfrac{\partial^{2}u}{\partial x_{1}^{2}}+\dfrac{\partial^{2}u}{\partial x_{2}^{2}}+\cdots+\dfrac{\partial^{2}u}{\partial x_{n}^{2}}=u$
Work Step by Step
We are given that $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$ for $u=e^{[a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{n}x_{n}]}$
Here, we have for each $x_{\iota}$,
$\dfrac{\partial}{\partial x_{i}}=a_{\iota} \times e^{[a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{n}x_{n}]}$
Take the second derivative as follows:
$ \dfrac{\partial^{2}}{\partial x_{i}^{2}}=a_{i}^{2} \times e^{[a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{n}x_{n}]}$.
Then, we have
$ \dfrac{\partial^{2}u}{\partial x_{1}^{2}}+\dfrac{\partial^{2}u}{\partial x_{2}^{2}}+\cdots+\dfrac{\partial^{2}u}{\partial x_{n}^{2}}=(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}) \times e^{[a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{n}x_{n}]}=e^{[a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{n}x_{n}]}=u$
Hence, the result has been verified.
