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
$\dfrac{\partial u }{\partial x_{i}}=a_{\iota} \times e^{[a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{n}x_{n}]}$ ...(1)
and
$ \dfrac{\partial^{2} u}{\partial x_{i}^{2}}=a_{i}^{2} \times e^{[a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{n}x_{n}]}$ ...(2)
From, the equations (1) and (2), 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}]}$
and $ \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}}=e^{[a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{n}x_{n}]}=u$
Hence, the result has been verified that $ \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$
