## Calculus 8th Edition

$\dfrac{\partial^{2}u}{\partial x_{1}^{2}}+\dfrac{\partial^{2}u}{\partial x_{2}^{2}}+\cdots+\dfrac{\partial^{2}u}{\partial x_{n}^{2}}=u$
$\dfrac{\partial u }{\partial x_{i}}=a_{\iota} \times e^{[a_{1}x_{1}+a_{2}x_{2}+\cdots+a_{n}x_{n}]}$ ...(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 $\dfrac{\partial^{2}u}{\partial x_{1}^{2}}+\dfrac{\partial^{2}u}{\partial x_{2}^{2}}+\cdots+\dfrac{\partial^{2}u}{\partial x_{n}^{2}}=u$