Question
If $ u = e^{a_1x_1 + a_2x_2 + \cdots + a_nx_n} $, where $ a_1^2 + a_2^2 + \cdots + a_n^2 = 1 $, show that $$ \dfrac{\partial ^2u}{\partial x_1^2} + \dfrac{\partial ^2u}{\partial x_2^2} + \cdots + \dfrac{\partial ^2u}{\partial x_n^2} = u $$
Step 1
Step 1: First, we differentiate $u$ with respect to $x_i$ to get $\frac{\partial u}{\partial x_i} = a_i e^{a_1x_1 + a_2x_2 + \cdots + a_nx_n} = a_i u$. Show more…
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