A) \[nu\]
B) \[n\,F(u)\]
C) \[\frac{n\,F(u)}{{F}'(u)}\]
D) None of these
Correct Answer: C
Solution :
Since \[F(u)\] is homogeneous in \[x,y,z\] of degree n. \[\therefore \] \[x\frac{\partial }{\partial x}(F(u))+y.\frac{\partial }{\partial y}(F(u))+z\frac{\partial }{\partial z}(F(u))=nF(u)\] Þ \[x.{F}'(u)\frac{\partial u}{\partial x}+y{F}'(u)\frac{\partial u}{\partial y}+z{F}'(u)\frac{\partial u}{\partial z}=nF(u)\] Þ \[\frac{{{\partial }^{2}}z}{\partial {{x}^{2}}}={{a}^{2}}{{\sec }^{3}}(y-ax)+{{a}^{2}}\sec (y-ax){{\tan }^{2}}(y-ax)\].
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