Question
If ${x^p}{y^q} = {(x + y)^{p + q}}$, then $\frac{{{d^2}y}}{{d{x^2}}} = $

Answer

a
(a) Taking $\log $ and differentiating, we get $x\frac{{dy}}{{dx}} - y = 0$
Differentiating again, we get $\frac{{{d^2}y}}{{d{x^2}}} = 0$.

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