Let xk + yk = ak, (a, k 0 ) and [dy dx + ( y x )^1 3 = 0], then k is:

Question

Let xk + yk = ak, (a, k > 0 ) and $${{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0$$, then k is:

Accepted Answer

xk + yk = ak

$$ \Rightarrow $$ kxk - 1 + kyk - 1$${{{dy} \over {dx}}}$$ = 0

$$ \Rightarrow $$ $${{{dy} \over {dx}} + {{\left( {{x \over y}} \right)}^{k - 1}}}$$ = 0 ...(1)

Given $${{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0$$ ...(2)

Comparing (1) and (2), we get

k - 1 = $$ - {1 \over 3}$$

$$ \Rightarrow $$ k = $${2 \over 3}$$

Solutions · 1