Let xᵏ + yᵏ = aᵏ, (a, k > 0) and dy/dx + (y/x)¹/³ = 0, then k is:
Let xᵏ + yᵏ = aᵏ, (a, k > 0) and dy/dx + (y/x)¹/³ = 0, then k is:
Option 1 -
3/2
Option 2 -
4/3
Option 3 -
1/3
Option 4 -
2/3
Asked by Shiksha User
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1 Answer
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Correct Option - 4
Detailed Solution:Given kx^ (k-1) + k * y^ (k-1) * dy/dx = 0.
dy/dx = - (kx^ (k-1) / (ky^ (k-1) = - (x/y)^ (k-1).
The provided solution has dy/dx + (x/y)^ (k-1) = 0.
It seems to relate to k-1 = -1/3, which implies k = 1 - 1/3 = 2/3.
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