The solution of the differential equation (x-1) d y+y d x=x(x-1) y^ 1 3 d x , is

Mathematics · JEE Main · NTA Exams — Differential Equations

1. \(y^{2 / 3}=C(x-1)^{-2 / 3}+\frac{1}{4}(x-1)^{2}+\frac{2}{5}(x-1)\)
2. \(y^{1 / 3}=C(x-1)^{-2 / 3}+\frac{1}{4}(x-1)^{2}+\frac{2}{5}(x-1)\)
3. \(y^{2 / 3}=C(x-1)^{2 / 3}+\frac{1}{4}(x-1)^{2}+\frac{2}{5}(x-1)\)
4. None of these

Answer

(A) y^ 2 / 3 =C(x-1)^ -2 / 3 + 1 4 (x-1)^ 2 + 2 5 (x-1)

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