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A function y=f(x) satisfies xf′(x)−2f(x)=x4f2(x), ∀ x>0 and f(1)=−6. Then the value of f′(31/5) is

Answer» A function y=f(x) satisfies xf′(x)−2f(x)=x4f2(x), ∀ x>0 and f(1)=−6. Then the value of f′(31/5) is


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