7. Solve the following initial value problem using the method of variables separable: $xy' = \frac{y^3 + 1}{3y^3}$ ; $y(1) = 3$. 7.
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We can do this by moving all the terms involving y and its derivatives to one side and all the terms involving x to the other side: 1 + 3y" J(-3x) = 0 Show more…
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