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The author of my textbook asks to verify that the function:

$ y = \sqrt{ \frac{2}{3} \ln{(1 + x^2)} + C} $

solves the differential equation

$ \frac{dy}{dx} = \frac{x^3}{y + yx^3}$

However, this is an error and this $y$ does not solve the differential equation. Is there a simple typo that makes the problem workable?

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    your book does have the wrong answer. Whatever answer you get, when you integrate you should back the original $y$ value2012-03-18

2 Answers 2

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$y = \sqrt{ \frac{2}{3} \ln{(1 + x^2)} + C}$ square both sides and differentiate and you get \displaystyle{2yy' = \frac{\frac{2}{3} \times 2x}{1+x^2}}

\begin{align*} yy' &= \frac{2}{3(1+x^2)}\\ \Rightarrow y' &= \frac{2}{3y(1+x^2)}\\ &= \frac{2}{3y+3yx^2} \end{align*}

(Checked with the answer above to be correct by Wolfram here )

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Hint:

Differentiating $y = \sqrt{ \frac{2}{3} \ln{(1 + x^2)} + C}$ With respect to x should lead you to

y'(x) = \frac{2x} {{\sqrt {6} \ {(x^2+1)}}\ \sqrt{ \ln{(1 + x^2)}}}

That should help you for $\frac{dy}{dx} = \frac{x^3}{y + yx^3}$

Edit: just saw your comment how the original had an $x^{3}$ term rather than $x^{2}$.

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    This is not correct answer, technically if you integrate both sides you should get back the $y$ value.2012-03-18