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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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I've solved my own question... Changing $ y = \sqrt{ \frac{2}{3} \ln{(1 + x^3)} + C} $ and $ \frac{dy}{dx} = \frac{x^2}{y + yx^3}$ seems to do the trick. –  Jonathan F. Mar 17 '12 at 19:03
    
Have you tried to differentiate y=23ln(1+x2)+C−−−−−−−−−−−−√ with respect to x? Make an attempt and add more detail to your question. –  Maxood Mar 17 '12 at 19:04
    
your book does have the wrong answer. Whatever answer you get, when you integrate you should back the original $y$ value –  Siddhi V Iyer Mar 18 '12 at 1:00
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2 Answers 2

$y = \sqrt{ \frac{2}{3} \ln{(1 + x^2)} + C}$ 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. –  Jeremy Carlos Mar 18 '12 at 0:51
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