Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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

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

share|cite|improve this answer


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 $x^{2}$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.