I was wondering if I can get a hint on the following question: Find a function $f:\mathbb{R}\to\mathbb{R}$ that satisfies:

$$2x^3f'(x)f(x) + (f(x))^2=2$$ and $$f(0.5)=1$$ I thought about integrating both sides to find my function but didn't understand where to go from there. We haven't studied differential equations yet so I need to solve it without it. Any hint will be helpful!



$$2x^3𝑓′(𝑥)𝑓(𝑥) + (𝑓(𝑥))^2=2$$

$$\int \frac {𝑓df}{2-f^2}=\int \frac {dx} {2x^3}=\frac {-1}{4x^2}+K$$

Then substitute $u=f^2$

$$\int \frac {du}{u-2}=\frac{1}{2x^2}+K$$


$$f^2=2+Ke^{\frac{1}{2x^2}}$$ $$.......$$

  • $\begingroup$ Thank you! :) I understand now. $\endgroup$ – Eliads Jan 14 '18 at 20:11
  • $\begingroup$ Wait, do I need absolute value for the integral? $\endgroup$ – Eliads Jan 14 '18 at 20:15
  • $\begingroup$ I don't understand how can I emit that absolute value after the integration. $\endgroup$ – Eliads Jan 14 '18 at 20:25
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    $\begingroup$ Yes, I understand that. I just don't understand why did you ignore the absolute after integrating. Thank you for explaining :) $\endgroup$ – Eliads Jan 14 '18 at 20:32

If $g(x) = f(x)^2$, this says $x^3 g'(x) + g(x) = 2$. Now if you can find a function $b(x)$ such that $b'(x)/b(x) = x^{-3}$, you could write this as $$(b(x) g(x))' = 2 x^{-3} b(x)$$ and then integrate the right side. To find $b(x)$, try $b(x) = \exp(c(x))$.

However, your result won't be defined on all of $\mathbb R$.


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