A continuous function $f:\mathbb R \to \mathbb R$ satisfying the differential equation $$f(x)=\left(1+x^2\right)\left(1+\int_{0}^{x} \frac{f^2(t)}{1+t^2} dt\right)$$ Then find the value of $f(1)$. The options are:

$a) -6$

$b) -4$

$c) -2$

$d) ~\text{None}$

Now, since this question is MCQ type, I just rejected options as follows:

$$f(1)=2\left(1+\int_{0}^{1} \frac{f^2(t)}{1+t^2} dt\right)$$ since $$\frac{f^2(t)}{1+t^2}\ge0$$ Hence $f(1) \gt 0$

Thereby rejecting options $a, b$, and $c$, I marked option $d$.

But to my surprise, the answer given was option $a$, i.e., $-6$

Their solution goes as follows:

$$\frac{f(x)}{ 1+x^2 }= 1+\int_{0}^{x} \frac{f^2(t)}{1+t^2} dt \implies \frac{dy}{dx}=\left( \frac{2x}{1+x^2}\right)y+y^2$$

$ \text{Let}~~ \dfrac{-1}{y}=t$

$$\therefore ~~f(x)=\frac{-3(1+x^2)}{x^3+3x-3}~ \text{(How? I don't know!)}~ \implies f(1)=-6$$

Can someone please explain what is going on here? (Which method is wrong and why?) Most probably they are doing something wrong. I know that this is probably poor framing of question, but both methods seem correct to me.


  • $\begingroup$ How did you conclude that $\dfrac{f^2(t)}{1+t^2}\ge0$? $\endgroup$
    – Kenny Lau
    May 15, 2017 at 15:23
  • 1
    $\begingroup$ @KennyLau Both numerator and denominator are non-negative. $\endgroup$ May 15, 2017 at 15:25
  • $\begingroup$ When you let $t(x) = -1/y$, you substitute into the ODE and you are left with a linear equation that can be solved using an integrating factor and arrived at the shown result. $\endgroup$
    – Moo
    May 15, 2017 at 15:27
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    $\begingroup$ Your answer is correct. They forgot to check whether their solution is continuous on $\mathbb R$, or even on $[0,1]$: it isn't. $\endgroup$ May 15, 2017 at 15:32

1 Answer 1


In the strict sense, as $1$ does not belong to the maximal domain of the solution through $(0,1)$, there is no function value $f(1)$.

The denominator $x^3+3x-3$ of the formal solution has value $-3$ at $x=0$ and value $1$ at $x=1$, thus a root inside that interval, so that the function itself has a pole in the interval. The solution of the ODE resp. integral equation ends at that pole.

For the solution I would substitute $g(x)=\frac{f(x)}{1+x^2}$ so that $$ g(x)=1+\int_0^x(1+t^2)g(t)^2dt $$ which is equivalent to the differential equation IVP $$ g'(x)=(1+x^2)g(x)^2,\;g(0)=1 $$ with solution $$ -\frac1{g(x)}+\frac1{g(0)}=x+\frac13x^3 $$ which seems a little more direct than the proposed solution.

  • $\begingroup$ Mathematica tells me that the root is at $-\sqrt[3]{\frac{2}{3 + \sqrt{13}}} + \sqrt[3]{\frac{ 3 + \sqrt{13}}{2}} \approx 0.817732...$ $\endgroup$ May 15, 2017 at 15:30
  • $\begingroup$ So the answer provided is wrong? $\endgroup$
    – Kenny Lau
    May 15, 2017 at 15:30
  • $\begingroup$ Already the question is wrong.The answer of "None" is right, but the reason should be formulated a bit more contradictory, "if there were a continuous (differentiable) real solution $f$ on $[0,1]$, then because of the positive integrand..." $\endgroup$ May 15, 2017 at 15:43
  • $\begingroup$ I want to confirm that finally, this question itself is wrong or incomplete.Right? $\endgroup$ May 15, 2017 at 16:02
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    $\begingroup$ All the answers are correct! Beginning with a false premise (let $f$ be a continuous function satisfying...), you can prove anything. en.wikipedia.org/wiki/Principle_of_explosion $\endgroup$ May 15, 2017 at 23:12

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