# Differential Equation : One question, two methods, both result in different answers

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.

Thanks!

• How did you conclude that $\dfrac{f^2(t)}{1+t^2}\ge0$? May 15, 2017 at 15:23
• @KennyLau Both numerator and denominator are non-negative. May 15, 2017 at 15:25
• 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.
– Moo
May 15, 2017 at 15:27
• Your answer is correct. They forgot to check whether their solution is continuous on $\mathbb R$, or even on $[0,1]$: it isn't. May 15, 2017 at 15:32

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.
• Mathematica tells me that the root is at $-\sqrt{\frac{2}{3 + \sqrt{13}}} + \sqrt{\frac{ 3 + \sqrt{13}}{2}} \approx 0.817732...$ May 15, 2017 at 15:30
• 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..." May 15, 2017 at 15:43
• 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 May 15, 2017 at 23:12