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This is a question from my calculus practice material.

$f(x)$ is defined for $x\ge0$ and has a continuous derivative. It satisfies $f(0)=1$, $f’(0)=0$ and $(1+f(x))f’’(x)=1+x$. Determine all possible functions $f(x)$.

I tried integrating both sides directly, but failed as I couldn’t figure out how to integrate $\left(f’(x)\right)^2$ since there is no term of $f’’(x)$ in the integrand.

$$\int_0^tf’’(x)\,\mathrm dx+f(x)f’(x)\bigg|_0^t-\int_0^t(f’(x))^2\,\mathrm dx=\int_0^t1+x\,\mathrm dx$$ $$\implies f’(t)+f(t)f’(t)-\int_0^t(f’(x))^2\mathrm dx=t+\frac{t^2}2$$

I am not able to think of any other approach to tackle this ODE. Any help is highly appreciated.

Rócherz
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Poonguzhali Annadurai
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    Try a solution $f(x) = \sum_{n=0}^\infty a_n x^n$ near 0 perhaps. You already have $a_0=1$ and $a_1=0$. – Gribouillis Mar 15 '25 at 14:01
  • By the way, Deepseek spent almost 20 minutes thinking about it and finally only gave a substitution $v(x)=1+f(x)$ and claimed(without proof) that $v(x)=\left(\frac{x^3+3x^2+32}{4}\right)^{1/3}.$ – Bowei Tang Mar 15 '25 at 14:22
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    @lurk Deepseek is wrong then (with no surprise). – NN2 Mar 15 '25 at 14:44
  • Maple can reduce the equation to a first-order one, but doesn't produce an explicit form. It's straightforward to compute the first terms of a series solution, $f(x) = 1 + \frac14 x^2 + \frac12 x^3 + \cdots$. – Travis Willse Mar 15 '25 at 16:14
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    There is the particular solution $f(x)=\frac43 (1+x)^{3/2}-1$, though it does not solve your IVP. – Eli Bartlett Mar 15 '25 at 18:08
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    Altough it doesn't answer the question, here are two other posts about the same ODE in case they may be useful:

    https://math.stackexchange.com/questions/3705582/differential-equation-monotonicity1fxfx-1x

    https://math.stackexchange.com/questions/3908368/a-function-satisfying-fx1fx-1x?noredirect=1&lq=1

     – stowo Mar 15 '25 at 20:11
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    @EliBartlett The coefficient should be $\frac2{\sqrt 3}$, I think. – Travis Willse Mar 15 '25 at 20:29
  • @TravisWillse You are correct. And more generally $\pm 2/\sqrt{3}$. – Eli Bartlett Mar 15 '25 at 21:07
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    If $f$ has a continuous derivative it's holomorphic thus having a series expansion. Your could differentiale both sides of the equation $n$ times using the general Leibniz rule to get $\sum\limits_{k = 0}^{n}\left( \binom{n}{k} \cdot ( \delta_{k,, 0} + f^{( n - k )}( x ) ) \cdot f^{( k + 2 )}( x ) \right) = \delta_{k,, 0} \cdot ( 1 + x ) + \delta_{k,, 1}$ and set $x = 0$, such that you get a recursive equation of the coefficients of the series. – The Art Of Repetition Mar 16 '25 at 06:04
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    Your ODE is equivalent to $y'' = x/y$, with $y(x) = 1 + f(x-1)$, which turns out to be an Emden-Fowler equation. Unfortunately, this precise case doesn't seem to be solvable (in closed form), apart from the particular monomial solution given by Eli Bartlett in his comment, that is why you shall be left with Taylor series, as mentioned by TheArtOfRepetition. Finally, it is to be noted that the considered equation may be reduced to a first-order ODE, namely an Abel differential equation of the second kind, with the help of a wise change of variables. See Polyanin & Zaitsev for further details. – Abezhiko Mar 16 '25 at 12:23
  • @Travis Willse There must be a typo, it should be $f(x) = 1 + \frac14 x^2 + \frac1{12} x^3 + \cdots$. –  Mar 25 '25 at 12:17
  • Yes, there was a typo, thanks. – Travis Willse Mar 25 '25 at 13:02
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    $y’’=\frac{ax+c}y+b$ is solved here – Тyma Gaidash Mar 25 '25 at 17:01

1 Answers1

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While closed form solutions are rare, Taylor series are a very useful tool for such problems. We can rewrite $y''(x)(1+y(x))=1+x, y(0)=1, y'(0)=0$ in integral form, with z=y': $$y(x)=1+\int^x_0z(x)\,dx\tag1$$ $$z(x)=\int^x_0\frac{1+x}{1+y(x)}\,dx.\tag2$$ If we have $k$ terms of the series for $y$, we can compute $k$ terms of the integrand in (2), and the integration gives $k+1$. From this, (1) gives $k+2$ terms of $y$, and so on. This is tedious and error-prone by hand, but simple with Pari/GP, say:

(08:09) gp > y=1+O(x)
%2 = 1 + O(x)
(13:04) gp > z=intformal((1+x)/(1+y))
%3 = 1/2*x + O(x^2)
(13:05) gp > y=1+intformal(z)
%4 = 1 + 1/4*x^2 + O(x^3)
(13:05) gp > z=intformal((1+x)/(1+y))
%5 = 1/2*x + 1/4*x^2 - 1/48*x^3 + O(x^4)
(13:05) gp > z=intformal((1+x)/(1+y))
%6 = 1/2*x + 1/4*x^2 - 1/48*x^3 + O(x^4)
(13:05) gp > y=1+intformal(z)
%7 = 1 + 1/4*x^2 + 1/12*x^3 - 1/192*x^4 + O(x^5)
(13:06) gp > z=intformal((1+x)/(1+y))
%8 = 1/2*x + 1/4*x^2 - 1/48*x^3 - 1/48*x^4 - 3/1280*x^5 + O(x^6)
(13:06) gp > y=1+intformal(z)
%9 = 1 + 1/4*x^2 + 1/12*x^3 - 1/192*x^4 - 1/240*x^5 - 1/2560*x^6 + O(x^7)
(13:06) gp > z=intformal((1+x)/(1+y))
%10 = 1/2*x + 1/4*x^2 - 1/48*x^3 - 1/48*x^4 - 3/1280*x^5 + 59/23040*x^6 + 109/129024*x^7 + O(x^8)
(13:07) gp > y=1+intformal(z)
%11 = 1 + 1/4*x^2 + 1/12*x^3 - 1/192*x^4 - 1/240*x^5 - 1/2560*x^6 + 59/161280*x^7 + 109/1032192*x^8 + O(x^9)

With those terms, we'd have $y(1)=759257/573440=1.32403913\ldots$, while a more precise value would be $y(1)=1.32399809566\ldots$. We don't know anything about the radius of convergence in advance, but we can calculate $y(x_0)$ and $z(x_0)$ for some reasonably small value of $x_0$, and repeat the same process to obtain a Taylor series around $x_0$.