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The given equation is: $$x^2\left(f^{\prime \prime}(x)+\left(f^{\prime}(x)\right)^2\right)=1$$ The full question actually requires you to find f(x) and then use it to evaluate $$\int_1^5 e^{f(x)} dx$$ However I am stuck on finding f(x) first.

First I attempted to guess what it could be using the boundaries given but that of course failed, so I am trying to get f(x) by manipulating the given equation by integrating both sides. Here is what I tried: $$ \begin{aligned} & x^{2}\left(f^{\prime \prime}(x)+\left(f^{\prime}(x)\right)^{2}\right)=1 . \\ & f^{\prime \prime}(x)+\left(f^{\prime}(x)\right)^{2}=\frac{1}{x^{2}} \end{aligned} $$

Integrate both sides:

$$ \int f^{\prime \prime}(x) d x+\int\left(f^{\prime}(x)\right)^{2} d x=\int \frac{1}{x^{2}} $$

let $u=f^{\prime}(x)$

$$ d u=f(x) d x \Rightarrow \frac{d u}{f(x)}=d x $$

then,

$$ \int \frac{u^{\prime}}{f(x)} d u+\int \frac{u^{2}}{f(x)} d u=\int \frac{1}{x^{2}} $$ Reaching here I am understanding that I am headed in the wrong direction however I tried a few other ways using the equation, but I cant seem to get it in a form that is more manageable. All help and hints are appreciated, Thank You.

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2 Answers 2

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Hint. Let $u(x):=e^{f(x)}$; then $u'=f'e^f$ and $u''=(f''+f'^2)e^f=\frac{u}{x^2},$ which implies that $u$ satisfies the Cauchy-Euler equation $$ x^2u''-u=0. \tag{1} $$

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    $\begingroup$ ...with the initial conditions $u(1) = 1$, $u(5) = 4$, $u'(1) = 2$, and $u'(5) = -8$. $\endgroup$
    – Dan
    Feb 23 at 20:19
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    $\begingroup$ Well noticed, @Dan. And this begs the question: since $u$ satisfies a second order ODE, are these four conditions consistent? $\endgroup$
    – Gonçalo
    Feb 23 at 21:08
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Since we have an $\ln$ in the initial conditions, assume that $f(x) = \ln g(x)$. Then $f'(x) = \frac{g'(x)}{g(x)}$ and $f''(x) = \frac{g''(x)g(x) - g'(x)^2}{g(x)^2}$. Plugging this into the given differential equation gives:

$$x^2\left(\frac{g''(x)g(x) - g'(x)^2}{g(x)^2}+\left(\frac{g'(x)}{g(x)}\right)^2\right)=1$$

which simplifies to

$$x^2 g''(x)=g(x)$$

As @Gonçalo has already pointed out, this is a second-order Cauchy-Euler equation.

We also need to reframe the initial conditions in terms of $g$ instead of $f$. Since $g(x) = e^{f(x)}$, and $g'(x) = f'(x)g(x) = f'(x)e^{f(x)}$. So,

$$f(1) = 0 \implies g(1) = e^0 = 1$$ $$f(5) = \ln 4 \implies g(5) = e^{\ln 4} = 4$$ $$f'(1) = 2 \implies g'(1) = 2g(1) = 2$$ $$f'(5) = -2 \implies g'(5) = -2g(5) = -8$$


Assume that $g(x) = ax^b + cx^d$. Then $g'(x) = abx^{b-1} + cdx^{d-1}$, and $g''(x) = ab(b-1)x^{b-2} + cd(d-1)x^{d-2}$. The differential equation then becomes:

$$ab(b-1)x^{b} + cd(d-1)x^{d} = ax^b + cx^d$$

For the coefficients to match up, we need $b(b-1) = d(d-1) = 1$, or equivalently, $b^2 - b - 1 = d^2 - d - 1 = 0$. By the quadratic formula,

$$\{b, d\} = \frac{1 \pm \sqrt{5}}{2}$$

So let $b = \frac{1 + \sqrt{5}}{2} \approx 1.618034$ and $d = \frac{1 - \sqrt{5}}{2} \approx -0.618034$. Thus,

$$g(x) = ax^{(1 + \sqrt{5})/2} + cx^{(1 - \sqrt{5})/2}$$ $$g'(x) = a\frac{1 + \sqrt{5}}{2}x^{(-1 + \sqrt{5})/2} + c\frac{1 - \sqrt{5}}{2}x^{(-1 - \sqrt{5})/2}$$

At $x = 1$, we have:

$$g(1) = a + c = 1$$ $$g'(1) = a\frac{1 + \sqrt{5}}{2} + c\frac{1 - \sqrt{5}}{2} = 2$$

Solving the first of these gives $c = 1 - a$. Plugging into the second gives:

$$a\frac{1 + \sqrt{5}}{2} + (1-a)\frac{1 - \sqrt{5}}{2} = 2$$

Solving for $a$ gives:

$$a = \frac{5 + 3\sqrt{5}}{10}$$ $$c = \frac{5 - 3\sqrt{5}}{10}$$

Thus,

$$g(x) = \frac{5 + 3\sqrt{5}}{10}x^{(1 + \sqrt{5})/2} + \frac{5 - 3\sqrt{5}}{10}x^{(1 - \sqrt{5})/2}$$

However, if you plug in $x = 5$, you get $g(5) \approx 15.765606$ and $g'(5) \approx 5.130111$, which do not match the given initial conditions. Not sure whether I made a math error, or if the original problem is misstated.

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    $\begingroup$ You made no mistake. You proved that the problem has no solution. $\endgroup$
    – Gonçalo
    Feb 28 at 23:09

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