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$$\int f(x) dx\cdot \int \frac{1}{f(x)}dx = c$$ where $c$ is a constant. Find $f(x)$.

Off first glance it seems that $f(x)$ is some form of $e^x$, but how does one go about doing this analytically (only 1 mark for guessing). I took the derivative of this but the resulting function was still ugly. I thought maybe treating it as a first order linear differential equation would work, but I wasn't sure how to put it in the form $y' + P(x)y = Q(x)$

bigfocalchord
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  • Are you assumimg $f>0$? – Math Lover Dec 14 '16 at 07:31
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    Let $F(x)$ be the antiderivative with the choice of constant of integration appearing in the equation. Then dividing both sides by $F(x)$ and differentiating, you obtain $-\frac{1}{c} = \frac{F'(x)^2}{F(x)^2}$. Can you proceed from here? – Sangchul Lee Dec 14 '16 at 07:33

2 Answers2

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Let $F(x)$ be the antiderivative of $f(x)$.

$$\int\frac1{F'(x)}dx=\frac c{F(x)},$$

$$\frac1{F'(x)}=-\frac{cF'(x)}{F^2(x)}.$$

Then

$$\left(F'(x)+\frac1{\sqrt{-c}}F(x)\right)\left(\,F'(x)-\frac1{\sqrt{-c}}F(x)\right)=0,$$

has the desired form (and of course the solutions are exponential). There are no real solutions for $c\ge0$.

  • (Independently of @Sangchul Lee's comment.) –  Dec 14 '16 at 08:13
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    The two solutions implied here may also be glued piece-wise. For instance, with $c=-1$, you could have $f(x)=\begin{cases}e^x&x<0\e^{-x}&x\geq0\end{cases}$. – 2'5 9'2 Dec 14 '16 at 08:47
  • @alex.jordan Mh, on second thoughts, this doesn't work, because $F$ would need to change sign at the same time, and be discontinuous. Then its derivative, $f$, would be undefined (though the left and right derivatives would coincide). If that was true, the function could freely alternate between the two exponentials, a pretty pathological situation. –  Dec 14 '16 at 09:26
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It is given that $$\int\frac{1}{f(x)}dx\cdot \int f(x)dx =c ...(1)$$

Now differentiating both side w.r.t $x$, We Get

$$\int \frac{1}{f(x)}dx\cdot f(x)+\frac{1}{f(x)}\cdot \int f(x)dx =0... (2)$$

Now from equation $(1)$, We get $$ \int\frac{1}{f(x)}dx = \frac{c}{\int f(x)dx}$$

and put into equation $(2)$, We get $$\frac{c\cdot f(x)}{\int f(x)dx}+\frac{\int f(x)dx}{f(x)} = 0$$

So $$ \left(\int f(x)dx\right)^2 = -c\cdot \left(f(x)\right)^2$$

Now Let $-c=\alpha^2\;,$ Then $$\left(\int f(x)dx\right)^2 = \alpha^2\cdot \left(f(x)\right)^2$$

So $$\int f(x)dx = \pm \alpha\cdot f(x)\;,$$ Differentiate both side w.r.t $x\;,$ we get

$$\Rightarrow f(x) = \pm \alpha\cdot f'(x)\Rightarrow \frac{f'(x)}{f(x)}=\pm k\;,$$ where $k = \pm \frac{1}{\alpha}$.

Now Integrate both side w.r.t $x\;,$ we get

$$\Rightarrow \int \frac{f'(x)}{f(x)} = \pm k\int dx\Rightarrow \ln\left|f(x)\right| = \pm k+\ln \left|c\right|\Rightarrow f(x)=c\cdot e^{\pm kx}$$ Hope it helps.