FTC saves the day!

Problem-1 Given

f(x) = \int\limits_{0}^{2x}f(\frac{t}{2})\;dt +\log(2)\quad\quad\quad(\star)

where f(x) is a continuous function, find f(x).

Solution

Let

p=2x \implies \frac{dp}{dx} = 2.\quad\quad\quad(1-1)

By (\star),

\frac{df(x)}{dx} =\frac{d}{dx} \int\limits_{0}^{p} f(\frac{t}{2})\;dt + \frac{d\log(2)}{dx} = \underline{\frac{d}{dp}\left(\int\limits_{0}^{p} f(\frac{t}{2})\;dt\right)} \cdot \frac{dp}{dx}\overset{\textbf{FTC}}{=}\underline{f(\frac{p}{2})}\cdot \frac{dp}{dx}\overset{(1-1)}{=}2f(x),

i.e.,

\frac{df(x)}{dx} = 2f(x).\quad\quad\quad(1-2)

Moreover, we see from (\star) that

f(0) = \int\limits_{0}^{0}f(\frac{t}{2})\;dt + \log(2) = 0 + \log(2) = \log(2).\quad\quad\quad(1-3)

Solving initial-value problem

\begin{cases} \frac{df(x)}{dx} = 2f(x)\\ f(0)=\log(2)\end{cases}

gives

f(x) = \log(2)\cdot e^{2x}.

We use Omega CAS Explorer to verify:

Fig. 1-1


Problem-2 Given

\int\limits_{0}^{1}f(u\cdot x) \;du = \frac{1}{2} f(x) +1\quad\quad\quad(\star\star)

where f(x) is a continuous function, find f(x).

Solution

Let p=u\cdot x,

u=\frac{p}{x} \implies \frac{du}{dp} = \frac{1}{x}\quad\quad\quad(2-1)

u=0\implies p=0; u=1\implies p=x.\quad\quad\quad(2-2)

\int\limits_{0}^{1}f(u\cdot x)\;du\overset{(2)}{=} \int\limits_{0}^{x}f(p)\cdot\frac{du}{dp}\cdot dp\overset{(1)}{=}\int\limits_{0}^{x}f(p)\frac{1}{x}\;dp=\frac{1}{x}\int\limits_{0}^{x}f(p)\;dp.\quad\quad\quad(2-3)

By (2-3), we express (\star\star) as

\frac{1}{x}\int\limits_{0}^{x}f(p)\;dp = \frac{1}{2}f(x)+1,

i.e.,

\int\limits_{0}^{x} f(p)\;dp = \frac{x}{2}f(x)+x.

It follows that

\underline{\frac{d}{dx}\left(\int\limits_{0}^{x}f(p)\;dp\right)}=\frac{d}{dx}\left(\frac{x}{2}f(x)+x\right)\overset{\textbf{FTC}}{\implies}\underline{f(x)}=\frac{1}{2}\left(f(x) + x\frac{d f(x)}{dx}\right)+1.\;(2-4)

Solving differential equation (2-4) (see Fig. 2-1) gives

f(x) = c x + 2.

Fig. 2-1

The solution is verified by Omega CAS Explorer:

Fig. 2-2


Exercise-1 Solving \begin{cases} \frac{df(x)}{dx} = 2f(x)\\ f(0)=\log(2)\end{cases} using a CAS.

Exercise-2 Solving (2-4) without using a CAS.

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