Joy of Mathematics

Differential equation (II)

May 4, 2021May 10, 2021 Tolaso

Let $f:(0, +\infty) \rightarrow \mathbb{R}$ be a differentiable function such that $f(1) = \frac{1}{2}$ and

$f'(x) + e^{f(x)} = x + \frac{1}{x} \quad \text{for all} \;\; x >0$

Find an explicit formula for $f$ .

Solution

Let us consider the function $g(x) = f(x) - \ln x - \frac{x^2}{2}$ which is clearly differentiable. Hence,

$\begin{align*} f'(x) + e^{f(x)} = x + \frac{1}{x} &\Leftrightarrow \left ( g(x) + \ln x + \frac{x^2}{2} \right )' + e^{g(x) + \ln x + \frac{x^2}{2}} = x + \frac{1}{x} \\ &\Leftrightarrow g'(x) + \frac{1}{x} + x + x e^{g(x)} e^{x^2/2}= x + \frac{1}{x} \\ &\Leftrightarrow g'(x) + x e^{g(x)} e^{x^2/2} =0 \\ &\Leftrightarrow g'(x) = - x e^{g(x)} e^{x^2/2} \\ &\Leftrightarrow e^{-g(x)} g'(x) = -x e^{x^2/2} \\ &\Leftrightarrow -e^{-g(x)} g'(x) = x e^{x^2/2} \\ &\Leftrightarrow \left ( e^{-g(x)} \right ) ' = \left ( e^{x^2/2} \right )' \\ &\Rightarrow e^{-g(x)} = e^{x^2/2} + c \\ &\!\!\!\!\!\!\!\!\!\! \overset{x=1 \Rightarrow c =1-\sqrt{e} }{=\! =\! =\! =\! =\! =\! =\! \Rightarrow} e^{-g(x)} = e^{x^2/2} + 1 - \sqrt{e} \\ &\Rightarrow g(x) = - \ln \left ( e^{x^2/2} + 1 - \sqrt{e} \right ) \end{align*}$

Thus,

$f(x) = \ln x + \frac{x^2}{2} - \ln \left ( e^{x^2/2} + 1 - \sqrt{e} \right )$

Differential equation (I)

May 4, 2021 Tolaso

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f(0)=1$ and

$f'(x) \sqrt{x^2+1} = f(x) \quad \text{for all} \;\; x \in \mathbb{R}$

Find an explicit formula for $f$ .

Solution

We have successively

$\begin{align*} f'(x)\sqrt{x^2+1}=f(x) &\Leftrightarrow f'(x)\sqrt{x^2+1}-f(x)=0 \\ &\Leftrightarrow f'(x)\sqrt{x^2+1}-xf'(x)+\frac{xf(x)}{\sqrt{x^2+1}}-f(x)=0 \\ &\Leftrightarrow f'(x)\left ( \sqrt{x^2+1}-x \right )+f(x)\left ( \frac{x}{\sqrt{x^2+1}}-1 \right )=0 \\ &\Rightarrow \left [ f(x)\left ( \sqrt{x^2+1}-x \right ) \right ]'=(c)'\\ &\!\!\!\!\!\!\!\!\!\!\!\!\overset{f(0)=1\Rightarrow c=1}{=\!\!=\!=\!=\!=\!=\!=\!=\! \Rightarrow}f(x)=x+\sqrt{x^2+1}, \; x \in \mathbb{R} \end{align*}$

which satisfies the given conditions.

Differential equation

April 4, 2021April 4, 2021 Tolaso

Let $f:(0, +\infty) \rightarrow \mathbb{R}$ be a twice differentiable function such that

$x^2 f''(x) + x f'(x) = x - 2 \quad \text{for all} \;\; x>0$

If $f(1) =0$ , $f'(1)=1$ find an explicit formula of $f$ .

Solution

We have successively

$\begin{align*} x^2 f''(x) + x f'(x) = x - 2 &\Rightarrow x f''(x) + f'(x) = 1 - \frac{2}{x} \\ &\Rightarrow \left ( x f'(x) \right )' = \left ( x - 2 \ln x \right )' \\ &\Rightarrow x f'(x) = x - 2 \ln x + c_1 \\ &\!\!\!\!\!\!\!\!\!\!\!\overset{x=1 \Rightarrow c_1=0}{=\! =\! =\!=\! =\! =\!\Rightarrow } x f'(x) = x - 2 \ln x \\ &\Rightarrow f'(x) = 1 - \frac{2 \ln x}{x} \\ &\Rightarrow \left ( f(x) \right ) ' = \left ( x - \ln^2 x \right )' \\ &\Rightarrow f(x) = x - \ln^2 x + c \\ &\!\!\!\!\!\!\!\!\!\!\!\overset{x=1 \Rightarrow c=-1}{=\! =\! =\!=\! =\! =\!\Rightarrow } f(x) = x - \ln^2 x - 1 \end{align*}$