Questions mathématiques diverses

SigMathS

Question 42:
Déterminer

f(x)

en fonction de

x

sachant que :

f(x)=x+\int_{0}^{\ln 2}{f(y)e^x(e^y+e^{-y})dy}

Déterminer $f(x)$ en fonction de $x$ sachant que :

$f(x)=x+\int_{0}^{\ln 2}{f(y)e^x(e^y+e^{-y})dy}$

SigMathS

Réponse 42:

\begin{align*} f(x)&=x+\int_{0}^{ln2}{f(y)e^x(e^y+e^{-y})dy}\\ &=x+e^x\int_{0}^{ln2}{f(y)(e^y+e^{-y})dy}\\ &=x+e^x(a+b) \\ \end{align*}

avec

a=\int_{0}^{ln2}{f(y)e^y\,dy}

b=\int_{0}^{ln2}{f(y)e^{-y}\,dy}

\begin{align*} a&=\int_{0}^{ln2}{f(y)e^y dy}\\ &=\int_{0}^{ln2}{(y+e^y(a+b))e^ydy}\\ &=\int_{0}^{ln2}{ye^ydy}+(a+b)\int_{0}^{ln2}{e^{2y}dy}\\ &=\left[{ye^y}\right]_0^{ln2}-\int_{0}^{ln2}{e^ydy}+(a+b)\left[{\dfrac{1}{2}e^{2y}}\right]_0^{ln2}\\ &=2\ln 2-1+\dfrac{3}{2}(a+b) \qquad\qquad (i)\\ b&=\int_{0}^{ln2}{f(y)e^{-y} dy}\\ &=\int_{0}^{ln2}{(y+e^y(a+b))e^{-y}dy}\\ &=\int_{0}^{ln2}{ye^{-y}dy}+(a+b)\int_{0}^{ln2}{dy}\\ &=\left[{-ye^{-y}}\right]_{0}^{ln2}+\int_{0}^{ln2}{e^{-y}dy}+(a+b)ln2\\ &=(a+b)\ln 2+\dfrac{1-\ln 2}{2} \qquad\qquad (ii) \end{align*}

(i)

(ii)\Longrightarrow

\left\{{\begin{aligned}&{a+3b=2-4\ln 2}\\&{2a\ln 2+2b(\ln 2-1=\ln 2-1)}\end{aligned}}\right.

\iff \left\{{\begin{aligned}&{(a+b)+2b=2-4\ln 2}\\&{2(a+b)\ln 2-2b=\ln 2-1}\end{aligned}}\right.

Donc

a+b=\dfrac{1-3\ln 2}{1+2\ln 2}

Alors

\boxed{f(x)=x+\dfrac{1-3\ln 2}{1+2\ln 2}e^x}