Question Number 163119 by tounghoungko last updated on 04/Jan/22
$$\:\:{f}\:'\left({x}\right)=\:{f}\left({x}\right)+\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\: \\ $$$$\:{f}\left(\mathrm{0}\right)=\mathrm{1}\:\Rightarrow{f}\left({x}\right)=? \\ $$
Answered by mr W last updated on 04/Jan/22
![y′=y+k with k=∫_0 ^1 f(x)dx (dy/(y+k))=dx ∫(dy/(y+k))=∫dx ln (y+k)=x+C_1 y+k=Ce^x y=f(x)=Ce^x −k ∫_0 ^1 f(x)dx=[Ce^x −kx]_0 ^1 =C(e−1)−k=^! k ⇒C=((2k)/(e−1)) f(x)=(((2e^x )/(e−1))−1)k f(0)=((2/(e−1))−1)k=^! 1 ⇒k=((e−1)/(3−e)) ⇒f(x)=(((e−1)/(3−e)))(((2e^x )/(e−1))−1)](https://www.tinkutara.com/question/Q163121.png)
$${y}'={y}+{k}\:{with}\:{k}=\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx} \\ $$$$\frac{{dy}}{{y}+{k}}={dx} \\ $$$$\int\frac{{dy}}{{y}+{k}}=\int{dx} \\ $$$$\mathrm{ln}\:\left({y}+{k}\right)={x}+{C}_{\mathrm{1}} \\ $$$${y}+{k}={Ce}^{{x}} \\ $$$${y}={f}\left({x}\right)={Ce}^{{x}} −{k} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}=\left[{Ce}^{{x}} −{kx}\right]_{\mathrm{0}} ^{\mathrm{1}} ={C}\left({e}−\mathrm{1}\right)−{k}\overset{!} {=}{k} \\ $$$$\Rightarrow{C}=\frac{\mathrm{2}{k}}{{e}−\mathrm{1}} \\ $$$${f}\left({x}\right)=\left(\frac{\mathrm{2}{e}^{{x}} }{{e}−\mathrm{1}}−\mathrm{1}\right){k} \\ $$$${f}\left(\mathrm{0}\right)=\left(\frac{\mathrm{2}}{{e}−\mathrm{1}}−\mathrm{1}\right){k}\overset{!} {=}\mathrm{1} \\ $$$$\Rightarrow{k}=\frac{{e}−\mathrm{1}}{\mathrm{3}−{e}} \\ $$$$\Rightarrow{f}\left({x}\right)=\left(\frac{{e}−\mathrm{1}}{\mathrm{3}−{e}}\right)\left(\frac{\mathrm{2}{e}^{{x}} }{{e}−\mathrm{1}}−\mathrm{1}\right) \\ $$
Commented by Tawa11 last updated on 04/Jan/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by tounghoungko last updated on 04/Jan/22
