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Question Number 163119 by tounghoungko last updated on 04/Jan/22

$$f{}^{\prime }\left(x\right)=f\left(x\right)+{\int }_0^{1}f\left(x\right)dx$$$$f\left(0\right)=1\Rightarrow f\left(x\right)=?$$

Answered by mr W last updated on 04/Jan/22

$$y^{\prime }=y+kwithk={\int }_0^{1}f\left(x\right)dx$$$$\frac{dy}{y+k}=dx$$$$\int \frac{dy}{y+k}=\int dx$$$$\ln (y+k)=x+C_1$$$$y+k={Ce}^x$$$$y=f\left(x\right)={Ce}^x-k$$$${\int }_0^{1}f\left(x\right)dx={[{Ce}^x-kx]}_0^1=C(e-1)-k\stackrel{!}{=}k$$$$\Rightarrow C=\frac{2k}{e-1}$$$$f\left(x\right)=(\frac{2e^x}{e-1}-1)k$$$$f\left(0\right)=(\frac{2}{e-1}-1)k\stackrel{!}{=}1$$$$\Rightarrow k=\frac{e-1}{3-e}$$$$\Rightarrow f\left(x\right)=\left(\frac{e-1}{3-e}\right)(\frac{2e^x}{e-1}-1)$$

Commented by Tawa11 last updated on 04/Jan/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by tounghoungko last updated on 04/Jan/22

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