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Question Number 206047 by universe last updated on 05/Apr/24

Answered by aleks041103 last updated on 06/Apr/24

$$y^{\prime }=y+const.$$$$\Rightarrow y=c_2{e}^x-c_1$$$$\Rightarrow y^{\prime }=y+c_1$$$$\Rightarrow c_1={\int }_0^{1}y\left(x\right)dx={\int }_0^1(c_2{e}^x-c_1)dx=$$$$=c_2(e^1-e^0)-c_1(1-0)=$$$$=(e-1)c_2-c_1$$$$\Rightarrow 2c_1=(e-1)c_2$$$$\Rightarrow f\left(x\right)=c_2(e^x-\frac{e-1}{2})=c(2e^x+1-e)$$$$f\left(0\right)=1\Rightarrow c(2+1-e)=1\Rightarrow c=\frac{1}{3-e}$$$$\Rightarrow f\left(x\right)=\frac{2e^x+1-e}{3-e}$$$$\left(a\right)f^{″}\left(x\right)=\frac{2}{3-e}{e}^x\Rightarrow f{}^{″}\left(0\right)=\frac{2}{3-e}$$$$\left(b\right)f\left(1\right)=\frac{1+e}{3-e}$$$$\left(c\right)\underset{x\to \mathrm{\infty }}{lim}\frac{f\left(x\right)-1}{x}=\underset{x\to \mathrm{\infty }}{lim}\frac{f\left(x\right)}{x}=\frac{2}{3-e}\underset{x\to \mathrm{\infty }}{lim}\frac{e^x}{x}=+\mathrm{\infty }$$$$\left(d\right)f{}^{\prime }\left(x\right)=\frac{2}{3-e}{e}^x\Rightarrow \frac{f{}^{\prime }\left(ln(3-e)\right)}{2}=1$$

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