Question Number 206047 by universe last updated on 05/Apr/24
Answered by aleks041103 last updated on 06/Apr/24
$${y}'={y}+{const}. \\ $$$$\Rightarrow{y}={c}_{\mathrm{2}} {e}^{{x}} −{c}_{\mathrm{1}} \\ $$$$\Rightarrow{y}'={y}+{c}_{\mathrm{1}} \\ $$$$\Rightarrow{c}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\:\mathrm{1}} {y}\left({x}\right){dx}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left({c}_{\mathrm{2}} {e}^{{x}} −{c}_{\mathrm{1}} \right){dx}= \\ $$$$={c}_{\mathrm{2}} \left({e}^{\mathrm{1}} −{e}^{\mathrm{0}} \right)−{c}_{\mathrm{1}} \left(\mathrm{1}−\mathrm{0}\right)= \\ $$$$=\left({e}−\mathrm{1}\right){c}_{\mathrm{2}} −{c}_{\mathrm{1}} \\ $$$$\Rightarrow\mathrm{2}{c}_{\mathrm{1}} =\left({e}−\mathrm{1}\right){c}_{\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)={c}_{\mathrm{2}} \left({e}^{{x}} −\frac{{e}−\mathrm{1}}{\mathrm{2}}\right)={c}\left(\mathrm{2}{e}^{{x}} +\mathrm{1}−{e}\right) \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1}\Rightarrow{c}\left(\mathrm{2}+\mathrm{1}−{e}\right)=\mathrm{1}\Rightarrow{c}=\frac{\mathrm{1}}{\mathrm{3}−{e}} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{2}{e}^{{x}} +\mathrm{1}−{e}}{\mathrm{3}−{e}} \\ $$$$\left({a}\right)\:{f}''\left({x}\right)=\frac{\mathrm{2}}{\mathrm{3}−{e}}{e}^{{x}} \Rightarrow{f}\:''\left(\mathrm{0}\right)=\frac{\mathrm{2}}{\mathrm{3}−{e}} \\ $$$$\left({b}\right)\:{f}\left(\mathrm{1}\right)=\frac{\mathrm{1}+{e}}{\mathrm{3}−{e}} \\ $$$$\left({c}\right)\:\underset{{x}\rightarrow\infty} {{lim}}\:\frac{{f}\left({x}\right)−\mathrm{1}}{{x}}\:=\:\underset{{x}\rightarrow\infty} {{lim}}\:\frac{{f}\left({x}\right)}{{x}}\:=\:\frac{\mathrm{2}}{\mathrm{3}−{e}}\underset{{x}\rightarrow\infty} {{lim}}\frac{{e}^{{x}} }{{x}}=+\infty \\ $$$$\left({d}\right)\:{f}\:'\left({x}\right)\:=\:\frac{\mathrm{2}}{\mathrm{3}−{e}}{e}^{{x}} \Rightarrow\frac{{f}\:'\left({ln}\left(\mathrm{3}−{e}\right)\right)}{\mathrm{2}}=\mathrm{1} \\ $$