Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 131183 by john_santu last updated on 02/Feb/21

Answered by liberty last updated on 02/Feb/21

$$\frac{\mathrm{dP}}{(32-\mathrm{P})\mathrm{P}}=0.0015\mathrm{dt}$$$$\frac{1}{32}\int [\frac{1}{32-\mathrm{P}}+\frac{1}{\mathrm{P}}]\mathrm{dP}=\int 0.0015\mathrm{dt}$$$$\frac{1}{32}\ln \mid \frac{\mathrm{P}}{32-\mathrm{P}}\mid =0.0015\mathrm{t}+\mathrm{c}$$$$\ln \mid \frac{\mathrm{P}}{32-\mathrm{P}}\mid =0.48\mathrm{t}+\mathrm{C};\frac{\mathrm{P}\left(\mathrm{t}\right)}{32-\mathrm{P}\left(\mathrm{t}\right)}=\lambda {\mathrm{e}}^{0.048\mathrm{t}}$$$$\iff \frac{32-\mathrm{P}\left(\mathrm{t}\right)}{\mathrm{P}\left(\mathrm{t}\right)}=\frac{1}{\lambda }{\mathrm{e}}^{-0.048\mathrm{t}}$$$$\Rightarrow \frac{32}{\mathrm{P}\left(\mathrm{t}\right)}-1=\frac{{\mathrm{e}}^{-0.048\mathrm{t}}}{\lambda };\frac{32}{\mathrm{P}\left(\mathrm{t}\right)}=\frac{\lambda +{\mathrm{e}}^{-0.048\mathrm{t}}}{\lambda }$$$$\iff \mathrm{P}\left(\mathrm{t}\right)=\frac{32\lambda }{\lambda +{\mathrm{e}}^{-0.048\mathrm{t}}}=\frac{32}{1+\frac{1}{\lambda }{\mathrm{e}}^{-0.048\mathrm{t}}}$$$$\mathrm{where}\mathrm{P}\left(0\right)=\frac{32}{1+\frac{1}{\lambda }}=6;1+\frac{1}{\lambda }=\frac{16}{3}$$$$\frac{1}{\lambda }=\frac{13}{3}=4.\stackrel{-}{3},\mathrm{Therefore}\mathrm{P}\left(\mathrm{t}\right)=\frac{32}{1+4.\stackrel{-}{3}{\mathrm{e}}^{-0.048\mathrm{t}}}$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com