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The population $\mathrm{p}(\mathrm{t})$ at time $t$ of a certain mouse species satisfies the differential equation $\frac{\mathrm{dp}(\mathrm{t})}{\mathrm{dt}}=0.5 \mathrm{~p}(\mathrm{t})$ $-450$. If $p(0)=850$, then the time at which the population becomes zero is
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The correct answer is:
$2 \ln 18$
$2 \ln 18$
$\frac{d(p(t))}{d t}=\frac{1}{2} p(t)-450$
$\frac{d(p(t))}{d t}=\frac{p(t)-900}{2}$
$2 \int \frac{d(p(t))}{p(t)-900}=\int d t$
$2 \ln |p(t)-900|=t+c$
$t=0 \quad \Rightarrow 2 \ln 50=0+c \quad \Rightarrow c=2 \ln 50$
$\therefore 2 \ln |\mathrm{p}(\mathrm{t})-900|=\mathrm{t}+2 \ln 50$
$P(\mathrm{t})=0 \quad \Rightarrow 2 \ln 900=\mathrm{t}+2 \ln 50$
$t=2(\ln 900-\ln 50)=2 \ln \left(\frac{900}{50}\right)=2 \ln 18$.
$\frac{d(p(t))}{d t}=\frac{p(t)-900}{2}$
$2 \int \frac{d(p(t))}{p(t)-900}=\int d t$
$2 \ln |p(t)-900|=t+c$
$t=0 \quad \Rightarrow 2 \ln 50=0+c \quad \Rightarrow c=2 \ln 50$
$\therefore 2 \ln |\mathrm{p}(\mathrm{t})-900|=\mathrm{t}+2 \ln 50$
$P(\mathrm{t})=0 \quad \Rightarrow 2 \ln 900=\mathrm{t}+2 \ln 50$
$t=2(\ln 900-\ln 50)=2 \ln \left(\frac{900}{50}\right)=2 \ln 18$.
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