Answer to Question #245776 in Differential Equations for Moses

Question #245776

In 2000, the population of a country was 4U million. I he population was growing at a rate of 5% per year and every year 30 000 people emigrated from the country. 6.1 Write down an initial value problem for the population P. (1 pt) 6.2 Solve the initial value problem. (2 pts) 6.3 Estimate the population in 2010


1
Expert's answer
2021-10-04T16:44:09-0400

"\\frac{dP}{dt}=0.05P-0.03, \\ P(0)=40"


6.2)

"\\frac{dP}{dt}=0.05P-0.03"

"\\frac{dP}{0.05P-0.03}=dt"

Let 0.05P-0.03=Q

"\\frac{dQ}{dP}=0.05"

"dP=\\frac{dQ}{0.05}"

Now,

"\\frac{1}{0.05Q}dQ=dt"

"\\frac{dQ}{Q}=0.05dt"

Integrate both sides;

"\\int \\frac{dQ}{Q}=\\int 0.05dt"

ln Q=0.05t+C

ln(0.05P-0.03)=0.05t+C

"0.05P-0.03=e^{0.05t+C}"

"P=\\frac{e^{0.05t+C}+0.03}{0.05}"

"P=\\frac{e^{0.05t}*e^{C}+0.03}{0.05}"

"P=\\frac{C_1e^{0.05t}+0.03}{0.05}" ,where "C_1=e^{C}"

"P=C_2e^{0.05t}+0.6" ,where "C_2=\\frac{C_1}{0.05}"

Using P(0)=40;

40=C2+0.6

C2=40-0.6=39.4

"P=39.4e^{0.05t}+0.6"


6.3)

We use the equation above,

"P=39.4e^{0.05t}+0.6"

t=10

"P=39.4e^{0.05*10}+0.6"

"=64.989618+0.6"

"=65.559618million"

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

Moses
04.10.21, 23:48

This was correct and helpful. Thank you so much

Leave a comment

LATEST TUTORIALS
New on Blog