Answer to Question #338365 in Differential Equations for Jas

Question #338365

Activity 1:


Solve the following problems applying the concepts learned above. Write your answer on a separate sheet of paper. (Show your solution)


1. The rate of change x is proportional to x. When t =0, x0 = 3 and when t =2, x=6. What is the value of x when t =4?


2. A certain plutonium isotope decays at a rate proportional to the amount present. Approximately 15% of the original amount decomposes in 100 years. How much amount of the substance has decayed after 600 years? Also, find the half - life t1/2 of this radioactive substance ; that is, find the time required for this substance to decay to one- half of its original amount.




1
Expert's answer
2022-05-09T16:08:29-0400

1.


dx/dt=kxdx/dt=kx

dxx=kdt\dfrac{dx}{x}=kdt

dxx=kdt\int \dfrac{dx}{x}=\int kdt

lnx=kt+lnC\ln |x|=kt+\ln C

x=Cektx=Ce^{kt}

Given x(0)=3x(0)=3


3=Ce0=>C=33=Ce^{0}=>C=3

x=3ektx=3e^{kt}

Given x(2)=6x(2)=6


6=3ek(2)6=3e^{k(2)}

2k=ln22k=\ln 2

k=ln22k=\dfrac{\ln 2}{2}

x=3(2)t/2x=3(2)^{t/2}

Then


x(4)=3(2)4/2x(4)=3(2)^{4/2}

x(4)=12x(4)=12

2.


N=1kdNdtN=\dfrac{1}{k}\dfrac{dN}{dt}

dNN=kdt\dfrac{dN}{N}=kdt

dNN=kdt\int \dfrac{dN}{N}=\int kdt

lnN=kN+lnC\ln |N|=kN+\ln C

N(t)=N0ektN(t)=N_0e^{kt}

Given N(100)=0.85N0N(100)=0.85N_0


0.85N0=N0e100k0.85N_0=N_0e^{100k}

k=0.01ln0.85k=0.01\ln0.85

N(t)=N0(0.85)t/100N(t)=N_0(0.85)^{t/100}

Then


N(600)=N0(0.85)600/100N(600)=N_0(0.85)^{600/100}

N(600)=0.37715N0N(600)=0.37715N_0

10037.715=62.285100-37.715=62.285

Approximately 62.285% of the original amount decomposes in 600 years.



N(t1/2)=N0/2=N0(0.85)t1/2/100N(t_{1/2})=N_0/2=N_0(0.85)^{t_{1/2}/100}

t1/2=100(ln0.5ln0.85)t_{1/2}=100(\dfrac{\ln0.5}{\ln0.85})

t1/2=426.5 yearst_{1/2}=426.5\ years


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