Answer to Question #338365 in Differential Equations for Jas

Question

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.

Answer 1

1.

$$dx/dt=kx$$

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

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

$$\ln |x|=kt+\ln C$$

$$x=Ce^{kt}$$

Given $x(0)=3$

$$3=Ce^{0}=>C=3$$

$$x=3e^{kt}$$

Given $x(2)=6$

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

$$2k=\ln 2$$

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

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

Then

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

$$x(4)=12$$

2.

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

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

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

$$\ln |N|=kN+\ln C$$

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

Given $N(100)=0.85N_0$

$$0.85N_0=N_0e^{100k}$$

$$k=0.01\ln0.85$$

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

Then

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

$$N(600)=0.37715N_0$$

$$100-37.715=62.285$$

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

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

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

$$t_{1/2}=426.5\ years$$