Differential Equation

Question

An initial amount α of a tracer (such as a dye or a radioactive isotope) is injected

into Compartment 1 of the two-compartment system shown in the Figure. At time

t > 0, let x1(t) and x2(t) denote the amount of tracer in Compartment 1 and

Compartment 2, respectively. Thus under the conditions stated, x1(0) = α and

x2

(0) = 0. The amounts are related to the corresponding concentrations ρ1(t)

and ρ2(t) by the equations

x1 = ρ1V1 and x2 = ρ2V2 (i)

where V1 and V2 are the constant respective

volumes of the compartments. The differential equations that describe the

exchange of tracer between the compartments (using the relations in (i),) are

dx1/dt = −L21x1 + L12x2

dx2/dt = L21x1 − L12x2

where L21 = k21 / V1 is the fractional turnover rate of Compartment 1 with respect

to 2 and L12 = k12 / V2 is the fractional turnover rate of Compartment 2 with

respect to 1. Solve the system of differential equations using elimination when

L21 = 2/25, L12 = 1/50 and α = 25.

Answer 1

The given system of differential equations is:

dx1dt=−L21x1+L12x2d x sub 1 over d t end-fraction equals negative cap L sub 21 x sub 1 plus cap L sub 12 x sub 2

𝑑𝑥1𝑑𝑡=−𝐿21𝑥1+𝐿12𝑥2

dx2dt=L21x1−L12x2d x sub 2 over d t end-fraction equals cap L sub 21 x sub 1 minus cap L sub 12 x sub 2

𝑑𝑥2𝑑𝑡=𝐿21𝑥1−𝐿12𝑥2

 

The given values are L21=225cap L sub 21 equals 2 over 25 end-fraction

𝐿21=225

, L12=150cap L sub 12 equals 1 over 50 end-fraction

𝐿12=150

, and α=25alpha equals 25

𝛼=25

.

The initial conditions are

x1(0)=α=25x sub 1 open paren 0 close paren equals alpha equals 25

𝑥1(0)=𝛼=25

and x2(0)=0x sub 2 open paren 0 close paren equals 0

𝑥2(0)=0

.The system becomes:

dx1dt=−225x1+150x2(1)d x sub 1 over d t end-fraction equals negative 2 over 25 end-fraction x sub 1 plus 1 over 50 end-fraction x sub 2 space open paren 1 close paren

𝑑𝑥1𝑑𝑡=−225𝑥1+150𝑥2(1)

dx2dt=225x1−150x2(2)d x sub 2 over d t end-fraction equals 2 over 25 end-fraction x sub 1 minus 1 over 50 end-fraction x sub 2 space open paren 2 close paren

𝑑𝑥2𝑑𝑡=225𝑥1−150𝑥2(2)

 

50d2x1dt2+5dx1dt=050 d squared x sub 1 over d t squared end-fraction plus 5 d x sub 1 over d t end-fraction equals 0

50𝑑2𝑥1𝑑𝑡2+5𝑑𝑥1𝑑𝑡=0

 

Using x1(0)=25x sub 1 open paren 0 close paren equals 25

𝑥1(0)=25

:

25=C1+C2e0=C1+C2(5)25 equals cap C sub 1 plus cap C sub 2 e to the 0 power equals cap C sub 1 plus cap C sub 2 space open paren 5 close paren

25=𝐶1+𝐶2𝑒0=𝐶1+𝐶2(5)

Differentiate x1(t)x sub 1 open paren t close paren

𝑥1(𝑡)

:

dx1dt=−110C2e−110td x sub 1 over d t end-fraction equals negative one-tenth cap C sub 2 e raised to the negative one-tenth t power

𝑑𝑥1𝑑𝑡=−110𝐶2𝑒−110𝑡

From equation (1)open paren 1 close paren

(1)

, at t=0t equals 0

𝑡=0

:

dx1dt(0)=−225x1(0)+150x2(0)d x sub 1 over d t end-fraction open paren 0 close paren equals negative 2 over 25 end-fraction x sub 1 open paren 0 close paren plus 1 over 50 end-fraction x sub 2 open paren 0 close paren

𝑑𝑥1𝑑𝑡(0)=−225𝑥1(0)+150𝑥2(0)

Since

x1(0)=25x sub 1 open paren 0 close paren equals 25

𝑥1(0)=25

and x2(0)=0x sub 2 open paren 0 close paren equals 0

𝑥2(0)=0

:

dx1dt(0)=−225(25)+150(0)=-2d x sub 1 over d t end-fraction open paren 0 close paren equals negative 2 over 25 end-fraction open paren 25 close paren plus 1 over 50 end-fraction open paren 0 close paren equals negative 2

𝑑𝑥1𝑑𝑡(0)=−225(25)+150(0)=−2

Now, use this in the differentiated

x1(t)x sub 1 open paren t close paren

𝑥1(𝑡)

expression:

-2=−110C2e0=−110C2negative 2 equals negative one-tenth cap C sub 2 e to the 0 power equals negative one-tenth cap C sub 2

−2=−110𝐶2𝑒0=−110𝐶2

C2=20cap C sub 2 equals 20

𝐶2=20

Substitute

C2=20cap C sub 2 equals 20

𝐶2=20

into equation (5)open paren 5 close paren

(5)

:

25=C1+20⟹C1=525 equals cap C sub 1 plus 20 ⟹ cap C sub 1 equals 5

25=𝐶1+20⟹𝐶1=5

So, x1(t)=5+20e−110tx sub 1 open paren t close paren equals 5 plus 20 e raised to the negative one-tenth t power

𝑥1(𝑡)=5+20𝑒−110𝑡

.dx1dt=−110(20)e−110t=-2e−110td x sub 1 over d t end-fraction equals negative one-tenth open paren 20 close paren e raised to the negative one-tenth t power equals negative 2 e raised to the negative one-tenth t power

𝑑𝑥1𝑑𝑡=−110(20)𝑒−110𝑡=−2𝑒−110𝑡

x2(t)=50(-2e−110t)+4(5+20e−110t)x sub 2 open paren t close paren equals 50 open paren negative 2 e raised to the negative one-tenth t power close paren plus 4 open paren 5 plus 20 e raised to the negative one-tenth t power close paren

𝑥2(𝑡)=50(−2𝑒−110𝑡)+4(5+20𝑒−110𝑡)

x2(t)=-100e−110t+20+80e−110tx sub 2 open paren t close paren equals negative 100 e raised to the negative one-tenth t power plus 20 plus 80 e raised to the negative one-tenth t power

𝑥2(𝑡)=−100𝑒−110𝑡+20+80𝑒−110𝑡

x2(t)=20−20e−110tx sub 2 open paren t close paren equals 20 minus 20 e raised to the negative one-tenth t power

𝑥2(𝑡)=20−20𝑒−110𝑡

 

Final