Respuesta :
Answer:
(a) Initial concentration of salt = 0.07 kg/L
(b) [tex]\therefore Y'(t)= 0.28 -\frac{y}{125}[/tex]
(c) [tex]Y(t)=35 +35e^{-\frac{1}{125} t}[/tex]
(d)Therefore the amount of salt after 5 hours is =38.18 kg
(e) The concentration of salt in the solution in the tank as time approaches infinity is = 0.035 kg/L
Step-by-step explanation:
Given that,
A tank contains 70 kg of salt and 1000 L of water.
(a)
[tex]\textrm{Concentration of salt }=\frac{\textrm{mass of salt}}{\textrm{volume of water}}[/tex]
[tex]=\frac{70}{1000} kg/L[/tex]
=0.07 kg/L
(b)
Let Y(t) be the amount of salt at any instant time t.
Therefore
[tex]Y'(t)= Y_{in}-Y_{out}[/tex]
[tex]Y_{in}=\textrm{concentration of salt} \times \textrm{rate of enter}[/tex]
=(8×0.035) kg/min
=0.28 kg/min
Since the rate of water in and out are same , the amount of solution remain constant.
[tex]Y_{out}= (\frac{y}{1000}\times 8) kg/min[/tex]
[tex]=\frac{y}{125} kg/L[/tex]
[tex]\therefore Y'(t)= 0.28 -\frac{y}{125}[/tex]
(c)
The above equation can be rewrite as
[tex]Y'(t) +\frac{y}{125} =0.28[/tex]
The coefficient of y is p(t) [tex]=\frac{1}{125}[/tex]
The integrating factor of the D.E is[tex]=e^{\int p(t) dt=[/tex] [tex]e^{\int \frac{1}{125} dt[/tex] [tex]=e^{\frac{1}{125} t[/tex]
Multiplying the integrating factor of both sides of D.E
[tex]e^{\frac{1}{125} t} \ \frac{dY}{dt} +e^{\frac{1}{125} t} .\frac{1}{125} Y=0.28 \ e^{\frac{1}{125} t}[/tex]
Integrating both sides
[tex]\int e^{\frac{1}{125} t} \ \ dY+\int e^{\frac{1}{125} t} .\frac{1}{125} Y \ dt=\int0.28 \ e^{\frac{1}{125} t}\ dt[/tex]
[tex]\Rightarrow e^{\frac{1}{125} t} \ Y= \frac{0.28e^{\frac{1}{125}t} }{\frac{1}{125}} +C[/tex]
[tex]\Rightarrow Y=35+Ce^{-\frac{1}{125}t}[/tex]
At initial when t= 0, y =70
Therefore
[tex]70=35+Ce^0[/tex]
⇒C= 70-35
⇒C=35
Therefore
[tex]Y(t)=35 +35e^{-\frac{1}{125} t}[/tex]
(d)
When t= 5 hour = 300 min
To find the amount of salt after 5 hour , we need to put the value of t in the general solution of D.E
Therefore the amount of salt after 5 hours is = [tex]Y(300)=35+35e^{-\frac{300}{125} }[/tex]
= 38.18 Kg
(e)
When t=∞
[tex]Y(\infty )= 35 +e^{-\infty}[/tex]
= 35 Kg [tex][e^{-\infty}=0][/tex]
Since the amount of water is remain same i.e 1000 L
Therefore the concentration of salt is [tex]=\frac{35}{1000}kg/L[/tex]
=0.035 kg/L
The concentration of salt in the solution in the tank as time approaches infinity is; 0.0035 kg/l
How to solve initial value differential equations?
A) Concentration of salt in the tank initially;
Concentration (kg/l) = mass of salt in kg/ volume of water in liter = 70kg/1000l = 0.07 kg/l
B) dy/dt = rate of salt in - rate of salt out
Rate of salt in = 0.035kg/l * 8l/min = 0.28 kg/min
Rate of salt out = 8y/1000
Thus;
dy/dt = 0.28 - 8y/1000
dy/dt = ⁸/₁₀₀₀(35 - y)
C) Collecting like terms from the above equation,
dy/(35 - y) = ⁸/₁₀₀₀dt
Integrating, we have;
-In(35 - y) = ⁸/₁₀₀₀t + C
Taking the exponential of both sides gives;
35 - y = Ce^(-⁸/₁₀₀₀t)
At initial condition of y = 0, t = 0, we have;
C = 35
Thus, initial value problem is;
y(t) = 35 - 35e^(-8t/1000)
D) After 5 hours,
5 hours * 60 mins = 300 mins
y(300 mins) = 35 - 35e^(-8*300/1000)
⇒ 35 - (35*0.0907)
⇒ 31.8255 kg of salt
E) As time approaches infinity, e^(∞) = 0,
y(t) = 35 - 35*0
y(t) = 35 kg
Concentration (kg/l) = 35/1000 = 0.0035 kg/l
Read more about initial value problems at; https://brainly.com/question/14531143
