Respuesta :
Answer:
The amounts of salt in the tank at any time prior to the instant when the solution begins to overflow is.
[tex]Q(t)=-\frac {4\times 10^6}{(200+t)^2}+200+t[/tex]
Step-by-step explanation:
Let Q(t) be the amount of salt in the tank at any time [tex]t[/tex].
Then, its time of change, [tex]Q'(t)[/tex] by (Balance law).
Since three gallons of salt water runs in the tank per minute, containing [tex]1[/tex] lb of salt, the salt inflow rate is
[tex]3\cdot 1=3[/tex]
The amount of water in the tank at any time [tex]t[/tex] is,
[tex]V(t)=200+(3-2)t=200+t,[/tex]
Now, the outflow is [tex]2[/tex] gal of the solution in a minute. That is [tex]\frac 2{200+t}[/tex] of the total solution content in the tank, hence [tex]\frac 2{200+t}[/tex] of the salt salt content [tex]Q(t)[/tex], that is [tex]\frac {2Q(t)}{200+t}[/tex],
Initially, the tank contains [tex]100[/tex] lb of salt,
Therefore, we obtain the initial condition [tex]Q(0)=100[/tex]
Thus, the model is
[tex]Q'(t)=3-\frac {2Q(t)}{200+t}, \;\;Q(0)=100[/tex]
[tex]\Rightarrow Q'(t)+\frac {2}{200+t}Q(t)=3, \;\;Q(0)=100[/tex]
here, [tex]p(t)=\frac 2{200+t}\;\; \text{and}\;\; q(t)=3[/tex] Linear ODE
So, the integrating factor is
[tex]e^{\int pdt}=e^{2\int \frac{dt}{200+t}=e^{\ln(200+t)^2}=(200+t)^2[/tex]
and the general solution is
[tex]Q(t)(200+t)^2=\int q(t)(200+t)^2dt+c[/tex]
[tex]\Rightarrow Q(t)=\frac 1{(200+t)^2}\int 3(200+t)^2dt+c[/tex]
[tex]=\frac c{(200+t)^2}+200+t[/tex]
using initial condition and find the value of constant c.
[tex]100=Q(0)=\frac c{(200+0)^2}+200+0\Rightarrow -100=\frac c{200^2}[/tex]
[tex]\Rightarrow c=-4000000=-4\times 10^6[/tex]
[tex]\Rightarrow Q(t)=-\frac {4\times 10^6}{(200+t)^2}+200+t[/tex]
Hence, is the amount of salt in the tank at any moment t.
