Answer:
[tex]y=20e^{-\frac{2}{3}t}[/tex]
Exponential decay
Step-by-step explanation:
We are given that
[tex]\frac{dy}{dt}=-\frac{2}{3}y[/tex]
y=20 when t=0
[tex]\frac{dy}{y}=-\frac{2}{3}dt[/tex]
Taking integration on both sides then we get
[tex]lny=-\frac{2}{3}t+C[/tex]
Using formula [tex]\int \frac{dx}{x}=lnx,\int dx=x[/tex]
[tex]y=e^{-\frac{2}{3}t+C}[/tex]
Using formula
[tex]lnx=y\implies x=e^y[/tex]
[tex]y=e^C\cdot e^{-\frac{2}{3}t}[/tex]
[tex]e^C=Constant=C[/tex]
[tex]y=Ce^{-\frac{2}{3}t}[/tex]
Substitute y=20 and t=0
[tex]20=C[/tex]
Substitute the value of C
[tex]y=20e^{-\frac{2}{3}t}[/tex]
When t tends to infinity then
[tex]\lim_{t\rightarrow\infty}=\lim_{t\rightarrow\infty}20e^{-\frac{2}{3}t}=0[/tex]
When time increases then the value of function decrease
Hence, the function is exponential decay.
