Answer:
![y=20e^{-(2)/(3)t}](https://img.qammunity.org/2021/formulas/mathematics/college/7qhrbf9ay2m1sed8kifqlrv2x3bqwokjc2.png)
Exponential decay
Explanation:
We are given that
![(dy)/(dt)=-(2)/(3)y](https://img.qammunity.org/2021/formulas/mathematics/college/4r20s4uiptbl2jkxf2618qvfh7nfwd8iom.png)
y=20 when t=0
![(dy)/(y)=-(2)/(3)dt](https://img.qammunity.org/2021/formulas/mathematics/college/gk5mtknuqib9co7st2s5402ar76qbyzkku.png)
Taking integration on both sides then we get
![lny=-(2)/(3)t+C](https://img.qammunity.org/2021/formulas/mathematics/college/ir3d5vgimh38l2pueq86lbfi1d5wmy44v0.png)
Using formula
![y=e^{-(2)/(3)t+C}](https://img.qammunity.org/2021/formulas/mathematics/college/ysey4uz3cy7b3cxjmgf0j68cd0v2squqba.png)
Using formula
![lnx=y\implies x=e^y](https://img.qammunity.org/2021/formulas/mathematics/college/d3fed2nejjnw2fva0vxo6qvn4b1nuc6y9m.png)
![y=e^C\cdot e^{-(2)/(3)t}](https://img.qammunity.org/2021/formulas/mathematics/college/2i43oymofs67qknhe91icdnqfsohqpdxrf.png)
![e^C=Constant=C](https://img.qammunity.org/2021/formulas/mathematics/college/jr4shbzzc9ix9jhej27ujz5se2pvvzpv5f.png)
![y=Ce^{-(2)/(3)t}](https://img.qammunity.org/2021/formulas/mathematics/college/vjcf1k91fwa6qdbb49iqljuk6a3ksfmuri.png)
Substitute y=20 and t=0
![20=C](https://img.qammunity.org/2021/formulas/mathematics/college/crz4fmnumjyg2ebze1l7as85gp3xv9kfvc.png)
Substitute the value of C
![y=20e^{-(2)/(3)t}](https://img.qammunity.org/2021/formulas/mathematics/college/7qhrbf9ay2m1sed8kifqlrv2x3bqwokjc2.png)
When t tends to infinity then
![\lim_(t\rightarrow\infty)=\lim_(t\rightarrow\infty)20e^{-(2)/(3)t}=0](https://img.qammunity.org/2021/formulas/mathematics/college/ajz0qw2wtcmivtxd4e58dujr0uhj1d16vz.png)
When time increases then the value of function decrease
Hence, the function is exponential decay.