Respuesta :
Solution :
Given :
The rate of change of y is proportional to y.
Therefore, [tex]$\frac{dy}{dt} \propto y$[/tex]
[tex]$\frac{dy}{dt} = k y$[/tex] , where k is the proportionality constant
[tex]$\frac{dy}{y}=k \ dt$[/tex]
Taking integration on both the sides,
[tex]$\int \frac{dy}{y}=\int kdt$[/tex]
[tex]$\int \frac{dy}{y}=k\int dt$[/tex]
[tex]$\ln |y| = kt+C$[/tex], where C is the integration constant
[tex]$y=e^{kt+C}$[/tex]
[tex]$y=e^{kT}e^C$[/tex]
[tex]$y(t)=C_1e^{kt}$[/tex] ....................(1)
When t = 0, y = 2, so
[tex]$2=c_1e^{k\times 0}$[/tex]
[tex]$2=c_1e^{ 0}$[/tex]
[tex]$2=C_1$[/tex]
From equation (1),
[tex]$y=2e^{kt}$[/tex] ........................(2)
when t = 2, y = 4
[tex]$4=2e^{k\times 2}$[/tex]
[tex]$e^{2k}=\frac{4}{2}$[/tex]
[tex]$e^{2k}=2$[/tex]
[tex]$2k= \ln(2)$[/tex]
[tex]$k=\frac{1}{2} \ln(2)$[/tex]
From (2),
[tex]$y=2e^{\frac{1}{2}\ln(2)t}$[/tex]
[tex]$y=2e^{\frac{t}{2}\ln(2)}$[/tex]
[tex]$y=2e^{\ln(2)^{t/2}}$[/tex]
[tex]$y=2.2^{t/2}$[/tex]
When t = 3,
[tex]$y = 2 . 2 ^{3/2}$[/tex]
[tex]$y=2[(\sqrt2)^2]^{3/2}$[/tex]
[tex]$y= 2. 2\sqrt2$[/tex]
[tex]$y=4 \sqrt2$[/tex]
