Question

The rate of change of y is proportional to y. When t = 0, y = 4, and when t = 2, y = 8. What is the value of y when t = 3?

Answer 1

Solution :

Given :

The rate of change of y is proportional to y.

Therefore, $\frac{dy}{dt} \propto y$

$\frac{dy}{dt} = k y$ ,             where k is the proportionality constant

$\frac{dy}{y}=k \ dt$

Taking integration on both the sides,

$\int \frac{dy}{y}=\int kdt$

$\int \frac{dy}{y}=k\int dt$

$\ln |y| = kt+C$,    where C is the integration constant

$y=e^{kt+C}$

$y=e^{kT}e^C$

$y(t)=C_1e^{kt}$  ....................(1)

When t = 0, y = 2, so

$2=c_1e^{k\times 0}$

$2=c_1e^{ 0}$

$2=C_1$

From equation (1),

$y=2e^{kt}$ ........................(2)

when t = 2, y = 4

$4=2e^{k\times 2}$

$e^{2k}=\frac{4}{2}$

$e^{2k}=2$

$2k= \ln(2)$

$k=\frac{1}{2} \ln(2)$

From (2),

$y=2e^{\frac{1}{2}\ln(2)t}$

$y=2e^{\frac{t}{2}\ln(2)}$

$y=2e^{\ln(2)^{t/2}}$

$y=2.2^{t/2}$

When t = 3,

$y = 2 . 2 ^{3/2}$

$y=2[(\sqrt2)^2]^{3/2}$

$y= 2. 2\sqrt2$

$y=4 \sqrt2$