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?

Answers

Answer 1

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]


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Step-by-step explanation:

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Step-by-step explanation:

[tex]1a+2p=14\\a+p=9\\\\a+p=9\\a=9-p\\\\1(9-p)+2p=14\\9-p+2p=14\\9+p=14\\p=5\\\\a=9-(5)\\a=4[/tex]

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Answers

Answer:

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