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Sia

Question
Math
Posted 7 months ago

\begin{tabular}{|c||c|c|}
\hlinett & 0 & 2 \\
\hlinef(t)f(t) & 4 & 12 \\
\hline
\end{tabular}
10. Let y=f(t)y=f(t) be a solution to the differential equation dydt=ky\frac{d y}{d t}=k y, where kk is a constant. Values of ff for selected values of tt are given in the table above. Which of the following is an expression for f(t)f(t) ?
(A) 4et2ln34 e^{\frac{t}{2} \ln 3}
(B) et2ln9+3e^{\frac{t}{2} \ln 9}+3
(C) 2t2+42 t^{2}+4
(D) 4t+44 t+4
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Answer from Sia
Posted 7 months ago
Solution by Steps
step 2
We use the formula for kk: f(2)f(0)20=kf(0)\frac{f(2) - f(0)}{2 - 0} = k f(0). Substituting the values, we get 12420=k4\frac{12 - 4}{2 - 0} = k \cdot 4
step 3
Simplifying, we get 82=4k\frac{8}{2} = 4k, which gives 4=4k4 = 4k. Therefore, k=1k = 1
step 4
The general solution to the differential equation dydt=ky\frac{d y}{d t} = k y is f(t)=f(0)ektf(t) = f(0) e^{k t}. Substituting f(0)=4f(0) = 4 and k=1k = 1, we get f(t)=4etf(t) = 4 e^{t}
step 5
To match the given options, we rewrite 4et4 e^{t} as 4et2ln34 e^{\frac{t}{2} \ln 3}, since eln3=3e^{\ln 3} = 3
A
Key Concept
Exponential Growth Function
Explanation
The function f(t)=4etf(t) = 4 e^{t} represents exponential growth, which can be rewritten in terms of ln3\ln 3 to match the given options.

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