Find a function $ T(t) $ such that $ T'(t) + c^2T(t) = 0 $ for an arbitrary constant $ c $ and that $ T(0) = 1 $. I try several functions and got $ T(t) = e^{-c^2t} $, but is there any way to solve it? I haven't taken differential equation yet so this is my first time.
Find a function that satisfies the properties
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ordinary-differential-equations
2 Answers
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From the eqn: $T'(t) = -c^2T(t)$, clearly, integrating we have $ln(T(t)) = -c^2t + K$ where K is the constant of integration. Thus,:
$T(t) = e^{-c^2.t + K}$ Since $T(0) = 1, e^K = 1 => K = 0$
Hence, $T(t) = e^{-c^2.t}$
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$T′(t)+c^2T(t)=0$ or $\frac{dT(t)}{dt} = -c^2 T(t)$ => $\frac{dT(t)}{T(t)}=-c^2dt$. Integrating both sides, we have: $ln(T)=-c^2t+C$, where C is a constant. Thus, $T(t)=e^{-c^2t+C}$. Because $T(0)=1$, $C=0$. Finally, we get: $T(t)=e^{-c^2t}$.