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Question

(A) Given the differential equation: $\frac{d	heta}{dt} = -k(	heta - 	heta_0)$,where $k$ is a constant. Separating the variables: $\frac{d	heta}{\

Vedclass · Accepted Answer

$	heta = 	heta_0 + a e^{-kt}$

Vedclass · Answer

(A) Given the differential equation: $\frac{d	heta}{dt} = -k(	heta - 	heta_0)$,where $k$ is a constant. Separating the variables: $\frac{d	heta}{	heta - 	heta_0} = -k dt$. Integrating both sides: $\int \frac{d	heta}{	heta - 	heta_0} = \int -k dt$. This gives: $\ln|	heta - 	heta_0| = -kt + C_1$. Taking the exponential of both sides: $	heta - 	heta_0 = e^{-kt + C_1} = e^{C_1} e^{-kt}$. Let $e^{C_1} = a$. Then,$	heta - 	heta_0 = a e^{-kt}$. Therefore,the solution is $	heta = 	heta_0 + a e^{-kt}$.

The solution of the differential equation $\frac{d\theta}{dt} = -k(\theta - \theta_0)$,where $k$ is a constant,is . . . . . .

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