HW3 - Physics2210.Homeworkassignment3. B 8.2.4. 2 The...

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1 Physics 2210. Homework assignment 3. B 8.2.4. The original equation, (1 + y 2 )dx + xydy = 0 , is separable and can be written as y (1 + y 2 ) dy = 1 x dx . Integrating this equation leads to ln(1 + y 2 ) 2 = lnx +C , where C is a constant and is determined to be ln5, using the given condition: y=0 when x=5. y = ± 25 x 2 1 , for | x | 5 . B 8.2.20. L dI dt + RI + q C = V . a) For L=0 and V=0, the equation is RI + q C = 0 or R dq dt + q C = 0 , where we consider I=dq/dt. This equation is separable and the solution is lnq = - t RC + D , where D is a constant and is determined to be lnq 0 using the condition that q=q 0 at t=0. Therefore, the final solution is q = q 0 e t RC . b) For RL circuit, and V=0, the equation is L dI dt + RI = 0 . This equation is separable and the solution is lnI = - R L t + D , where D is a constant and is determined to be lnI 0 using the condition that I=I 0 at t=0. Therefore, the final solution is I = I 0 e R L t . c) For a), the time constant τ =RC. For b), the time constant τ =L/R. Clearly, the time constant τ has a unit of time, and at t= τ , either q or I is 1/e of its initial value. Half-life or half-value thickness is defined as the time when the initial value is halved, and τ 1/2 = τ ln2.
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2 B 8.2.27. dT dt = k(T T s ) describes the law of cooling, where k is a proportional constant and T s is
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HW3 - Physics2210.Homeworkassignment3. B 8.2.4. 2 The...

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