326_pdfsam_math 54 differential equation solutions odd

326_pdfsam_math 54 differential equation solutions odd - I...

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Chapter 5 we see that a particular solution is given by I 2 p ( t )=1 . Thus, the current, I 2 , will satisfy the equation I 2 ( t )= c 1 e 3 t/ 2 + c 2 e 2 t/ 3 +1 . As we noticed above, I 3 can now be found from the Frst equation I 3 ( t )= (2 D +1)[ I 2 ]+1= 2 ± 3 2 c 1 e 3 t/ 2 2 3 c 2 e 2 t/ 3 ² ³ c 1 e 3 t/ 2 + c 2 e 2 t/ 3 +1 ´ +1 I 3 ( t )=2 c 1 e 3 t/ 2 + 1 3 c 2 e 2 t/ 3 . To Fnd I 1 , we will use the equation I 1 = I 2 + I 3 . Therefore, we have I 1 ( t )= c 1 e 3 t/ 2 + c 2 e 2 t/ 3 +1+2 c 1 e 3 t/ 2 + 1 3 c 2 e 2 t/ 3 I 1 ( t )=3 c 1 e 3 t/ 2 + 4 3 c 2 e 2 t/ 3 +1 . We will use the initial condition I 2 (0) = I 3 (0) = 0 to Fnd the constants c 1 and c 2 .Thu s ,w e have I 2 (0) = c 1 + c 2 +1=0 and I 3 (0) = 2 c 1 + 1 3 c 2 =0 . Solving these two equations simultaneously yields c 1 =1 / 5and c 2 = 6
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Unformatted text preview: I 1 ( t ) = 3 5 e − 3 t/ 2 − 8 5 e − 2 t/ 3 + 1 , I 2 ( t ) = 1 5 e − 3 t/ 2 − 6 5 e − 2 t/ 3 + 1 , I 3 ( t ) = 2 5 e − 3 t/ 2 − 2 5 e − 2 t/ 3 . 13. In this problem, there are three loops. Loop 1 is through a 0 . 5 H inductor and a 1 Ω resistor. Loop 2 is through is through a 0 . 5 H inductor, a 0 . 5 ± capacitor, and a voltage source supplying the voltage cos 3 t V at time t . Loop 3 is through a 1 Ω resistor, a 0 . 5 ± capacitor, and the 322...
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