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325_pdfsam_math 54 differential equation solutions odd

# 325_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Exercises 5.6 Therefore, applying Kirchhoﬀ’s second law to this network yields the three equations given by Loop 1 : Loop 2 : Loop 3 : dI2 = 10, dt 10I1 + 5I3 + 30q3 = 10, dI2 5I3 + 30q3 − 20 = 0. dt 10I1 + 20 Since the equation for Loop 2 minus the equation for Loop 1 yields the remaining equation, we will use the ﬁrst and second equations above for our calculations. By examining a junction point, we see that we also have the equation I1 = I2 + I3 . Thus, we have I1 = I2 + I3 . We begin by dividing the equation for Loop 1 by 10 and the equation for Loop 2 by 5. Diﬀerentiating the equation for Loop 2 yields the system I1 + 2 dI2 = 1, dt dI1 dI3 2 + + 6I3 = 0, dt dt where I3 = q3 . Since I1 = I2 + I3 and I1 = I2 + I3 , we can rewrite the system using operator notation in the form (2D + 1)[I2 ] + I3 = 1, (2D )[I2 ] + (3D + 6)[I3 ] = 0. If we multiply the ﬁrst equation above by (3D + 6) and then subtract the second equation, we obtain {(3D + 6)(2D + 1) − 2D } [I2 ] = 6 ⇒ 6D 2 + 13D + 6 [I2 ] = 6. This last diﬀerential equation is a linear equation with constant coeﬃcients whose associated equation, 6r 2 +13r +6 = 0, has roots −3/2, −2/3. Therefore, the solution to the homogeneous equation corresponding to the equation above is given by I2h (t) = c1 e−3t/2 + c2 e−2t/3 . By the method of undetermined coeﬃcients, the form of a particular solution to the diﬀerential equation above will be I2p (t) = A. By substituting this function into the diﬀerential equation, 321 ...
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