Question

Consider the equation for the charge on a capacitor in an LRC circuit da + dt2 +79 = E dt which is linear with constant coefficients. , and find the auxiliary equation (using m as your First we will work on solving the corresponding homogeneous equation. Divide through the equation by the coefficient on variable) = 0 which has roots The solutions of the homogeneous equation are Now we are ready to solve the nonhomogeneous equation + 16 + 634 = 9E. We will use e-A to do reduction of order (it doesn't matter which one of the homogenous solutions we choose), which will give us a particular solution of the nonhomogeneous equation. With this choice the particular solution has the form y = ue- Then (using the prime notation for the derivatives) y' = y = So, plugging y, into the left side of the differential equation, and reducing, we get + 16y', +63y, = Using this reduced form, moving the term e-9 to the right side of the equation, we can rewrite y, + 16y, +63y, = 9E in terms of u" and u' as At this point to solve for u we want to integrate both sides, but to do that we need to integrate E. We will look at two special cases for E so that you get the idea how to proceed.

Answer 1

SOLUTION O + I d2q 16 da 9 dt² +79 = 9 at Complemeut ary solution I d2q + 16 h 16 da +7q co 9 at CD2 +16D +63)9 = 0 Auxiliay m2 + 16m +63 = 6 0 Dad at equation. m2 tam+im+6320 (m+q) (m+7)=0 Hence its roots are m, -9 + m2=-7 Hence which are Real t distinct solution of the homogeneous equation are -7t te e y ca ci è at Particular Solution

Particular solution choose "p=uelt » Typ = -queet tulegt yp - -96 olu egt -quient 9 8 lutolat olueet_18ulegt tulle at => 9p yo Now y +164p +63 Yp = 810eet_iguent + cheat + 16 (-90 ēqt tu!ēqt) +63 vēqt -20'è attulē at " +169p+634p (-20'+0") 29t Hence yp" +10yp +639p = 9€ (-20'+u") e 9t=9€ lull – 20' qe eat →