Question

(1 point) Consider the equation for the charge on a capacitor in an LRC circuit which is linear with constant coefficients d2 dt2 First we will work on solving the corresponding homogeneous equation. Divide through the equation by the coefficient on and find the auxiliary equation (using m as your variable) mA2+16m+63 - 0 which has roots The solutions of the homogeneous equation are e^(-7t),engr d2 dt2 di dt Now we are ready to solve the nonhomogeneous equation 4 + 16"4 + 63q- 3E. We will use e 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 9t Then (using the prime notation for the derivatives) yp-ue(-9t)-9u'e (-9t) So, plugging v into the left side of the differential equation, and reducing, we get Using this reduced form, moving the term e-9t to the right side of the equation, we can rewrite yg + 1бу + 63y,-3E in terms of u" and u' as u"-2u' 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 Case 1: E is a constant (DC voltage) Then integrating both sides (we don't use a constant when we do this integration, it turns out to be redundant) This is a linear equation with integrating factor eng2t) Solving this equation, using b as our constant, we get that Plugging this back into yp we finally find that Making the general solution yaeyp yp-ae-9't be-T + be- E . Notice that we already knew that e-9t and e-T where 3E 63 solutions to the homogeneous equation, we just did all that work to figure out that the constantis a solution of the nonhomogeneous equation, which is kind of obvious when you look at it. Now lets do an example where the answer will not be so obvious Case 2: E sin(pt) (AC voltage) You might want to break out your TI-89, Maple, Mathematica, etc for this one. Or you can practice up your integration by parts, a lot. Starting with the equation from above,2 -3Ee9', integrating both sides (again without using a constant) we get u' _ 211 3(ел(9t)/(81+pA2))(9*sin(pt)-p"cos(pt)) This is a linear equation with integrating factor e(-2t) using b as our constant, we get u3/(81+p 2)(49+p 2))*(63sin(pt)-9*cos(pt)-7 p'cos(pt)-p 2*sin(pt) Plugging this back into yp we finally find that , just like the example above. Solving this equation, Making the general solution 63 sinkpt) - 9pcos(p)-7p cos(pt) - P sin(pt) 7t (81+p2X49+p2) Since we already knew that eand e7 were solutions to the homogeneous equation, what we really worked out here is that the particular soluon-9pcos()+63 sin(r) +(7)pcos(pr) - p sin(pr)). Going back to look at our (81+p2X49+p2) calculations, we could have done the integrations above without using constants, and pieced the solution together using superposition

Answer 1

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