% Notes for Edwards-Penney BVP textbook, sections 3.7 and 7.6 F2010 ================================================================ EPbvp3.7-4 The RL-circuit equation for th ecurrent I(t) is L I'(t) + R I(t) = E(t) Then L=2, R=40, E=100 exp(-10 t), initial current zero, gives 2 I'(t) + 40 I(t) = 100 exp(-10 t), I(0)=0 which in standard form is I'(t) + 20 I(t) = 50 exp(-10 t), I(0)=0 =============== CHAPTER 1 METHOD. This is a first order linear differential equation with constant coefficients, first treated in section 1.5 of Edwards-Penney. The answer from the superposition principle is I(t) = I_h + I_p where I_h = homogeneous solution = constant/integrating factor. A particular solution I_p can be found by undetermined coefficients. =========================== LAPLACE METHOD (RECOMMENDED). The isolated equation for L(I(t)) is 50 L(I(t)) = ------------ (s+20)(s+10) Then partial fractions and backward table methods give the solution I(t) = 5 exp(-10 t) - 5 exp(-20 t) MAXIMUM. To find the maximum on t >= 0, differentiate the formula for I(t), set it equal to zero, and solve for t. Substitute this special value of t back into the formula for I(t), to obtain the maximum current I. ANSWER CHECK. The maple answer check appears below. Symbol % means the answer from the previous line. de:=2*diff(i(t),t)+40*i(t)=100*exp(-10*t); ic:=i(0)=0; dsolve({de,ic},i(t)); u:=unapply(rhs(%),t); diff(u(t),t); # find the derivative of the current solve(%=0,t); # set derivative equal to zero and solve for t u(%); # evaluate current at this value of t ================================================================ EPbvp3.7-12 The equation to be solved is 5 Q'' + 30 Q' + 50 Q = 50 sin(2 t) 5 I'' + 30 I' + 50 I = 100 cos(2t) [diff prev eq] The steady-state solution is the undetermined coefficients solution I(t) of the current equation, I(t) = d1 cos(2t) + d2 sin(2t) Using formulas in the textbook is easier, and this gives rise to the phase-amplitude equation directly: I(t) = (E0 / Z) cos(2t - alpha) E0 = 50 Z = impedance from formula (11) in the text alpha = arctan(...) from formula (9) in the text ================================================================ EPbvp7.6-8 The differential equation is solved routinely with laplace theory, in these steps: x'' + 2x' + x = Dirac(t) - Dirac(t-2) x(0)=2, x'(0)=2 -x'(0) -s x(0) + (s+1)^2 L(x) = 1 - exp(-2s) L(x) = (3+2s-exp(-2s))/(s+1)^2 Write x =x1 + x2, then L(x1) = (3+2s)/(s+1)^2 L(x2) = (-1/(s+1)^2)exp(-2s) Backwards table methods with the first shifting theorem work directly to find L(x1) = (1+u)/u^2 |u=s+1 = L((t+1)exp(-t)) Backwards table methods with the second shifting theorem work directly to find L(x2) = L(-t exp(-t)) exp(-2s) = by the second shift theorem, in words, (1) Remove exp(-2s) (2) Time shift -t exp(-t) by t -> t+2 (3) Multiply answer from (2) by switch u(t-2) = L( (-t exp(-t) | t->t+2) u(t-2)) = L( (-2-t)exp(-t-2) u(t-2)) Then x = x1 + x2 = (t+1)exp(-t) - (t+2)exp(-t-2)u(t-2) Graphing can be done in maple, or by hand, using the piecewise definition of x(t): x(t) = (t+1)exp(-t) for t < 2 x(t) = (t+1)exp(-t) - (t+2)exp(-t-2) for t > 2 ANSWER CHECK. The maple code uses a special version of dsolve. See 7.6-22 below for more complete syntax rules. dsolve({de,ic},x(t),method=laplace); To get the answer in readable format, convert(%,piecewise,t); ================================================================ EPbvp7.6-22 The problem can be solved by considering this example: x'' + x = Dirac(t) + Dirac(t - 2 Pi) + Dirac(t - 4 Pi) x(0)=0, x'(0)=0 Answer check to the example using Maple code. de:=diff(x(t),t,t) + x(t) = Dirac(t) + Dirac(t - 2*Pi) + Dirac(t - 4*Pi); ic:=x(0)=0,D(x)(0)=0; dsolve({de,ic},x(t),method=laplace); convert(%,piecewise,t); The answer to 7.6-22 from maple. This is not a solution, but an answer check. sin(t), t < 2*Pi undefined, t = 2*Pi x(t) = 2*sin(t), t < 4*Pi undefined, t = 4*Pi 3*sin(t), 4*Pi < t Solution steps to 7.6-22 using Laplace theory. (s^2+1)L(x) = 1 + exp(-2 Pi s) + exp(-4 Pi s) x = x1 + x2 + x3 L(x1) = 1/(s^2+1) = L(sin t) L(x2) = (1/(s^2+1))exp(-2 Pi s) = L(sin t) exp(-2 Pi s) = fill in using 2nd shift theorem = L(sin(t+2 Pi)u(t-2 Pi)) L(x3) = (1/(s^2+1))exp(-4 Pi s) = L(sin t) exp(-4 Pi s) = fill in using 2nd shift theorem = L(sin(t+4 Pi)u(t-4 Pi)) Then x1, x2, x3 are found from Lerch's theorem [or inverse laplace]. Because the sine is 2Pi-periodic, all sine terms reduce to sin(t). There is an answer published in the book, which is correct. =======================end=========================