{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

diffeqs - First—Order Differential Equations Review We...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: First—Order Differential Equations Review We consider first-order differential equations of the form: d x t 1 0 xtf)=f(t) (1) + Whereflt) is the forcing function. In general, the differential equation has two solutions: dt ? 1. complementary (or natural or homogeneous) solution, xdt) (When/(t) = 0), and 2. particular (or forced or non-homogeneous) solution, x;(t) (whenf(t‘) at D). In our problems, fl!) is often a constant, and therefore, the overall solution to the differential equation is J«7(t‘)=xc(f)+xp(f)=K1P7”r +K2 (2) The time constant, 1:, for first-order electrical circuits is either T = RC or ‘E = UK Complementary Solution The complementary solution is found by considering the homogeneous equation: dx(t) l _ dr +rx(t)—0 (3) The complementary solution is the system’s natural response, which is xc (t) = K1 2"” This result can be verified by substituting this answer into the differential equation, Eq. (3). Particular Solution The particular solution is found by considering the full (non-homogeneous) differential equation, that is, Eq. (1). If the forcing function is a constant, then xP(t) is a constant (K2) also, and hence “yd, = 0. Substituting xP into the original differential equation, Eq. (1), yields: dxp l l — + —x = 0 + —x = dt 2' P 1' P f Reducing the above expression pr'OVides the particular solution, which is the forced response: x; = K2 = t f Note that, alternatively, K3 could have been found using Eq. (2) while considering the fnral value reached by the variable of interest, and hence, in this case is also the steady-state response: x(°0) =Kl e‘” +K2 =K2 Total Solution The total solution is the sum of the complementary and particular solutions: xr(t)=xc(t)+xp =K1e_’” +1'f (4) The coefficient term, K1, is found from the initial conditions (at t= 0). Initial Conditions Typically, the initial conditions are determined by considering that for dc steady-state conditions: (a) a capacitor is like an open circuit (ic = C ”V4, = 0 ), and (b) an inductor appears as a short circuit ("1. = L (ii/‘1’ = 0 ). Further, the cruient through an inductor and the voltage across a capacitor cannot change instantaneously, for example, i,_ (0—) = iL (0+) and vc (0—) = vc (0+) . Second-Order Differential Equations Review The second-order differential equations of interest are of the form: d 2 d ‘ _d12y +2gw°7i+w§ y(t)=f(t) (5) wheref(t) is the forcing function. Like the first-order case, this differential equation has two solutions: (1) complementary (or natural) solution, ydt) when fit) = 0, and (2) particular (or forced) solution, yp(t) whenflt) # 0. In our problems, fit) is often a constant, and therefore, the overall solution to the differential equation is typically y(t)=yc(t)+y,,(t)=K1e"'+K2 e‘='+K3 (6) Complementary Solution The complementary solution is found by considering the homogeneous equation: ’1 %+a:—f+by(t)=0 (7) where a = 2 gwo and b = mg. The complementary solution is the system’s natural response: J’c(t)=Kr er't‘iKz 95" (8) The natural fi'equencies, :1 and 52, are the roots of the characteristic equation: 52 + a s +b = 0 (9) These roots lead to three possible cases: Roots (51, 32) Damping Real and unequal Overdamped Real and equal Critically damped Complex conjugates Under-donated Natural Response Sum of two decaying exponentials Two faster decaying exponentials Exponentially damped sinusoid Particular Solution The particular solution is found by considering the full (non-homogeneous) differential equation, Eq. (5). If the forcing function is a constant, then yp(t) is also a constant (K3), and hence 2 dy" d t = 0 = d y /dt 2 . Substituting yp into the differential equation yields: d 2 y dy —+a—+b 't =O+a0+bt = W dr y( ) r p f Reducing the above expression provides the particular solution, which is the forced response f = K = _ yr 3 b Alternatively, K3 could have been determined from Eq. (6) and the final value y(oo)t Initial Conditions Initial conditions for y and its derivative are necessary to find K1 and K2; specifically needed are y(0) and 33L). A numerical value for f? M is generally found by reapplying KV'L or KCL at t=0r Practically speaking, as the KVL or KCL is performed, the inductor voltage and capacitor dv and C C A, is then equated to an analytical derivative of y(!) determined from Eq. (6). The initial conditions for y(0) are found in the same manner as described for first-order differential equations. current, respectively, are expressed as L d”- dt . The numerical value of ,, [=0 {=0 ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern