Review of ordinary differential equations with constantcoefficients2nd-order homogeneous equationsI.e., equati ons of the form ay00+ by0+ cy = 0. This problem reduces to factor-ing the characteristic equation ar2+ br + c = 0. There are three cases:(1) Distinct real roots r1, r2. General solution is y = C1er1x+ C2er2x.(2) Repeated real ro ot r. General solution is y = C1erx+ C2xerx.(3) Distinct complex conjugate roots a±bi. General solution i s y = C1eaxcos bx+C2eaxsin bx.Method of undetermined coefficientsThis method applies for finding a particular solution to the non-homogeneousdifferential equation ay00+ by0+ cy = g(x ), with g(x) either exponential, sinu-soidal, polynomial, or any product and sum of these types. Recall that a par-ticular solution is all we need: we can then write the general solution as y =C1y1+ C2y2+ yp, where y1, y2are linearly independent solutions to the homoge-neous equation ay00+ by0+ cy = 0, and ypis our particula r solution.The method of undetermined coefficients is the ‘method of g uessing’. We guessa particular form for the solution yp, with some unknown coefficients, then plug ypinto the DE to figure out what the coefficients have to be. We start by making thefollowing initial guesses:• If g(x) is a poly nomia l of degree n, then we guess yp= Anxn+An−1xn−1+...+A1x+A0, a polynomial of degree n with n+1 undetermined coefficients.• If g(x) = P (x)eaxfor some polyno mial of degree n, again we guess yp=Anxneax+ ... + A1xeax+ A0eax.• If g(x) = P (x)eaxsin bx + Q(x)eaxcos bx, with polynomials P (x), Q(x)of degree n and m, respectively, we guess yp= Akxkeaxsin bx + ... +A0eaxsin bx + Bkxkeaxcos bx + ... + B0eaxcos bx, where k is the maximumof m and n.After ma king our initial guess for yp, we must check that our guess satisfies thefollowing condition: NO PIECE of the solution (a piece means a function multipli edby a single paramter, like Aixieax) can be a solution to the homogeneous differen-tial equation ay00+ by0+ cy = 0. If it is, multiply the ENTIRE guessed solution byx. If necessary, repeat the process.Higher-order homogeneous DEsWe consider equations of the form any(n)+ an−1y(n−1)+ ... + a1y0+ a0= 0 .Finding solutions again amo unts to factoring the characteristic equation anrn+an−1rn−1+ ... + a1r + a0. We get n roots, counting multiplicity, so we will getn linearly independent solutions to the equation. We can use the roots to findsolutions as follows:(1) if r is a real root with multiplicity k, then erx, xerx, ..., xk−1erxare k linearlyindependent solutions coming from r.12(2) If r = a + bi is a complex root of multiplicity k , then its complex conjugateis also a root, with the same multiplicity k, giving 2k total roots (countingmultiplicity) coming from a±bi. From these 2k roots we get the 2k linearlyindependent solutions eaxsin bx, xeaxsin bx, ..., xk−1eaxsin bx, eaxcos bx,xeaxcos bx, ..., xk−1eaxcos bx.In total, the general solution will by the sum of n linearly independent f unc-tions, each multipli ed by an arbitrary parameter to be determined by the initialconditions.True or False questions(1) If the Wronskian W [y1, ..., yn](t) 6= 0 for some t ∈ I, then y1, ..., ynarelinearly independent on I.(2) If the Wronskian W [y1, ..., yn](t) = 0 for some t ∈ I, then y1, ..., ynarelinearly dependent on I.(3) If x2eaxis a solution to the di fferential equation any(n)+ an−1y(n−1)+ ... +a1y0+ a0y = 0, then so is xeax.(4) If ypand yqare two solutions to the nonhom ogeneous differential equationay00+ by0+ cy = g(x), then so is ayp+ byqfor any numbers a and b.(5) If f1, ..., fm: R → R are linearly independent functions on the interval[−1, 1], then they are linearly independent on the interval [−2, 2] as
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