Math 2280 - Lecture 8Dylan ZwickSpring 20091 2nd-Ord er Linear Homogeneous DifferentialEquations with Constan t CoefficientsWe’ll finish some of what we started last time. A second-order linear ho-mogeneous differential equation is a differential equation of the form:ay′′+ by′+ cy = 0where a, b, c ∈ R are constants. Now, we know that if we can find twolinearly independent solutions to the above equations, then we’re done. Inother words, for any given initial conditions, we can find the solution as alinear combination of our two linearly indep endent solutions. So, supposewe “guess” that our solution is of the form y = erx. If we plug this solutioninto our differential equation we get:ar2erx+ brerx+ cerx= 0Now, erxis never 0, and so we can divide both sides of this equation byit to get:ar2+ br + c = 0So, if r is a root of the quadratic equation:1ax2+ bx + cthen erxis a solution to the above differential equation.Now, remembering back to junior high school, the roots of the quadraticequation are:r =−b ±√b2− 4ac2a.There are three possibilities, depend in g upon whether b2− 4ac, calledthe descriminant, is positive, negative, or 0.If b2− 4ac > 0 then we get two distinct real roots, call them r1, r2. andso we get two linearly independent solutions, er1xand er2x, and all of oursolutions are of the form:y = c1er1x+ c2er2x.On the other hand, if b2− 4ac = 0 then we only have one (repeated)root, call it r, and our solutions are of the form:y = c1erx+ c2xerxNow, how do we know xerxis a solution? Well, try it out:y′(x) = rxerx+ erxy′′(x) = r2xerx+ 2rerxand plugging these into our differential equation we get:a(r2xerx+ 2rerx) + b(rxerx+ erx) + cxerx= xerx(ar2+ br + c) + erx(2ar + b)Now, the first term is 0 by definition, and if b2− 4ac = 0 then we haver = −b/2a, and so the second term is 0 as well. So, it’s also a solution, andwe have a second linearly independent solution.Now, sorry to put this off some more, but we’ll deal with the other case,b2− 4ac < 0, later.22 General Solutions o f Linear EquationsLast lecture we saw what ha ppens whe n we have 2nd-order linear ODEsof the form:A(x)y′′+ B(x)y′+ C(x)y = F (x)Now, it only takes a little imagination to generalize to nth-order linearODEs:P0(x)y(n)+ ··· + Pn−1(x)y′+ Pn(x)y = F (x)Now, if we assume P0(x) 6= 0 on our interval of interest I then we candivide both sides by P0(x) to get:y(n)+ p1(x)y(n−1)+ ··· + pn−1(x)y′+ pn(x)y = f (x)which has a corresponding homogeneous equation:y(n)+ p1(x)y(n−1)+ ··· + pn−1(x)y′+ pn(x)y = 0.As with the second-order case, for a homogeneous nth-order ODE wehave any linear combination of solutions:y = c1y1+ ···ckykis also a solution.33 General Theory of n th-Order ODEsTheorem - If p1, . . . , pnare continuous on I and f (x) is too then:y(n)+ p1y(n−1)+ ··· + pn−1y′+ pny = f(x)has a unique solution satisfying the initial conditions:y(a) = b0, y′(a) = b1, . . . , y(n−1)(a) = bn−1for bk∈ R and a ∈ I.Now, suppose we ha ve a homogeneous nth-order ODE:y(n)+ p1y(n−1)+ ···pn−1y′+ pny = 0and we have n solutions y1, . . . , yn. Can we get all solutions by solu-tions of the form:y = c1y1+ ··· + cnyn?Well, as with 2nd-order linear homogeneous ODEs the answer is yes,if the ykare linearly independent. Again, as with 2nd-order linear homo-geneous ODEs we can check for linear independence by using the Wron-skian.DefinitionA set of functions f1, . . . , fnare linearly independent on an interval Iprovided:c1f1+ ···cnfn= 0has no solutions on I except the trivial solution.Now, how do we tell if they are linearly independent? As I said, welook at the Wronskian.4W (x) =f1f2··· fnf′1f′2··· f′n............f(n−1)1f(n−1)2··· f(n−1)nNow, again, the if everything is continuous on our interval of interest,then the Wronskian is either never 0 or always 0. If it’s never 0 then thefunctions are linearly independent, if it’s 0 then the functions are linearlydependent.The general proof uses the uniqueness theorem in the same way a swith 2nd-order linear homogeneous diffe rential equations.4 Nonhom o geneous SolutionsSo far we’ve just bee n discussing homogeneous differential equations, butwhat happens if we have have a non-homogeneous differential equation?Well, suppose we have a non-homogeneous differential equation:y(n)+ p1y(n−1)+ ··· + pn−1y′+ pny = f(x)where f(x) 6= 0. Now, if we have a solution, which we’ll for rightnow call yp1, and another solution which we’ll call yp2, then if we takethe difference of these two solutions yp1− yp2the difference will solve thehomogeneous equation:y(n)+ p1y(n−1)+ ··· + pn−1y′+ pny = 0and so will be of the form discussed above. In other words:yp2= yp1+ c1y1+ c2y2+ ···cnynwhere yp1is a particular solution to the given nonhomogeneous equa-tion, and the yiare n linearly independent solutions to the attendant ho-mogeneous equation.5OK, so what does this mea n? It means that if we can fine just onesolution to the above nonhomogeneous equation, then we’ve found allthe other solutions if we can solve the attendant homogeneous equation.Finding one solution can be hard, and finding the solutions of the atten-dant homogeneous equation can also be hard, but in many situations we’llfind out it’s possible and not that hard.ExampleFind the solution to the initial value problem:y′′− 4y = 12y(0) = 0, y′(0) = 10Noting that a solution to the given differential equation is yp= −3.5 Homogeneous E quations with Constant Coef-ficientsSo far we’ve larned how to solve 2nd-order homogeneous equations withconstant coefficients, as long as the roots of the characteristic polynomialare not complex.Well, we’re now going to use this as a starting point for our study ofmore general equations, and we ’re going to figure out what to do if theroots are complex.65.1 General Form of the EquationThe more general version of an nth order homogeneous linear diffe rentialequation with constant coefficients is:any(n)+ an−1y(n−1)+ ··· + a1y′+ a0y = 0The first thing to notice is that if we try what we did in the second ordercase, namely plugging in the solution y = erxand seeing if it works, weget:anrnerx+ an−1rn−1erx+ ···a1rerx+ a0erx= 0Now, if we divide both sides by erx, which we know we can do for anyvalue of x as erxis never 0, we get the equation:anrn+ an−1rn−1+ ··· + a1r + a0= 0which