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There will be some further material related to this posted elsewhere.Without drawing the graph, we can also show how solutions will behaveclose to an equilibrium. Please see section 8.2 of your book for more infor-mation. This section might be expanded later.4.4Second-Order Linear Homogeneous Dif-ferential EquationsConsider a differential equation of the formay00+by0+cy= 0.(4.2)Notice that the coefficients are all constant and the left-hand side is zero.Such equations turn out to have relatively simple solutions. However, sinceit is second order, our answer will have two arbitrary constants in it.We proceed by guessingy=erx(4.3)and solving forrthat work.This process of guessing a general form ofa solution and then solving for parameters that you introduce is called an
78CHAPTER 4.ORDINARY DIFFERENTIAL EQUATIONSAnsatz.a(erx)00+b(erx)0+c(erx)=0ar2(erx) +br(erx) +c(erx)=0ar2+br+c=0r=-b±b2-4ac2awhere we applied the quadratic formula in the last step.Ifb2-4ac >0,then we’re in business, i.e we have two real values ofr.If not, specialconsiderations are necessary. We’ll get to that later.Example:y00+y0-6y= 0, y(0) = 0, y0(0) = 5.Our Ansatz ofy=erxgives usr2+r-6 = 0r=-3,2.So we have two solutions,y=e2xandy=e-3x.Notice that any linearcombination of them is also a solution:(Ae2x+Be-3x)00+ (Ae2x+Be-3x)0-6(Ae2x+Be-3x) = 0.To findAandBfor our problem, we use the initial conditionsAe2(0)+Be-3(0)=0A+B=02Ae2(0)-3Be-3(0)=52A-3(-A)=5A=1B=-1y(x)=e2x-e-3x.Now, let’s return to the question of what happens whenb2-4ac= 0.In that case, the equation is calleddegenerate, and we need to find a second
4.4. SECOND-ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS79solution. Let’s look at an example.y00+ 6y0+ 9y=0r2+ 6r+ 9=0(r+ 3)2=0y=Ae-3x.We only got one solution, but we need a second one for a complete answer.There is actually a relatively clear logical path to a second answer, but itinvolves some tricky limits. Since we’re already pulling solutions out of hats,why not another one? Try the same solution, only multiplied byx:(xe-3x)00+6(xe-3x)0+9(xe-3x) =e-3x(9x-6)+6(e-3x(1-3x))+9xe-3x=e-3x(9x-6+6-18x+9x) = 0.The calculation almost ran off the end of the page, but a little product ruleshowshowit works. As usual, all you need to do is rememberthatit works.Let’s turn our example into an honest initial value problem. With our newmathematical hammer firmly in hand, we will have no problem smashing itinto easy bits.y00+ 6y0+ 9y=0,y(0) = 2,y0(0) = 1y=Ae-3x+Bxe-3xy(0)=A= 2y0=-3e-3x(A+Bx) +Be-3xy0(0)=-3A+BB=7y=e-3x(2 + 7x).Basically, keep an eye out for a perfect square. It’s a special case, and it getsa special solution.There’s one more situation that we’ve been avoiding. What in the worlddo we do ifb2-4ac <0? This means that the quadratic has complex roots.
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