Ma 221. The material below was covered during the lecture given last Wed. (1/29/14). Homogeneous Linear Equations with Constant Coefficients

Transcription

1 Ma 1 The material below was covered during the lecture given last Wed. (1/9/1). Homogeneous Linear Equations with Constant Coefficients We shall now discuss the problem of solving the homogeneous equation ay by cy 0 where a,b and c are real constants and a 0. Possible candidates for a solution are x and powers of x. These are no good. lnx is also no good. We shall try e x. Ify e x is a solution of (*) a e x be x ce x e x a b c 0. This is to be a solution x. a b c 0. This equation for is called the auxiliary or characteristic equation. It has the solution b b ac b Δ Δb ac There are three possibilities: (1) Δ0 two real, distinct roots () Δ0 one real root, repeated (3) Δ0 two imaginary roots which are the complex conjugates of each other, i.e. if 1 i i We shall now discuss the three cases in detail. Case 1. Δ0. There are two real distinct roots 1,, where 1 1 b b ac b b ac e 1x and e x are both solutions of the differential equation. These functions are LI, general solution is y c 1 e 1x c e x. where 1 and are both real and 1. y y 3y or 1

2 y c 1 e x c e 3 x. Case. Δ0. There is one real, repeated root 1 b e1x is a solution. We need a second LI solution. To find it we shall use the method of variation of parameters. We seek a solution of the form y vxe 1x, where vx is a function to be determined. Now y v e 1x v 1 e 1x and y v e 1x v 1 e 1x v 1 e 1x av xe 1x v 1 e 1x av 1 e 1x bv e 1x bv 1 e 1x cve 1x 0 av 1 bv a 1 b 1 cv 0 v 0 (why?) v c 1 c x y ve 1x c 1 e 1x c xe 1x is a solution of the differential equation in the case where there exists one repeated root 1.Sincee 1x and xc 1x are LI this is the general solution. y y y 0 0 or 0 one real, repeated root. y c 1 e x c xe x. Section.3 Auxiliary (Characteristic) Equation with Complex Roots Case 3. If Δb ac 0 Then there are two complex roots b b ac b i ac b 1 i b i c a b a i b i c a b a

3 where and are real numbers. Thus we have two complex solutions. e ix e x cosx isinx and e ix e x cosx isinx Since the differential equation has real coefficients, from an earlier theorem that the real and imaginary parts of above are solutions, i.e., e x cosx and e x sin x are both solutions in this case. These are LI functions. a general solution in this case is y h e x Acosx Bsinx where A,B real constants. Solve Solution: The characteristic equation is So y y y 0 pr r r 0 r 1 i The complex solutions corresponding to the roots i are e ix e x cosx isinx and e ix e x cosx isin x From an earlier theorem we know that the real and imaginary parts of these solutions are also solutions. Thus we have two LI solutions e x cosx and e x sinx For this equation 1and 1. Thus y h e x C 1 sin x C cosx 3y 0y 17y

4 Thus 0 0 i i i and i y h e 5 8 x Acos 3 8 x Bsin 3 8 x Write down a second order homogeneous linear differential equation with real constant coefficients whose solutions are 1 e x cos3x and 3e x sin3x. 3sothat 1 3i and 3i. p 3i 3i 3i 3i 3i 3i 9 13 (Check: bi 3i ) equation is y y 13y 0 Note: There is a slide show dealing with the case of complex roots at with corresponding PDF file at Undetermined Coefficients Let us now consider the problem of solving ay by cy fx a 0 a,b,c real constants. We know that the general solution is y y h y p, where

5 and y h the solution of the homogeneous equation y p a particular solution of the equation We know how to find y h. We shall now discuss ways of finding y p for certain special functions fx. 1. fx Ke x K constant, constant. Thus we seek y p for ay by cy Ke x. Due to the exponential form of fx we seek y p of the form A? The differential equation (*) y p Ae x a b cae x Ke x A y p K a b c Ke x a b c The above is a particular solution provided the denominator is non-zero. Note that the denominator is p a b c with. This is the characteristic polynomial with. If p 0, we do not have a y p. However, p 0 is a root of characteristic equation. e x is solution of the homogeneous equation, and therefore Ae x cannot be a solution of the nonhomogeneous equation. If p 0, we try y p Axe x y p Axe x Ae x and y p A xe x Ae x Ae x A xe x Ae x Substitution into the differential equation Axe x a b c Ae x b Ke x A K b y p Kxex if p 0 b provided, of course, that b 0. Note that p a b c p b 5

6 y p Kxex p when p 0andp 0 If p 0andp 0 above y p is no good. But p 0 b 0 b is a double (repeated) root of a b c 0. Hence both e x and xe x are solutions of the homogeneous equation if p p 0, and these cannot therefore be solutions of the nonhomogeneous equation. If p p 0wetry y p Ax e x. Differentiating and substituting into the equation leads to A K K p since p b p Thus y p K p x e x if p p 0. p 0 since a 0byassumption.. Summary: A particular solution of Ly ke x is y p Kex p if p 0 y p Kxex p if p 0,p 0 y p K p x e x if p p Solve y 5y y e x We first solve the homogeneous equation y 5y y 0 For this equation p 5 1, 1 y h c 1 e x c e x Now we find a particular solution for e x. Here 1 and p1 0 Since p 5and p Using above since p1 0andp 1 0 y p1 kxex p xe x 3 y y h y p1 c 1 e x c e x 3 xex Solve y 5y y 3 e x 6

7 From above y h c 1 e x c e x and p 5 Consider y 5y y 3 Then 3 ke x with k 3 0so p0 0 y p 3 (Note that by observation one can see the for 3 y p 3 works.) Thus the solution to y 5y y 3 e x is yx y h y p1 y p c 1 e x c e x 3 3 xex Let us now consider an IVP for this DE, namely y 5y y 3 e x y0 1 y 0 0 Substituting y0 0into we have Taking the derivation of we have The initial condition y 0 0 implies y0 c 1 c 3 1 y x c 1 e x c e x 3 xex 3 ex y 0 c 1 c 3 0 Thus we have the two equations below for c 1 and c : c 1 c 1 c 1 c 3 The solution to these equations is: c 1 1,c 9 5 so 36 yx 1 9 ex xex Note: There is a slide show at with corresponding PDF file at 7