3.4.1 Distinct Real Roots

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1 Math CONSTANT COEFFICIENT EQUATIONS 34 Assume that P(x) = p (const.) and Q(x) = q (const.), so that we have y + py + qy = 0, or L[y] = 0 (where L := d2 dx 2 + p d + q). (3.7) dx Guess a solution of the form y = e rx. The equation becomes e rx (r 2 + pr + q) = 0. But e rx 0, which means that y = e rx is a solution iff r satisfies C(r) := r 2 + pr + q = 0. (3.8) Definition The polynoial C(r) is called the characteristic polynomial of the linear operator L, and Eq. (3.8) is called the characteristic equation (or auxiliary equation) of the homogeneous differential equation (3.7). The solution to the characteristic equation (3.8) is r = 1 ( q ± ) p 2 2 4q. There are three cases to consider: > 0 p 2 4q = 0. < Distinct Real Roots If p 2 4q > 0, we get two distinct real roots: r 1 = 1 2 ( p ) p 2 4q, r 2 = 1 ( p + ) p 2 2 4q. Lemma If p 2 4q > 0, then the general solution to (3.7) is y = c 1 e r1x + c 2 e r2x. Proof We already know that e r1x and e r2x are solutions. We have W[e r1x,e r2x ] = er 1x e r 2x r 1 e r 1x r 2 e r 2x = (r 2 r 1 )e (r 1+r 2 )x = p 2 4q e px 0. Therefore e r1x and e r2x are linearly independent and the result follws from theorem Example Consider the differential equation y 5y +6y = 0. Substituting y = e rx into the equation leads to the auxiliary equation r 2 5r +6 = 0. This has distinct real roots r 1 = 2 and r 2 = 3, so the general solution is y(x) = c 1 e 2x + c 2 e 3x. (compare with example 1.4.)

2 Math CONSTANT COEFFICIENT EQUATIONS Equal Real Roots If p 2 4q = 0, we get one real root: r = p/2. One solution is ϕ 1 (x) = e px/2. We need another linearly independent solution. To get one we use the reduction of order technique of section 3.3. Let ϕ 2 (x) = v(x)ϕ 1 (x) = v(x) e px/2. Then the equation for v becomes: v + ( p + p)v 1 2 (p2 4q)v = 0, which reduces to v = 0. Therefore v(x) = ax + b, and the second linearly independent solution is ϕ 2 (x) = x ϕ 1 (x) = x e px/2. Lemma If p 2 4q = 0, then the general solution to (3.7) is y = (c 1 + c 2 x) e px/2. Example Consider the differential equation y +4y +4y = 0. Substituting y = e rx into the equation leads to the auxiliary equation r 2 + 4r + 4 = 0. This has a double root r = 2. The general solution is y(x) = (c 1 + c 2 x)e 2x Complex Roots If p 2 4q < 0, we get a pair of complex conjugate roots: We have complex solutions: We want real solutions, so take linear combinations: r = α ± iβ, where α = p 2, β = 1 2 4q p2. e r 1x = e (α+iβ)x = e αx e iβx = e αx (cosβx + isin βx), e r 2x = e (α iβ)x = e αx e iβx = e αx (cosβx isin βx). 1 2 (er 1x + e r2x ) = e αx cos βx, (real function) 1 2i (er 1x e r2x ) = e αx sin βx. (real function) Lemma If p 2 4q < 0, then the general solution to (3.7) is where α = p 2, and β = 1 2 4q p2. y = e αx (c 1 cos x + c 2 sin x),

3 Math CAUCHY EULER EQUATION 36 Proof One can easily verify by direct substitution that ϕ 1 (x) = e αx cosβx and ϕ 2 (x) = e αx sin βx are solutiuons to the differential equation. To show that they are linearly independent, consider the wronskian: W[ϕ 1,ϕ 2 ] = e αx cos βx e αx sin βx e αx (αcosβx β sin βx) e αx (αsin βx + β cos βx) = e 2αx ( α sin βx cosβx + β cos 2 βx α cosβx sin βx + β sin 2 βx ) = βe 2αx = 1 2 4q p2 e px 0. Thus, ϕ 1 and ϕ 2 are linearly independent and the result follows. Example Consider the differential equation y +2y +4y = 0. Substituting y = e rx into the equation leads to the auxiliary equation r 2 + 2r + 4 = 0. This has a pair of complex conjugate roots r = 1 ± 3i, so the general solution is y(x) = e x (c 1 cos 3x + c 2 sin 3x). 3.5 Cauchy Euler Equation The second order, homogeneous equation x 2 y + pxy + qy = 0, (3.9) where p and q are constant, is called the Cauchy-Euler equation. Since this equation does not have constant coefficients, the method of the previous section will not work. However, this equation can be transformed to a constant coefficient equation by means of the transformation Then we have u (t) = dx dt which, upon rearrangement gives x = e t, d dx (y(x)) = xy (x), xy (x) = u (t), y(x) = u(t). u (t) = dx dt x 2 y (x) = u (t) u (t). d dx (xy (x)) = x 2 y + xy, The transformed equation becomes u + (p 1)u + q = 0. (3.10) This is a constant coefficient equation whose characteristic equation is r 2 + (p 1)r + q = 0. The solution to the constant coefficient equation (3.10) is c 1 e r1t + c 2 r r2t, for distinct real roots; u(t) = (c 1 + c 2 t)e rt, for equal real roots; e αt (c 1 cosβt + c 2 sin βt), for complex roots.

4 Math NONHOMOGENEOUS EQUATIONS 37 Using y(x) = u(t) = u(ln x) leads to the solution of Eq. (3.9): c 1 x r 1 + c 2 x r 2, for distinct real roots; y(x) = (c 1 + c 2 ln x)x r, for equal real roots; x α (c 1 cos(β lnx) + c 2 sin(β ln x)), for complex roots. Example Consider the differential equation 3x 2 y + 11xy 3y = 0. Substituting y = x r into the equation leads to the auxiliary equation 3r 2 +8r 3 = 0. This has distinct real roots r 1 = 1/3 and r 2 = 3, so the general solution is y(x) = c 1 x 1/3 + c 2 x Nonhomogeneous Equations Consider the nonhomogeneous differential equation: or, more compactly where L is the differential operator defined by (3.2). d 2 y dx 2 + P(x)dy + Q(x) y = g(x). (3.11) dx L[y] = g, Theorem If (i) ϕ p (x) is any particular solution of the nonhomogeneous equation L[y] = g on (a,b); and (ii) ϕ 1 and ϕ 2 are linearly independent solutions to the homogeneous equation L[y] = 0 on (a,b), then the general solution of the nonhomogeneous equation L[y] = g on (a, b) is given by y(x) = c 1 ϕ 1 (x) + c 2 ϕ 2 (x) + ϕ p (x). Proof Hypothesis (i) implies that L[ϕ p ] = g. Let y(x) be any solution of (3.13), so that L[y] = g. Now define another function v(x) = y(x) ϕ p (x). Letting L act ov v we get L[v] = L[y ϕ p ] = L[y] L[ϕ p ] = g g = 0. Thus v, being a solution of the homogeneous equation, satisfies for some constants c 1 and c 2. The result now follows. v(x) = c 1 ϕ 1 (x) + c 2 ϕ 2 (x) Example Find the general solution of y y = 2 x 2, given that ϕ p (x) = x 2 is a particular solution. One easily checks that ϕ p is indeed a solution of the equation, as ϕ p ϕ p = 2 x 2. The corresponding homogeneous equation is y y = 0. Substituting y = e rx into the homogeneous equation yields r 2 1 = 0 = r = ±1 = ϕ 1 (x) = e x, ϕ 2 (x) = e x. Thus, the general solution to the original nonhomogeneous equation is y(x) = c 1 e x + c 2 e x + x 2.

5 Math UNDETERMINED COEFFICIENTS 38 In order to solve nonhomogeneous equations, we require a method to determine a particular solution to the nonhomogeneous equation. Methods for determining particular solutions will be developed in the next two sections. 3.7 Undetermined Coefficients In this section we develop a method that will only work for constant coefficient equations. The method involves guessing the form of the particular solution and then just determining the value of a constant, hence the name: method of undetermined coefficients. The method is best illustrated by means of eamples. We begin with examples in which the nonhomogeneous term is an exponential. Example Find a particular solution to the equation y + 3y + 2y = e 3x. Given the nature of the nonhomogeneous term, a reasonable guess for a particular solution is ϕ p (x) = Ae 3x, where the constant A is to be determined. Letting the linear operator L (defined by the left side of the DE) act on ϕ p we get L[ϕ p ] = ϕ p + 3ϕ p + 2ϕ p = 9Ae 3x + 3(3Ae 3x ) + 2Ae 3x = 20Ae 3x. Thus, from the DE we have L[ϕ p ] = e 3x = A = 1 20 = ϕ p (x) = 1 20 e3x. Example Find a particular solution to the equation y 5y + 6y = e 2x. Again, given the nature of the nonhomogeneous term, a reasonable guess for a particular solution would seem to be ϕ p (x) = Ae 2x, where the constant A is to be determined. Letting the linear operator L (defined by the left side of the DE) act on ϕ p we get L[ϕ p ] = ϕ p 5ϕ p + 6ϕ p = 4Ae 2x 5(2Ae 2x ) + 6Ae 2x = 0. Thus L[ϕ p ] e 2x for any value of the constant A! The guessed form of particular solution doesn t work. What went wrong? Let us examine this situation more closely. Consider the more general equation y + py + qy = ae λx for which the linear operator is L := d2 dx 2 + p d dx + q. If we guess a solution of the form y = Ae rx and plug into the equation, we get L[Ae rx ] = ae λx = AC(r)e rx = ae λx, where C(r) is the characteristic polynomial for the operator L. We can see that our guess can only be a solution if r = λ, in which case we get A = a/c(λ). However, if e λx is a solution to the homogeneous equation, then C(λ) = 0, so y = Ae λx can t possibly be a solution to the nonhomogeneous equation. Suppose we go back to the expression L[Ae rx ] = AC(r)e rx

6 Math UNDETERMINED COEFFICIENTS 39 and differentiate with respect to r. This leads to Now, setting r = λ into this we get Plugging this into the equation we get L[Axe rx ] = A[C (r) + xc(r)]e rx. (3.12) L[Axe λx ] = AC (λ)e λx, L[Axe λx ] = ae λx = AC (λ)e λx = ae λx = A = a C (λ), provided C (λ) 0. What if C (λ) = 0, in addition to C(λ) = 0 (i.e. λ is a double root of C(r), i.e. xe λx is also a solution to the homogeneous equation)? Then return to (3.12), and differentiate with respect to r again to get L[Ax 2 e rx ] = A[C (r) + 2xC (r) + x 2 C(r)]e rx. Now, setting r = λ into this we get Plugging this into the equation we get L[Ax 2 e λx ] = AC (λ)e λx. L[Ax 2 e λx ] = ae λx = AC (λ)e λx = ae λx = A = a C (λ), provided C (λ) 0. But, C(r) = r 2 +pr+q, which means that C (r) = 2 0, so that A = a/2 and Ax 2 e λx is a particular solution. To summarize, for equations of the form y +py +qy = ae λx, the following particular solution can be found: Ae λx, if L[e λx ] 0, (i.e. if C(λ) 0) ϕ p (x) = Axe λx, if L[e λx ] = 0, L[xe λx ] 0, (i.e. if C(λ) = 0, C (λ) 0) Ax 2 e λx, if L[e λx ] = L[xe λx ] = 0. (i.e. if C(λ) = C (λ) = 0) Example Find the general solution to y 5y + 6y = e 2x. First look at the homogeneous equation: y 5y + 6y = 0. Substituting y = e rx yields a characteristic equation r 2 5r + 6 = 0 which has solutions r = 2 and r = 3. The solution to the homogeneous equation is: ϕ h (x) = c 1 e 2x + c 2 e 3x The nonhomogeneous term e 2x is a solution to the homogeneous equation (i.e. L[e 2x ] = 0), so try a particular solution of the form ϕ p (x) = Axe 2x. Now, letting the linear operator L (defined by the left side of the DE) act on ϕ p we get L[ϕ p ] = ϕ p 5ϕ p + 6ϕ p = AC(2)xe 2x + AC (2)e 2x = Ae 2x, so, inserting this into the original equation yields: L[ϕ p ] = e 2x = A = 1 = ϕ p (x) = xe 2x. Therefore the general solution to the original equation is y(x) = c 1 e 2x + c 2 e 3x xe 2x. Now we look at an equation with a triginometric inhomogeneity.

7 Math UNDETERMINED COEFFICIENTS 40 Example Find the general solution to y y y = sin x. From the nature of the nonhomogeneous term, a reasonable guess for a particular solution to this equation is ϕ p (x) = A cosx +B sin x. Applying the linear operator L (defined by the left side of the DE) to this gives L[ϕ p ] = ϕ p ϕ p ϕ p = ( 2A + B) cosx + (A 2B) sin x. Inserting this into the DE yields: L[ϕ p ] = sinx = { 2A + B = 0 A 2B = 1 = A = 1 5, B = 2 5. Therefore the particular solution is ϕ p (x) = 1 (cosx 2 sin x). 5 In general, the nonhomogeneous equation L[y] = acos λx + b sin λx will have a particular solution of the form { A cosλx + B sin λx, if L[cosλx] 0, ϕ p (x) = x(a cosλx + B sin λx), if L[cosλx] = 0. Now we look at an example with a polynoial inhomogeneity. Example Find a particular solution to the equation y + 3y + 2y = 2x. Try ϕ p (x) = Ax. Applying the linear operator L (defined by the left side of the DE) to this gives L[ϕ p ] = 0 + 3A + 2Ax = A(3 + 2x). Thus L[ϕ p ] 2x for any value of the constant A. Instead try ϕ p (x) = Ax + B. Then we have L[ϕ p ] = 0 + 3A + 2(Ax + B) = 2Ax + 3A + B, so that Therefore ϕ p (x) = x 3. L[ϕ p ] = 2x = { 2A = 2 3A + B = 0 = A = 1, B = 3. In general, for a differential equation of the form y + py + qy = a 0 + a 1 x + a 2 x a n x n, try a particular solution of the form { A 0 + A 1 x + A 2 x A n x n, if q 0, ϕ p (x) = x(a 0 + A 1 x + A 2 x A n x n ), if q = 0. We have found particular solutions for nonhomogeneous equations where the nonhomogeneous term g(x) is an exponential, sine or cosine, and a polynoial. We can also find a particular solution if g(x) is a product of these.

8 Math UNDETERMINED COEFFICIENTS 41 Let r n (x) = a 0 + a 1 x + a 2 x a n x n, R n (x) = A 0 + A 1 x + A 2 x A n x n, s m (x) = b 0 + b 1 x + b 2 x b m x m, S m (x) = B 0 + B 1 x + B 2 x B m x m. The following table summarizes the form of the particular solution for a given form of g(x): g(x) r n (x) ae λx acos βx + b sin βx r n (x)e λx r n (x) cosβx + s m (x) sinβx e λx [acos βx + b sin βx] e λx [r n (x) cosβx + s m (x) sinβx] ϕ p (x) x σ R n (x) x σ Ae λx x σ (A cosβx + B sin βx) x σ R n (x)e λx x σ (R N (x) cosβx + S N (x) sin βx) x σ e λx [A cosβx + B sin βx] x σ e λx [R N (x) cosβx + S N (x) sin βx] In the right column of the above table σ is the smallest integer which guarantees that no term in ϕ p (x) is a solution to the corresponding homogeneous equation, and N = max{n, m}. Example Find the general solution to the differential equation y y = (2 4x)e x + 10 cos2x. We first solve the homogeneous equation y y = 0, or L[y] = 0, where L := d2 dx 2 1. Setting y = e rx, we get r 2 1 = 0 = r = ±1 = {e x,e x } are linearly independent solutions. Therefore the solution to the homogeneous equation is ϕ h (x) = c 1 e x + c 2 e x. Now look for a particular solution to the nonhomogeneous equation. Given the form of the nonhomogeneous term, we try a particular solution of the form ϕ p (x) = x σ 1 (A 0 + A 1 x)e x + x σ 2 (A cos2x + B sin 2x). Since A 0 e x is a solution to the homogeneous equation, we take σ 1 = 1. We can take σ 2 = 0 and the particular solution takes the form Applying the differential operator to ϕ p yields ϕ p (x) = x(a 0 + A 1 x)e x + A cos2x + B sin 2x. L[ϕ p ] = 4A 1 xe x + (2A 1 2A 2 )e x 5A cos2x 5B sin 2x. Inserting this into the equation yields L[ϕ p ] = (2 4x)e x + 10 cos2x = 4A 1 = 4 A 0 = 0 2A 1 2A 0 = 2 A 1 = 1 = 5A = 10 A = 2 5B = 0 B = 0 Therefore a particular solution is ϕ p (x) = x 2 e x 2 cos2x and the general solution is: y(x) = c 1 e x + (c 2 + x 2 )e x 2 cos2x..

9 Math VARIATION OF PARAMETERS Variation of Parameters Consider the nonhomogeneous equation y + P(x)y + Q(x)y = g(x). (3.13) In the previous section the method of undetermined coefficients was used to find particular solutions to equations for which the linear operator had constant coefficients. Even then, it only works for a certain type of nonhomogeneous term, one for which the form of the particular solution can be guessed. In this section we discuss a method more general that the previous one. The method called variation of parameters, that we will discuss here, has one major advantage over the method of undetermined coefficients. It always works, even for general coefficients P(x) and Q(x), and it requires no guessing, provided that two linearly independent solutions to the homogeneous equation are known. So why did we bother with the method of undetermined coefficients? Because the method of variation of parameters can sometimes be difficult or tedious to implement, and if one can guess the form of a particular solution, it is sometimes easier to use the method of undetermined coefficients. We now derive the procedure known as variation of parameters. First consider the homogeneous counterpart to (3.13), namely y + P(x)y + Q(x)y = 0. (3.14) Suppose that {ϕ 1 (x),ϕ 2 (x)} are linearly independent solutions to (3.14), so that the general solution to (3.14) is y(x) = c 1 ϕ 1 (x) + c 2 ϕ 2 (x). The idea behind this method is to look for a particular solution to (3.13) of the form ϕ p (x) = v 1 (x)ϕ 1 (x) + v 2 (x)ϕ 2 (x). (3.15) This is the homogeneous solution in which the constants, or parameters, c 1 and c 2 are allowed to vary, hence the name variation of parameters. By looking for a particular solution of this form, we have introduced two new unknown functions, v 1 and v 2, so we will require two conditions to solve for them. One condition will, of course, come from requiring that ϕ p satisfy (3.13). But what about a second condition? Since there is no obvious way to get one, we are at liberty to arbitrarily impose one. What we will do is impose a condition at the appropriate point in the calculation so as to simplify things as much as possible. Differentiate ϕ p to get ϕ p (x) = v 1 ϕ 1 + v 2 ϕ 2 + v 1 ϕ 1 + v 2ϕ 2. We now impose our second condition. We will require that v 1 and v 2 satisfy v 1ϕ 1 + v 2ϕ 2 = 0. (3.16) Why did we choose this condition? This condition simplifies the expression for ϕ p, which now reduces to Now differentiate again to get Insert these quantities into (3.13) to get ϕ p = v 1ϕ 1 + v 2ϕ 2. ϕ p = v 1 ϕ 1 + v 2 ϕ 2 + v 1ϕ 1 + v 2ϕ 2. v 1ϕ 1 + v 2ϕ 2 + v 1 ϕ 1 + v 2 ϕ 2 + P(x)(v 1 ϕ 1 + v 2 ϕ 2) + Q(x)(v 1 ϕ 1 + v 2 ϕ 2 ) = g(x).

10 Math VARIATION OF PARAMETERS 43 Re-arranging this equation we get v 1 ϕ 1 + v 2 ϕ 2 + v 1[ϕ 1 + P(x)ϕ 1 + Q(x)ϕ 1] + v 2 [ϕ 2 + P(x)ϕ 2 + Q(x)ϕ 2] = g(x). Notice that the terms in square brackets are zero since ϕ 1 and ϕ 2 are solutions to the homogeneous equation (3.14), so that the above equation reduces to v 1 ϕ 1 + v 2 ϕ 2 = g(x). (3.17) Equations (3.16) and (3.17) form a pair of equations for the two unknown functions v 1 and v 2, which can be written in matrix form as follows: [ ] [ ] [ ϕ1 ϕ 2 v 1 0 ϕ 1 ϕ =. 2 g] This matrix equation has a solution provided its coefficient matrix is non-singular. But this is in fact true since the determinant of the coefficient matrix is just the Wronskian W[ϕ 1,ϕ 2 ], which by Theorem 3.20, is non-zero for linearly independent solutions of the homogeneous equation (3.14). Thus, this equation can be solved by Cramer s rule yielding v 1 = 0 ϕ 2 g ϕ 2 ϕ 1 ϕ 2, v 2 = ϕ 1 0 ϕ 1 g ϕ 1 ϕ ϕ 1 ϕ 2, 2 ϕ 1 ϕ 2 which can be written as v 2 v 1(x) = ϕ 2(x) g(x) W[ϕ 1,ϕ 2 ](x), v 2(x) = ϕ 1(x) g(x) W[ϕ 1,ϕ 2 ](x). Thus, v 1 and v 2 are obtained by integration, and the particular solution (3.15) becomes ϕ p (x) = x ϕ 1 (ξ)ϕ 2 (x) ϕ 1 (x)ϕ 2 (ξ) x 0 W[ϕ 1,ϕ 2 ](ξ) g(ξ) dξ. Example Find the general solution to the equation y + y = tanx on the interval ( π/2,π/2). The homogeneous equation y + y = 0 has two linearly independent solutions given by ϕ 1 (x) = cosx, ϕ 2 (x) = sin x. The Wronskian for these is W[ϕ 1,ϕ 2 ] = cosx sin x sin x cosx = cos2 x + sin 2 x = 1. We look for a particular solution of the form The functions v 1 and v 2 satisfy ϕ p (x) = v 1 (x)ϕ 1 (x) + v 2 (x)ϕ 2 (x) = v 1 (x) cosx + v 2 (x) sinx. v 1 (x) = ϕ 2(x) g(x) = sin x tanx, W[ϕ 1,ϕ 2 ](x) v 2 (x) = ϕ 1(x) g(x) = cosx tanx. W[ϕ 1,ϕ 2 ](x)

11 Math APPLICATIONS 44 Integration yields v 1 (x) = = sin 2 x cos 2 sin x tanx dx = cosx dx = x 1 dx cosx (cosx sec x) dx = sin x ln(secx + tanx) + c 1, and v 2 (x) = cosx tanx dx = sin x dx = cos x + c 2. We only need one particular solution, so choose c 1 = c 2 = 0. The particular solution is ϕ p (x) = v 1 (x)ϕ 1 (x) + v 2 (x)ϕ 2 (x) = [sin x ln(secx + tanx)]cosx cosx sin x = cosx ln(secx + tanx) and the general solution is y(x) = c 1 cosx + c 2 sin x cosx ln(secx + tanx). 3.9 Applications As an application of the various material covered in this chapter, we study the mechanical vibrations of a spring-mass system. Consider an block of mass m suspended from the ceiling by a spring as depicted in Figure 3.1. Let x = 0 be the equilibrium point of the mass spring combination. The forces acting on the mass are: gravity: spring: The total force acting on the mass is: F g = mg; F s = mg; restoring force: F r = kx; (Hooke s Law) friction: F f = b dx dt ; external forcing: F e = mf(t). F = F g + F s + F r + F f + F e = kx b dx dt + mf(t). k m x Figure 3.1: Mass spring combination.

12 Math APPLICATIONS 45 Apply Newton s law F = ma to get m d2 x = kx bdx dt2 dt + mf(t), or, divide by m and re-arrange to put the equation into a more convenient form: where β = b 2m, and ω2 = k m. Some terminology. The equation is said to be: undamped if b = 0 (i.e. if β = 0); damped if b > 0 (i.e. if β > 0); free if f(t) 0; forced if f(t) 0. x + 2βx + ω 2 x = f(t), (3.18) Case 1. (undamped, free vibrations: b = 0, f(t) 0) Eq. (3.18) reduces to x + ω 2 x = 0, the general solution of which is: We can re-write this as follows: x(t) = { c c2 2 x(t) = c 1 cosωt + c 2 sin ωt. c 1 cosωt + c c 2 2 c 2 sin ωt c c 2 2 = A(sinϕcos ωt + cosϕsin ωt) = A sin(ωt + ϕ), where A := c c2 2, sin ϕ = c 1 A, cosϕ = c 2 A. For undamped, free vibration the motion, called simple harmonic motion, is simply a sine wave. } Case 2. (damped, free vibrations: b > 0, f(t) 0) Eq. (3.18) reduces to x + 2βx + ω 2 x = 0. Letting x = e rt, we get r 2 + 2βt + ω 2 = 0 = r = β ± β 2 ω 2. Case 2(a) (underdamped motion: β < ω). In this case we have r = β ± iν, where ν = ω 2 β 2. The solution is: x(t) = e βt (c 1 cosνt + c 2 sin νt) or x(t) = Ae βt sin(νt + ϕ). (3.19) Note that lim x(t) = lim t t Ae βt sin(νt + ϕ) = 0.

13 Math APPLICATIONS 46 Case 2(b) (critially damped motion: β = ω). In this case the characteristic equation has a double root r = β, so the solution is: Note that If we differentiate we see that which means that x(t) = (c 1 + c 2 t)e βt. lim x(t) = lim (c 1 + c 2 t)e βt = 0. t t x (t) = (c 2 c 1 β c 2 βt)e βt x (t) = 0 = t = c 2 c 1 β c 2 β so that there is at most one maximum point or one minimum point. Case 2(c) (overdamped motion: β > ω). In this case the characteristic equation has two real roots: r 1 = β + β 2 ω 2 and r 2 = β β 2 ω 2 and the solution is x(t) = c 1 e r 1t + c 2 e r 2t. Note that both r 1 and r 2 are negative so that If we differentiate we see that lim x(t) = 0. t x (t) = e r 1t (c 1 r 1 + c 2 r 2 e (r 2 r 2 )t ) = e r 1t (c 1 r 1 + c 2 r 2 e 2 β 2 ω 2t ) which means that there is at most one maximum point or one minimum point. The graphs look much the same as they do for Case 2(b). Case 3. (forced vibrations: f(t) 0) We consider a periodic forcing term of the form f(t) = F 0 cosγt, where F 0, γ > 0. Eq. (3.18) becomes We look for a particular solution of the form Plug this into the equation to get x + 2βx + ω 2 x = F 0 cosγt. x p (t) = A 1 cos γt + B 1 sin γt. [(ω 2 γ 2 )A 1 + 2ββA 2 ]cosγt + [(ω 2 γ 2 )A 2 2βγA 1 ]sin γt = F 0 cosγt. This leads to two simultaneous equations for A 1 and A 2 : which are easily solved yielding (ω 2 γ 2 )A 1 + 2ββA 2 = F 0, (ω 2 γ 2 )A 2 2βγA 1 = 0, A 1 = F 0 (ω 2 γ 2 ) (ω 2 γ 2 ) 2 + 4β 2 γ 2, A 2F 0 βγ 2 = (ω 2 γ 2 ) 2 + 4β 2 γ 2,

14 Math APPLICATIONS 47 provided that either β 0 or ω γ. Thus, a particular solution is where x p (t) = F 0 (ω 2 γ 2 ) 2 + 4β 2 γ 2[(ω2 γ 2 ) cosγt + 2βγ sin γt] = F 0 M sin(γt + θ), M := 1 (ω2 γ 2 ) 2 + 4β 2 γ 2, tanθ := ω2 γ 2 2βγ. For the homogeneous solution we will only consider the underdamped case (i.e. β < ω), so that the homogeneous solution is given by (3.19): x h (t) = Ae βt sin(νt + ϕ). Therefore the general solution is: x(t) = x h (t) + x p (t) = Ae βt sin(νt + ϕ) + F 0 M sin(γt + θ). Figure 3.2: Forced vibrations with transient. Some terminology: x h (t) is called a transient solution, since lim t x h (t) = 0; x p (t) is called the steady state solution or the forced response. Let us examine the amplitude of the steady state solution. Fix β and ω and consider the amplitude as a function of γ: 1 M(γ) = (ω2 γ 2 ) 2 + 4β 2 γ 2. Figure 3.3: Amplitude M(γ).

15 Math APPLICATIONS 48 We have M(0) = 1/ω and lim M(γ) = 0. To find the maximum we differentiate: γ and see that M (γ) = 2γ(ω2 γ 2 2β 2 ) [(ω 2 γ 2 ) 2 + 4β 2 γ 2 ] 3/2 M (γ) = 0 = γ = 0, γ = γ c := ω 2 2β 2. (if β < ω/ 2) The frequency γ c is called the resonance frequency and the maximum amplitude is Notice that M(γ c ) = 1 2β ω 2 β 2, (for β < ω/ 2). lim M(γ c) = +, β 0 What happens when β = 0? Setting β = 0 yields a solution x(t) = A sin(ωt + ϕ) + lim γ c = ω. β 0 F 0 ω 2 γ 2 sin(γt + θ). For the case of pure resonance, with β = 0 and γ = ω, we can not simply plug γ = 0 into the above formulas. We need to go back to the original ODE: x + ω 2 x = F 0 cos ωt. Since the forcing term is a solution of the homogeneous equation, we look for a particular solution of the form x p (t) = t(a 1 cosωt + A 2 sin ωt), which leads to We see that the solution is unbounded. x p (t) = F 0 t sin ωt. 2ω Figure 3.4: Pure resonance. The conclusion: if β 1 and γ ω, the system is subject to large oscillations. These resonance vibrations have been known to cause airplane wings to snap, bridges to collapse, and wine glasses to shatter.