The general solution of the differential equation is thus y = c 1 cos x + c 2 sin x + x. Applying the two initial conditions gives us the equations c 1 = 1, c 2 + 1 = 0. The solution subject to the initial conditions is y = cos x − sin x + x. Solution 21.5 Solve x 2 y  (x) − xy  (x) + y(x) = x. The homogeneous equation is x 2 y  (x) − xy  (x) + y(x) = 0. Substituting y = x λ into the homogeneous differential equation yields x 2 λ(λ − 1)x λ−2 − xλx λ + x λ = 0 λ 2 − 2λ + 1 = 0 (λ − 1) 2 = 0 λ = 1. The homogeneous solutions are y 1 = x, y 2 = x log x. The Wronskian of the homogeneous solutions is W [x, x log x] =     x x log x 1 1 + log x     = x + x log x −x log x = x. 1134 Writing the inhomogeneous equation in the standard form: y  (x) − 1 x y  (x) + 1 x 2 y(x) = 1 x . Using variation of parameters to find the particular solution, y p = −x  log x x dx + x log x  1 x dx = −x 1 2 log 2 x + x log x log x = 1 2 x log 2 x. Thus the general solution of the inhomogeneous differential equation is y = c 1 x + c 2 x log x + 1 2 x log 2 x. Solution 21.6 1. First we find the homogeneous solutions. We substitute y = e λx into the homogeneous differential equation. y  + y = 0 λ 2 + 1 = 0 λ = ±ı y =  e ıx , e −ıx  We can also write the solutions in terms of real-valued functions. y = {cos x, sin x} 1135 The Wronskian of the homogeneous solutions is W [cos x, sin x] =     cos x sin x −sin x cos x     = cos 2 x + sin 2 x = 1. We obtain a particular solution with the variation of parameters formula. y p = −cos x  e x sin x dx + sin x  e x cos x dx y p = −cos x 1 2 e x (sin x − cos x) + sin x 1 2 e x (sin x + cos x) y p = 1 2 e x The general solution is the particular solution plus a linear combination of the homogeneous solutions. y = 1 2 e x + cos x + sin x 2. y  + λ 2 y = sin x, y(0) = y  (0) = 0 Assume that λ is positive. First we find the homogeneous solutions by substituting y = e αx into the homogeneous differential equation. y  + λ 2 y = 0 α 2 + λ 2 = 0 α = ±ıλ y =  e ıλx , e −ıλx  y = {cos(λx), sin(λx)} 1136 The Wronskian of these homogeneous solution is W [cos(λx), sin(λx)] =     cos(λx) sin(λx) −λ sin(λx) λ cos(λx)     = λ cos 2 (λx) + λ sin 2 (λx) = λ. We obtain a particular solution with the variation of parameters formula. y p = −cos(λx)  sin(λx) sin x λ dx + sin(λx)  cos(λx) sin x λ dx We evaluate the integrals for λ = 1. y p = −cos(λx) cos(x) sin(λx) − λ sin x cos(λx) λ(λ 2 − 1) + sin(λx) cos(x) cos(λx) + λ sin x sin(λx) λ(λ 2 − 1) y p = sin x λ 2 − 1 The general solution for λ = 1 is y = sin x λ 2 − 1 + c 1 cos(λx) + c 2 sin(λx). The initial conditions give us the constraints: c 1 = 0, 1 λ 2 − 1 + λc 2 = 0, For λ = 1, (non-resonant forcing), the solution subject to the initial conditions is y = λ sin(x) − sin(λx) λ(λ 2 − 1) . 1137 Now consider the case λ = 1. We obtain a particular solution with the variation of parameters formula. y p = −cos(x)  sin 2 (x) dx + sin(x)  cos(x) sin x dx y p = −cos(x) 1 2 (x − cos(x) sin(x)) + sin(x)  − 1 2 cos 2 (x)  y p = − 1 2 x cos(x) The general solution for λ = 1 is y = − 1 2 x cos(x) + c 1 cos(x) + c 2 sin(x). The initial conditions give us the constraints: c 1 = 0 − 1 2 + c 2 = 0 For λ = 1, (resonant forcing), the solution subject to the initial conditions is y = 1 2 (sin(x) − x cos x). Solution 21.7 1. A set of linearly independent, homogeneous solutions is {cos t, sin t}. The Wronskian of these solutions is W (t) =     cos t sin t −sin t cos t     = cos 2 t + sin 2 t = 1. We use variation of parameters to find a particular solution. y p = −cos t  g(t) sin t dt + sin t  g(t) cos t dt 1138 The general solution can be written in the form, y(t) =  c 1 −  t a g(τ) sin τ dτ  cos t +  c 2 +  t b g(τ) cos τ dτ  sin t. 2. Since the initial conditions are given at t = 0 we choose the lower bounds of integration in the general solution to be that point. y =  c 1 −  t 0 g(τ) sin τ dτ  cos t +  c 2 +  t 0 g(τ) cos τ dτ  sin t The initial condition y(0) = 0 gives the constraint, c 1 = 0. The derivative of y(t) is then, y  (t) = −g(t) sin t cos t +  t 0 g(τ) sin τ dτ sin t + g(t) cos t sin t +  c 2 +  t 0 g(τ) cos τ dτ  cos t, y  (t) =  t 0 g(τ) sin τ dτ sin t +  c 2 +  t 0 g(τ) cos τ dτ  cos t. The initial condition y  (0) = 0 gives the constraint c 2 = 0. The solution subject to the initial conditions is y =  t 0 g(τ)(sin t cos τ − cos t sin τ) dτ y =  t 0 g(τ) sin(t − τ) dτ 3. The solution of the initial value problem y  + y = sin(λt), y(0) = 0, y  (0) = 0, is y =  t 0 sin(λτ) sin(t − τ) dτ. 1139 Figure 21.5: Non-resonant Forcing For λ = 1, this is y = 1 2  t 0  cos(t − τ − λτ) − cos(t − τ + λτ )  dτ = 1 2  − sin(t − τ − λτ) 1 + λ + sin(t − τ + λτ) 1 − λ  t 0 = 1 2  sin(t) − sin(−λt) 1 + λ + −sin(t) + sin(λt) 1 − λ  y = − λ sin t 1 − λ 2 + sin(λt) 1 − λ 2 . (21.6) The solution is the sum of two periodic functions of period 2π and 2π/λ. This solution is plotted in Figure 21.5 on the interval t ∈ [0, 16π] for the values λ = 1/4, 7/8, 5/2. 1140 Figure 21.6: Resonant Forcing For λ = 1, we have y = 1 2  t 0  cos(t − 2τ) − cos(tau)  dτ = 1 2  − 1 2 sin(t − 2τ) − τ cos t  t 0 y = 1 2 (sin t − t cos t) . (21.7) The solution has both a periodic and a transient term. This solution is plotted in Figure 21.5 on the interval t ∈ [0, 16π]. Note that we can derive (21.7) from (21.6) by taking the limit as λ → 0. lim λ→1 sin(λt) − λ sin t 1 − λ 2 = lim λ→1 t cos(λt) − sin t −2λ = 1 2 (sin t − t cos t) 1141 Solution 21.8 Let y 1 , y 2 and y 3 be linearly independ ent homogeneous solutions to the differential equation L[y] = y  + p 2 y  + p 1 y  + p 0 y = f(x). We will look for a particular solution of the form y p = u 1 y 1 + u 2 y 2 + u 3 y 3 . Since the u j ’s are undetermined functions, we are free to impose two constraints. We choose the constraints to simplify the algebra. u  1 y 1 + u  2 y 2 + u  3 y 3 = 0 u  1 y  1 + u  2 y  2 + u  3 y  3 = 0 Differentiating the expression for y p , y  p = u  1 y 1 + u 1 y  1 + u  2 y 2 + u 2 y  2 + u  3 y 3 + u 3 y  3 = u 1 y  1 + u 2 y  2 + u 3 y  3 y  p = u  1 y  1 + u 1 y  1 + u  2 y  2 + u 2 y  2 + u  3 y  3 + u 3 y  3 = u 1 y  1 + u 2 y  2 + u 3 y  3 y  p = u  1 y  1 + u 1 y  1 + u  2 y  2 + u 2 y  2 + u  3 y  3 + u 3 y  3 Substituting the expressions for y p and its derivatives into the differential equation, u  1 y  1 + u 1 y  1 + u  2 y  2 + u 2 y  2 + u  3 y  3 + u 3 y  3 + p 2 (u 1 y  1 + u 2 y  2 + u 3 y  3 ) + p 1 (u 1 y  1 + u 2 y  2 + u 3 y  3 ) + p 0 (u 1 y 1 + u 2 y 2 + u 3 y 3 ) = f(x) u  1 y  1 + u  2 y  2 + u  3 y  3 + u 1 L[y 1 ] + u 2 L[y 2 ] + u 3 L[y 3 ] = f(x) u  1 y  1 + u  2 y  2 + u  3 y  3 = f(x). 1142 [...]... u3 y 3 = 0 u 1 y 1 + u2 y 2 + u3 y 3 = 0 u1 y1 + u2 y2 + u3 y3 = f (x) We solve for the uj using Kramer’s rule u1 = (y2 y3 − y2 y3 )f (x) , W (x) u2 = − (y1 y3 − y1 y3 )f (x) , W (x) u3 = (y1 y2 − y1 y2 )f (x) W (x) Here W (x) is the Wronskian of {y1 , y2 , y3 } Integrating the expressions for uj , the particular solution is yp = y1 (y2 y3 − y2 y3 )f (x) dx + y2 W (x) (y3 y1 − y3 y1 )f (x) dx + y3... the homogeneous solution sin(3x) and no solution otherwise π (1 + αx) sin(3x) dx = 0 πα + 2 3 The problem has a solution only for α = −2/π For this case the general solution of the inhomogeneous differential equation is 1 2x y= 1− + c1 cos(3x) + c2 sin(3x) 9 π The one-parameter family of solutions that satisfies the boundary conditions is y= 1 9 1− 2x − cos(3x) + c sin(3x) π 3 For λ = n2 , n ∈ Z+ , y =... problem for u into the two problems: Lv ≡ (pv ) + qv = f (x), a < x < b, v(a) = 0, v(b) = 0 Lw ≡ (pw ) + qw = 0, a < x < b, w(a) = α, w(b) = β and note that the solution for u is u = v + w The problem for v has the solution, b v= g(x; ξ)f (ξ) dξ, a with the Green function, g(x; ξ) = v1 (x< )v2 (x> ) ≡ p(ξ)W (ξ) v1 (x)v2 (ξ) p(ξ)W (ξ) v1 (ξ)v2 (x) p(ξ)W (ξ) for a ≤ x ≤ ξ, for ξ ≤ x ≤ b Here v1 and v2... satisfy the left and right homogeneous boundary conditions 1157 Since g(x; ξ) is a solution of the homogeneous equation for x = ξ, gξ (x; ξ) is a solution of the homogeneous equation for x = ξ This is because for x = ξ, L ∂ ∂ ∂ g = L[g] = δ(x − ξ) = 0 ∂ξ ∂ξ ∂ξ If ξ is outside of the domain, (a, b), then g(x; ξ) and gξ (x; ξ) are homogeneous solutions on that domain In particular gξ (x; a) and gξ (x; b)... 1 2 3 4 5 -0.1 -0.2 -0 .3 -0.4 -0.1 -0.2 -0 .3 -0.4 -0.5 1 2 3 4 5 Figure 21.8: G(x; 1) and G(x; −1) The solutions that respectively satisfy the left and right boundary conditions are u1 = sinh(ax), u2 = cosh(a(x − L)) The Wronskian of these solutions is W (x) = sinh(ax) cosh(a(x − L)) a cosh(ax) a sinh(a(x − L)) = −a cosh(aL) Thus the Green function is g(x; ξ) = − sinh(ax) cosh(a(ξ−L)) a cosh(aL) for. .. know a closed-form formula for the an we can calculate the an in order by substituting into the difference equation The first few an are 1, 1, 2, 3, 5, 8, 13, 21, We recognize this as the Fibonacci sequence 22.2 Exact Equations Consider the sequence a1 , a2 , Exact difference equations on this sequence have the form D[F (an , an+1 , , n)] = g(n) We can reduce the order of, (or solve for first order),... = cosh(1) 1150 The Green function for the problem is then G(x|ξ) = cosh(x< ) sinh(x> − 1) , cosh(1) cosh(x) sinh(ξ−1) cosh(1) cosh(ξ) sinh(x−1) cosh(1) G(x|ξ) = for 0 ≤ x ≤ ξ, for ξ ≤ x ≤ 1 Solution 21.14 The differential equation for the Green function is G − G = δ(x − ξ), G(0|ξ) = G(∞|ξ) = 0 Note that sinh(x) and e−x are homogeneous solutions that satisfy the left and right boundary conditions, respectively... qw = 0, a < x < b, w(a) = α, w(b) = β, is w = −αp(a)gξ (x; a) + βp(b)gξ (x; b) Therefore the solution of the problem for u is b g(x; ξ)f (ξ) dξ − αp(a)gξ (x; a) + βp(b)gξ (x; b) u= a Solution 21.18 Figure 21.7 shows a plot of G(x; 1) and G(x; −1) for k = 1 1159 -4 -2 2 4 -0.1 -0.2 -0 .3 -0.4 -0.5 Figure 21.7: G(x; 1) and G(x; −1) First we consider the boundary condition u(0) = 0 Note that the solution... satisfy the homogeneous boundary conditions, it has the form G(x|ξ) = 0 cx + d/x for x < ξ for x > ξ From the continuity condition, 0 = cξ + d/ξ The jump condition yields c − d/ξ 2 = 1 Solving these two equations, we obtain G(x|ξ) = 0 1 x 2 − ξ2 2x 1146 for x < ξ for x > ξ Thus the solution is ∞ G(x|ξ)ξ 2 dξ u(x) = 0 x 1 ξ2 x− 2 2x 0 1 1 = x4 − x4 6 10 x4 = 15 = ξ 2 dξ Now to solve the homogeneous differential... −a + e−x (a cosh(ax) + sinh(ax)) + 2−1 a a+1 −ax −x e −e = 2−1 a For a = 1, we have 1 1 e −x −1 + 2x + e−2x + e−2x sinh(x) y=− 4 2 1 = − x e−x 2 Thus the solution of the problem is =− 1 a y= e−ax − e−x a2 −1 1 − 2 x e−x for a = 1, for a = 1 We note that this solution satisfies the differential equation and the boundary conditions 11 63 21. 13 Quiz Problem 21.1 Find the general solution of y − y = f (x), . u  3 y  3 + u 3 y  3 = u 1 y  1 + u 2 y  2 + u 3 y  3 y  p = u  1 y  1 + u 1 y  1 + u  2 y  2 + u 2 y  2 + u  3 y  3 + u 3 y  3 Substituting the expressions for. u 3 y  3 ) + p 0 (u 1 y 1 + u 2 y 2 + u 3 y 3 ) = f(x) u  1 y  1 + u  2 y  2 + u  3 y  3 + u 1 L[y 1 ] + u 2 L[y 2 ] + u 3 L[y 3 ] = f(x) u  1 y  1 + u  2 y  2 + u  3 y  3 =. u  2 y  2 + u  3 y  3 = 0 Differentiating the expression for y p , y  p = u  1 y 1 + u 1 y  1 + u  2 y 2 + u 2 y  2 + u  3 y 3 + u 3 y  3 = u 1 y  1 + u 2 y  2 + u 3 y  3 y  p = u  1 y  1 +