What would happen if we continued this method? Since y = x + 1 c−x is a solution of the Ricatti equation we can make the substitution, y = x + 1 c − x + 1 u(x) , (18.7) which will lead to a solution for y which has two constants of integration. Then we could repeat the process, substituting the sum of that solution and 1/u(x) into the Ricatti equation to find a solution with three constants of integration. We know that the general solution of a first order, ordinary differential equation has only one constant of integration. Does this method for Ricatti equations violate this theorem? There’s only one way to find out. We substitute Equation 18.7 into the Ricatti equation. u  = −  −2x + 2  x + 1 c − x  u − 1 u  = − 2 c − x u − 1 u  + 2 c − x u = −1 The integrating factor is I(x) = e 2/(c−x) = e −2 log(c−x) = 1 (c − x) 2 . Upon multiplying by the integrating factor, the equation b ecome s exact. d dx  1 (c − x) 2 u  = − 1 (c − x) 2 u = (c − x) 2 −1 c − x + b(c − x) 2 u = x − c + b(c − x) 2 Thus the Ricatti equation has the solution, y = x + 1 c − x + 1 x − c + b(c − x) 2 . 1014 It appears that we we have found a solution that has two constants of integration, but appearances can be deceptive. We do a little algebraic simplification of the solution. y = x + 1 c − x + 1 (b(c − x) −1)(c − x) y = x + (b(c − x) −1) + 1 (b(c − x) −1)(c − x) y = x + b b(c − x) −1 y = x + 1 (c − 1/b) −x This is actually a solution, (namely the solution we had before), with one constant of integration, (namely c−1/b). Thus we see that repeated applications of the procedure will not produce more general solutions. 3. The substitution y = − u  au gives us the second order, linear, homogeneous differential equation, u  −  a  a + b  u  + acu = 0. The solution to this linear equation is a linear combination of two homogeneous s olutions, u 1 and u 2 . u = c 1 u 1 (x) + c 2 u 2 (x) The solution of the Ricatti equation is then y = − c 1 u  1 (x) + c 2 u  2 (x) a(x)(c 1 u 1 (x) + c 2 u 2 (x)) . 1015 Since we can divide the numerator and denominator by either c 1 or c 2 , this answer has only one constant of integration, (namely c 1 /c 2 or c 2 /c 1 ). Exchanging the Dependent and Independent Variables Solution 18.6 Exchanging the dependent and independent variables in the differential equation, y  = √ y xy + y , yields x  (y) = y 1/2 x + y 1/2 . 1016 This is a first order differential equation for x(y). x  − y 1/2 x = y 1/2 d dy  x exp  − 2y 3/2 3  = y 1/2 exp  − 2y 3/2 3  x exp  − 2y 3/2 3  = −exp  − 2y 3/2 3  + c 1 x = −1 + c 1 exp  2y 3/2 3  x + 1 c 1 = exp  2y 3/2 3  log  x + 1 c 1  = 2 3 y 3/2 y =  3 2 log  x + 1 c 1  2/3 y =  c + 3 2 log(x + 1)  2/3 Autonomous Equations *Equidimensional-in-x Equations *Equidimensional-in-y Equations *Scale-Invariant Equations 1017 Chapter 19 Transformations and Canonical Forms Prize intensity more than extent. Excellence resides in quality not in quantity. The best is always few and rare - abundance lowers value. Even among men, the giants are usually really dwarfs. Some reckon books by the thickness, as if they were written to exercise the brawn more than the brain. Extent alone never rises above mediocrity; it is the misfortune of universal geniuses that in attempting to be at home everywhere are so nowhere. Intensity gives eminence and rises to the heroic in matters sublime. -Balthasar Gracian 19.1 The Constant Coefficient Equation The solution of any second order linear homogeneous diffe rential equation can be written in terms of the solutions to either y  = 0, or y  − y = 0 Consider the general equation y  + ay  + by = 0. 1018 We can solve this differential equation by making the substitution y = e λx . This yields the algebraic equation λ 2 + aλ + b = 0. λ = 1 2  −a ± √ a 2 − 4b  There are two cases to consider. If a 2 = 4b then the solutions are y 1 = e (−a+ √ a 2 −4b)x/2 , y 2 = e (−a− √ a 2 −4b)x/2 If a 2 = 4b then we have y 1 = e −ax/2 , y 2 = x e −ax/2 Note that regardless of the values of a and b the solutions are of the form y = e −ax/2 u(x) We would like to write the solutions to the general differential equation in terms of the solutions to simpler differential equations. We make the substitution y = e λx u The derivatives of y are y  = e λx (u  + λu) y  = e λx (u  + 2λu  + λ 2 u) Substituting these into the differential equation yields u  + (2λ + a)u  + (λ 2 + aλ + b)u = 0 In order to get rid of the u  term we choose λ = − a 2 . The equation is then u  +  b − a 2 4  u = 0. There are now two cases to consider. 1019 Case 1. If b = a 2 /4 then the differential equation is u  = 0 which has solutions 1 and x. The general solution for y is then y = e −ax/2 (c 1 + c 2 x). Case 2. If b = a 2 /4 then the differential equation is u  −  a 2 4 − b  u = 0. We make the change variables u(x) = v(ξ), x = µξ. The derivatives in terms of ξ are d dx = dξ dx d dξ = 1 µ d dξ d 2 dx 2 = 1 µ d dξ 1 µ d dξ = 1 µ 2 d 2 dξ 2 . The differential equation for v is 1 µ 2 v  −  a 2 4 − b  v = 0 v  − µ 2  a 2 4 − b  v = 0 We choose µ =  a 2 4 − b  −1/2 1020 to obtain v  − v = 0 which has solutions e ±ξ . The solution for y is y = e λx  c 1 e x/µ +c 2 e −x/µ  y = e −ax/2  c 1 e √ a 2 /4−b x +c 2 e − √ a 2 /4−b x  19.2 Normal Form 19.2.1 Second Orde r Equations Consider the second order equation y  + p(x)y  + q(x)y = 0. (19.1) Through a change of dependent variable, this equation can be transformed to u  + I(x)y = 0. This is known as the normal form of (19.1). The function I(x) is known as the invariant of the equation. Now to find the change of variables that will accomplish this transformation. We make the substitution y(x) = a(x)u(x) in (19.1). au  + 2a  u  + a  u + p(au  + a  u) + qau = 0 u  +  2 a  a + p  u  +  a  a + pa  a + q  u = 0 To eliminate the u  term, a(x) must satisfy 2 a  a + p = 0 a  + 1 2 pa = 0 1021 a = c exp  − 1 2  p(x) dx  . For this choice of a, our differential equation for u becomes u  +  q − p 2 4 − p  2  u = 0. Two differential equations having the same normal form are called equivalent. Result 19.2.1 The change of variables y(x) = exp  − 1 2  p(x) dx  u(x) transforms the differential equation y  + p(x)y  + q(x)y = 0 into its normal form u  + I(x)u = 0 where the invariant of the equation, I(x), is I(x) = q − p 2 4 − p  2 . 19.2.2 Higher Order Differential Equations Consider the third order differential equation y  + p(x)y  + q(x)y  + r(x)y = 0. 1022 [...]... dependent variable y = u exp − 1 3 p(x) dx 1 1 y = u − pu exp − p(x) dx 3 3 1 1 2 p(x) dx y = u − pu + (p2 − 3p )u exp − 3 9 3 1 1 1 y = u − pu + (p2 − 3p )u + (9p − 9p − p3 )u exp − 3 27 3 p(x) dx yields the differential equation 1 1 u + (3q − 3p − p2 )u + (27r − 9pq − 9p + 2p3 )u = 0 3 27 Result 19.2.2 The change of variables y(x) = exp − 1 n pn−1 (x) dx u(x) transforms the differential equation y... transformed to a constant coefficient equation Hint, Solution Exercise 19.5 Solve Bessel’s equation of order 1/2, 1 1 y + y + 1− 2 x 4x Hint, Solution 1 033 y = 0 19.6 Hints The Constant Coefficient Equation Normal Form Hint 19.1 Transform the equation to normal form Transformations of the Independent Variable Integral Equations Hint 19.2 Transform the equation to normal form and then apply the scale transformation... transformation x = λξ + µ Hint 19 .3 Transform the equation to normal form and then apply the scale transformation x = λξ Hint 19.4 Make the change of variables x = et , y(x) = u(t) Write the derivatives with respect to x in terms of t x = et dx = et dt d d = e−t dx dt d d x = dx dt Hint 19.5 Transform the equation to normal form 1 034 19 .7 Solutions The Constant Coefficient Equation Normal Form Solution 19.1 1 4... pn−1 (x) dx u(x) transforms the differential equation y (n) + pn−1 (x)y (n−1) + pn−2 (x)y (n−2) + · · · + p0 (x)y = 0 into the form u(n) + an−2 (x)u(n−2) + an 3 (x)u(n 3) + · · · + a0 (x)u = 0 10 23 19 .3 Transformations of the Independent Variable 19 .3. 1 Transformation to the form u” + a(x) u = 0 Consider the second order linear differential equation y + p(x)y + q(x)y = 0 We make the change of independent... 4x2 y = 0 3 9 To transform the equation to normal form we make the substitution y = exp − = e−x−x 1 2 2 /3 4 2+ x 3 dx u u The invariant of the equation is 1 1 24 + 12x + 4x2 − 9 4 = 1 I(x) = 4 2+ x 3 2 − 1 d 2 dx The normal form of the differential equation is then u +u=0 which has the general solution u = c1 cos x + c2 sin x Thus the equation for y has the general solution y = c1 e−x−x 2 /3 cos x +... H(x|ξ) = H(x|ξ)y(ξ) dξ, a (x−a)(ξ−b) , for x ≤ ξ b−a (x−b)(ξ−a) , for x ≥ ξ, b−a (x−a) (x−a)(ξ−b) [p (ξ) − q(ξ)] b−a p(ξ) + b−a (x−b) (x−b)(ξ−a) [p (ξ) − q(ξ)] b−a p(ξ) + b−a 1 031 for x ≤ ξ for x ≥ ξ 19.5 Exercises The Constant Coefficient Equation Normal Form Exercise 19.1 Solve the differential equation 1 4 24 + 12x + 4x2 y = 0 y + 2+ x y + 3 9 Hint, Solution Transformations of the Independent Variable... solutions of the differential equation that satisfy the left and right boundary conditions are c1 (x − a) and c2 (x − b) Thus the Green’s function has the form G(x|ξ) = c1 (x − a), c2 (x − b), for x ≤ ξ for x ≥ ξ Imposing continuity of G(x|ξ) at x = ξ and a unit jump of G(x|ξ) at x = ξ, we obtain G(x|ξ) = (x−a)(ξ−b) , b−a (x−b)(ξ−a) , b−a for x ≤ ξ for x ≥ ξ Thus the solution of the (19.2) is y(x) = α +... )u + qu = 0 For this to be a constant coefficient equation we must have (f )2 = c1 q, and 1025 f + pf = c2 q, for some constants c1 and c2 Solving the first condition, √ f = c q, q(x) dx f =c The second constraint becomes f + pf = const q 1 −1/2 cq q + pcq 1/2 2 = const q q + 2pq = const q 3/ 2 Result 19 .3. 2 Consider the differential equation y + p(x)y + q(x)y = 0 If the expression q + 2pq q 3/ 2 is a constant... + c2 e−x−x 1 035 2 /3 sin x 4 2+ x 3 Transformations of the Independent Variable Integral Equations Solution 19.2 The substitution that will transform the equation to normal form is y = exp − = e−ax−bx 1 2 2(a + bx) dx u 2 /2 u The invariant of the equation is 1 1 d I(x) = c + dx + ex2 − (2(a + bx))2 − (2(a + bx)) 4 2 dx = c − b − a2 + (d − 2ab)x + (e − b2 )x2 ≡ α + βx + γx2 The normal form of the differential... differential equation v + (ξ 2 + A)v = 0 where 1 A = γ −1/2 − βγ 3/ 2 4 1 0 37 Solution 19 .3 The substitution that will transform the equation to normal form is 1 b 2 a+ 2 x −b −ax =x e u y = exp − dx u The invariant of the equation is 2 e 1 b 1 d d 2 a+ − I(x) = c + + 2 − x x 4 x 2 dx 2 d − 2ab e + b − b = c − ax + + x x2 β γ ≡ α + + 2 x x The invariant form of the differential equation is u + α+ β γ + 2 x x u . exp  − 2y 3/ 2 3  = y 1/2 exp  − 2y 3/ 2 3  x exp  − 2y 3/ 2 3  = −exp  − 2y 3/ 2 3  + c 1 x = −1 + c 1 exp  2y 3/ 2 3  x + 1 c 1 = exp  2y 3/ 2 3  log  x + 1 c 1  = 2 3 y 3/ 2 y =  3 2 log  x. exp  − 1 3  p(x) dx  y  =  u  − 1 3 pu  exp  − 1 3  p(x) dx  y  =  u  − 2 3 pu  + 1 9 (p 2 − 3p  )u  exp  − 1 3  p(x) dx  y  =  u  − pu  + 1 3 (p 2 − 3p  )u  + 1 27 (9p  −. 3p  )u  + 1 27 (9p  − 9p  − p 3 )u  exp  − 1 3  p(x) dx  yields the differential equation u  + 1 3 (3q − 3p  − p 2 )u  + 1 27 (27r −9pq − 9p  + 2p 3 )u = 0. Result 19.2.2 The change