Lecture 1: Review of methods to solve Ordinary Differential Equations
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1 Introductory lecture notes on Partial Differential Equations - c Anthony Peirce Not to be copied, used, or revised without explicit written permission from the copyright owner 1 Lecture 1: Review of methods to solve Ordinary Differential Equations (Compiled 6 January 018) In this lecture we will briefly review some of the techniques for solving First Order ODE and Second Order Linear ODE, including Cauchy-Euler/Equidimensional Equations Key Concepts: First order ODEs: Separable and Linear equations; Second Order Linear ODEs: Constant Coefficient Linear ODE, Cauchy-Euler/Equidimensional Equations 1 First Order ordinary Differential Equations: 11 Separable Equations: Example 1: = P (x)q(y) (11) dx Q(y) = P (x) dx + C dx = 4y x(y 3) ( ) y 3 = 4 y x dx y 3 ln y = 4 ln x + C (1) y = ln(x 4 y 3 ) + C Ax 4 y 3 = e y 1 Linear First Order equations - The Integrating Factor: y (x) + P (x)y = Q(x) (13) Can we find a function F (x) to multiply (43) by in order to turn the left hand side into a derivative of a product: F y + F P y = F Q (14) (F y) = F y + F y = F Q (15)
2 So let F = F P which is a separable Eq df df = P (x) dx F (x) F = P (x) dx + C Therefore ln F = P (x) dx + C (16) or F = Ae P (x) dx F = e P (x) dx choose A = 1 integrating factor Therefore e P (x) dx y + e P (x) dx P (x)y = e P (x) dx Q(x) (e P (x) dx y) = e P (x) dx Q(x) y(x) = e { P (x) dx x } (17) e P (t) dt Q(x) dx + C Example : y + y = 0 (18) F (x) = e x e x y + e x y = (e x y) = 0 e x y = c y(x) = Ce x Example 3: Solve dx + cot(x)y = 5ecos x, y(π/) = 4 (19) P (x) = cot x Q(x) = 5e cos x F (x) = e cot x dx = e ln(sin x) = sin x (110) Therefore sin(x)y + cos(x)y = (sin(x)y) = 5e cos x sin x sin(x)y = 5e cos x + C y(x) = 5ecos x C sin x (111) 4 = y(π/) = 5 C 1 C = 1 Therefore y(x) = 1 5ecos x sin x Second Order Constant Coefficient Linear Equations: Ly = ay + by + cy = 0 Guess y = e rx y = re rx y = r e rx Ly = [ar + br + c]e rx = 0 provided [] = 0 Indicial Eq: g(r) = ar + br + c = 0 r 1, = b± b 4ac a or g(r) = a(r r 1 )(r r ) = 0 (1) Case I: = b 4ac > 0, r 1 r, y(x) = c 1 e r1x + c e rx is the general solution
3 Review of methods to solve Ordinary Differential Equations 3 Case II: = 0, r 1 = r, repeated roots Ly = a(r r 1 ) e rx = 0 In this case obtain only one solution y(x) = e r1x How do we get a second solution? g(r) = ar + br + c r 1 ϵ r 1 = b a r 1 + ϵ = b 4ac < 0 = b 4ac = 0 = b 4ac > 0 r c c 1 c = c 1 = 1 ϵ Figure 1 Left Figure: Roots of the characteristic polynomial g(r) = ar + br + c for the different cases of the discriminant = b 4ac We consider special solution, in which g(r) = a(r (r 1 ϵ))(r (r + ϵ)) = a[(r r 1) ϵ ] a(r r 1) Right Figure: We consider the special solution (3) for the case in which the two parameters c 1 and c have been chosen to be c = c 1 = 1, which represents a straight line in the two-parameter c1 c space ϵ First Method: Perturbation of the double root: Consider a small perturbation (see figure 1 a) to the double root case, such that g(r) = a(r (r 1 ϵ))(r (r 1 + ϵ)) = a[(r r 1 ) ϵ ] a(r r 1 ) In this case the two, very close but distinct, roots of g(r) = 0 are given by: r = r 1 + ϵ and r = r 1 ϵ () Now since we still have two distinct roots in this perturbed case, the general solution is: y(x) = c 1 e (r1+ϵ)x + c e (r1 ϵ)x (3) Now choosing a special solution by selecting c 1 = 1 ϵ = c, and we obtain a family of solutions that depend on the small parameter ϵ (see figure 1 b): y(x, ϵ) = e(r1+ϵ)x e (r1 ϵ)x ϵ r erx (4) r=r1 Now taking the limit as ϵ 0 by making use of L Hospital s Rule, we obtain the following limiting solution: ( y(x, ϵ) = e r 1x e ϵx e ϵx ) ϵ 0 xe r1x = ϵ r erx (5) r=r1 Second Method: taking the derivative with respect to r: From (4) and (5) we see that the new solution xe r1x was obtained by taking the derivative of y(x, r) = e rx with respect to r and then making the substitution r = r 1 This
4 4 is, in fact, a general procedure that we will use later in the course To see why this procedure works, let Therefore [ L y y(r, x) = e rx Ly(r, x) = a(r r 1 ) e rx r ]r=r (r, x) = [ a(r r 1 )e rx + a(r r 1 ) xe rx] r=r 1 = 0 [ 1 y r ]r=r (r, x) = xe r1x is also a solution 1 Thus, to summarize, the general solution for the case of a double root is: y(x) = c 1 e r 1x + c xe r 1x (6) (7) Case III: Complex Conjugate Roots: = b 4ac < 0 r ± = b a ± i 4ac b = λ ± iµ a y(x) = c 1 e (λ+iµ)x + c e (λ iµ)x (8) = e λx [A cos µx + B sin µx] Example 4: Ly = y + y 6y = 0 y = e rx (r + r 6) = (r + 3)(r ) = 0 (9) y(x) = c 1 e 3x + c e x Example 5: Ly = y + 6y + 9y = 0 y = e rx (r + 3) = 0 (10) y(x) = c 1 e 3x + c xe 3x Example 6: Ly = y 4y + 13y = 0 y = e rx : r 4r + 13 = 0 r = 4± 16 5 = ± 3i Therefore y(x) = e x [A cos 3x + B sin 3x] (11) 3 Cauchy/Euler/Equidimensional Equations: Ly = x y + αxy + βy = 0 (31) Aside: Note if we let t = ln x or x = e t then d dx = d dt dt dx d dt = x d dx d dt = x d ( x d ) = x d dx dx dx + x d d x dx dx = d dt d (3) dt Therefore ÿ ẏ + αẏ + βy = 0 ÿ + (α 1)ẏ + βy = 0 (33) y = e rt r + (α 1)r + β = 0 Characteristic Eq
5 Review of methods to solve Ordinary Differential Equations 5 Back to (31): Guess y = x r, y = rx r 1, and y = r(r 1)x r Therefore {r(r 1) + αr + β} x r = 0 f(r) = r + (α 1)r + β = 0 as above (34) r ± = 1 α ± (α 1) 4β Case 1: = (α 1) 4β > 0 Two Distinct Real Roots r 1, r (35) y = c 1 x r1 + c x r (36) If r 1 or r < 0 then y as x 0 Case : = 0 Double Root (r r 1 ) = 0 We obtain only one solution in this case: y = c 1 x r1 (37) To get a second solution we use second method introduced above, in which we differentiate with respect to the parameter r: r L[xr ] = L [ r xr] = L[x r log x] r {f(r)xr } = f (r)x r + f(r)x r log x = 0 since f(r) = (r r 1 ) (38) General Solution: y(x) = (c 1 + c log x)x r 1 Check: L(x r 1 log x) = x (x r log x) + αx(x r log x) + β(x r log x) = x [ r(r 1)x r log x + rx r + (r 1)x r ] (39) + αx [ rx r 1 log x + x r 1] + β(x r log x) = { r + (α 1)r + β } x r log x + {r 1 + α} x r = 0 Case 3: = (α 1) 4β < 0 Observations: (1 α) r ± = ± i [4β (α 1) ] 1/ = λ ± iµ y(x) = c 1 x (λ+iµ) + c x (λ iµ) x r = e r ln x = c 1 e (λ+iµ) ln x + c e (λ iµ) ln x (310) = x λ { c 1 e iµ ln x iµ ln + c e x} = A 1 x λ cos(µ ln x) + A x λ sin(µ ln x) If x < 0 replace by x
6 6 The two solutions are linearly independent as we can verify by applying the Wronskian test, as follows: w(y 1, y ) = y 1 y y 1 y = y 1y y 1y (look up the definition of the Wronskian) = { x λ cos(µ ln x) } { log xx λ sin(µ ln x) + x λ 1 cos(µ ln x)µ } { x λ log x cos(µ ln x) x λ 1 sin(µ ln x)µ } { x λ sin(µ ln x) } = µx λ 1 independent for x 0 Example 7: x y xy y = 0, y(1) = 0, y (1) = 1 y = x r r(r 1) r = 0 r r = 0 (r 1) = 3 r = 1 ± 3 (311) y = c 1 x c x 1 3 y(1) = c 1 + c = 0 c = c 1 y(x) = c 1 (x 1+ 3 x 1 3 ) (31) Example 8: y (x) = c 1 [ (1 + 3 ) x 3 ( 1 3 ) x 3 ] x=1 = c 1 3 = 1 Therefore y(x) = 1 3 ( x 1+ 3 x 1 3 ) (313) x y 3xy + 4y = 0 y(1) = 1 y (1) = 0 y = x r = r(r 1) 3r + 4 = r 4r + 4 = 0 (r ) = 0 (314) y(x) = c 1 x + c x log x y(1) = c 1 = 1 y (1) = [x + c (x log x + x)] x=1 (315) = + c = 0 Therefore y(x) = x x log x
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