SOLUTION SET VI FOR FALL [(n + 2)(n + 1)a n+2 a n 1 ]x n = 0,
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1 4. Series Solutios of Differetial Equatios:Special Fuctios 4.. Illustrative examples.. 5. Obtai the geeral solutio of each of the followig differetial equatios i terms of Maclauri series: d y (a dx = xy, d y dy (b dx + x dx y = 0. Solutio. (a Try the Maclauri series y = =0 a x to get + xy = a x = a x, a = 0, =0 =0 d y = ( a x = ( + ( + a + x. dx = =0 The differetial equatio yields [( + ( + a + a ]x = 0, =0 which is satisfied by all x i some eighborhood of x 0 = 0. Hece, the recurrece formula (relatio for the coefficiets a reads ( + ( + a + = a ; a = 0, = 0,,, 3,.... Fid the coefficiets explicitly for various : = 0 : a = 0 = : 3 a 3 = a 0 = : 4 3a 4 = a = 3 : 5 4a 5 = a = 4 : 6 5a 6 = a 3 = 5 : 7 6a 7 = a 4 = 6 : 8 7a 8 = a 5,.... Notice that a 0 ad a are idepedet ad arbitrary, while all coefficiets a, a 5, a 8,... a = 0. Date: October, 00.
2 The correspodig power series for y(x reads as x4 x 7 x3+ y(x = a x (3 4(6 7 (3 4( [3 (3 + ] x3 x 6 x3 +a ( 3(5 6 ( 3( [(3 + (3 + 3] (b Oce agai, we try the Maclauri series y(x =0 a x to get dy d y x = a x, = ( + ( + a + x, dx dx =0 =0 which i tur lead to the equatio [( + ( + a + + ( a ]x = 0, =0 satisfied by all x i some eighborhood of x 0 = 0. It follows that ( + ( + a + = ( a, = 0,,, 3,.... Write the esuig coefficiets explicitly: = 0 : a = a 0, = : 3 a 3 = 0 a = 0, = : 4 3a 4 = a, = 3 : 5 4a 5 = a 3 = 0, = 4 : 6 5a 6 = 3a 4, = 5 : 7 6a 7 = 4a 5 = 0. It follows that a 0 ad a are idepedet ad arbitrary. Further, all coefficiets with odd idex are zero, with the exceptio of a (sice the right-had side of the equatio for = vaishes. The fial Maclauri series for y(x reads as x x 4 3x 6 3 5x 8 y(x = a ! 4! 6! 8! +( 3... ( x a x. (! Notice that the idepedet solutio ivolvig a is u(x = x. 6. For each of the followig equatios, obtai the most geeral solutio which is represetable by a Maclauri series: d y d y (a dx + y = 0, (b dx (x 3y = 0, ( d y dy (d x d y dy (c x + x dx dx y = 0, dx dx + y = 0,
3 3 y (e (x + x d dy dx (x dx (x + y = 0. Obtai three ovaishig terms i each ifiite series ivolved. Solutio. (a With y(x = =0 A x, the recurrece formula for the coefficiets A is Specifically, ( + ( + A + + A = 0, = 0,,, 3,.... Specifically, = 0 : 3 A + 3 = 0 A =, = : A 3 A 3 + 3A = 0 A 3 =, 3 = : A A A 3 4 3A 4 A + 3A = 0 A 4 = = +, It follows that A 3 A 3 4 y(x = + A x 3 x + x + + x x x 4 = 3 x 3 + x... + A x (c With y(x = =0 A x, we get x dy = A x, x d y = ( A x, dx dx =0 =0 = 0 : A + = 0 A =, = : 3 A 3 + A = 0 A 3 =, 3 A = : 4 3A 4 + A = 0 A 4 = =, 3 4 4! A 3 A = 3 : 5 4A 5 + A 3 = 0 A 5 = =, ! It follows that x x4 x3 x5 y(x = A x +....! 4! 3! 5! (b Agai, start with y(x = A x =0 ad xy(x = =0 A x, where A = 0, to arrive at the recurrece formula ( + ( + A + A + 3A = 0; A = 0, = 0,,,.... ad we fid the recurrece formula ( + ( + A + ( ( A = 0. A
4 4 Try differet values of : It follows that all coefficiets A with 3 vaish! Hece, x y(x = + + A x. (d Clearly, The recurrece formula is Specifically, Hece, (e Clearly, dy = ( + A + x, dx d y x = ( A x. (x + y = A x + A x, A = 0, dy (x = ( A x ( + A + x, dx (x + x d y = ( A x + ( + A + x. dx =0 =0 By puttig all these terms together, the recurrece formula reads = 0 : A = 0 A =, = : 3 A 3 0 = 0 A 3 = 0, = : 4 3A 4 = 0, = 3 : 5 4A 5 = A 3 = 0, = 4 : 6 5A 6 3A 4 = 0 A 6 = 0, = 5 : 7 6A 7 3A 5 = 0 A 7 = 0 etc. =0 dx =0 [( + ]A = ( + A +, = 0,,,..... = 0 : = A, = : A = A A =, = : 3A = 3A 3 A 3 = etc. x y(x = + x =0 =0 =0 =0 ( ( + A + ( + ( + A + A = 0; A = 0, = 0,,,....
5 5 Specifically, Fially, = 0 : + A = 0 = A, = : A + 3A = 0 A = etc. x y(x = + x Sigular poits of liear, secod-order differetial equatios.. 8. Locate ad classify the sigular poits of the followig differetial equatios: (a (x y + xy = 0 (x 0, (b y + y log x + xy = 0 (x 0, (c xy + y si x = 0, (d y x y = 0, (e y + y cos x = 0 (x 0. Solutio. (a The sigular poits are x = ad x = 0. x = is a regular sigular poit sice (x x x (x = (x x has a Taylor expasio ear x =. Sice (x (x x=0 x does ot exist, x (x does ot have a Taylor expasio ear x = 0. So x = 0 is a irregular sigular poit. (b The sigular poit is x = 0, which is irregular sice x log x is ot differetiable at x = 0. x (c There are o sigular poits. (Note that si x x = = ( (!. (d The sigular poits are x = ad x =. Sice either ((x x x= or ((x + x x= is well defied, both sigular poits are irregular. (e There are o sigular poits. (Note that cos x x = =0 ( (! The Method of Frobeius... Use the method of Frobeius to obtai the geeral solutio of each of the followig differetial equatios, valid ear x = 0: (a xy + ( xy y = 0, (b x y + xy + (x 4 y = 0, (c xy + y + xy = 0, (d x( xy y + y = 0. Solutio. (arewrite the equatio as x y + ( xy + y = 0. x x (
6 6 The we ca see that P 0 = /, P =,Q = /,ad all other P s, Q s ad R s are zeros. So f (s = s s, g (s = s + /, ad g (s = 0 if =. f (s = 0 has two roots: A s = ad s = 0. Take s = 0, the = A, for all. Hece, by iductio, A =! for all 0. Therefore x y = A = A0 e x 0! = A Now, take s = /, the A = +, for all. Therefore y = x A x =0 Here ( +!! = ( +. The geeral solutio is the of the form: x + C x y(x = C e x. ( +!! (b Rewrite the equatio as A x = x x. ( +!! =0 =0 y + x y + x (x 4 y = 0. The we ca see that P 0 =, Q 0 = 4, Q =, ad all other P s, Q s ad R s are zeros. So f (s = s 4, g (s =, ad g (s = 0 if =. f (s = 0 has two roots: s = ad s =. For s = we have A = ( A for all. From this, it easy to check by ( iductio that A = ( ad A + = (+! A for all 0. So, i this case, y = (! x =0 ( ( + = x ( x + A x ( x (! ( +! = =0 =0 x cos x + A x si x. The geeral solutio is the of the form y = c 0 x cos x + c x si x. (c Rewrite the equatio as x y + y + x x y = 0. The we ca see that P 0 =, Q =, ad all other P s, Q s ad R s are zeros. So f (s = s + s, g (s =, ad g (s = 0 if =. f (s = 0 has two roots: s = ad s = 0.
7 7 ( For s =, we have A = A. So A = ( A0 ad A + = ( (! (+! A for all 0. The y = x A x =0 ( x + = x ( (A x + ( +! (! =0 =0 = x (A si x + cos x. The geeral solutio is the of the form (d Rewrite the equatio as y = x (c si x + c 0 cos x. x ( xy y + x x y = 0. The we ca see R =, P 0 =, Q =, ad all other P s, Q s ad R s are zeros. So f (s = s 3s, g (s = s + 3s, ad g (s = 0 for all >. f (s has two roots: s = 3 ad s = 0. g For s = 0, A = ( f ( A = A for all, = 3. Thus, A = A =, ad A 3 = A 4 = A 5 =. So, i this case, y = x 0 A x =0 = ( + x + x + A 3 x 3 x x 3 = x + A 3 x. The geeral solutio is the of the form x 3 x 3 y = c 0 + c. x x =0. Use the method of Frobeius to obtai the geeral solutio of each of the followig differetial equatios, valid ear x = 0: (a x y xy + ( x y = 0, (b (x y xy + y = 0, (c xy y + 4x 3 y = 0, (d ( cos xy si xy + y = 0. Solutio. (arewrite the equatio as y y + ( x y = 0. x x
8 8 The we ca see that P 0 =, Q 0 =,Q = ad all other P s, Q s ad R s are zeros. So f (s = s 3s +, g (s =, ad g (s = 0 if =. f (s = 0 has two roots: s = ad s =. For s =, we have A A = ( A for. From this, it s easy to check by iductio that A = ( 0! ad A + = (+! for all 0. So + y = x x + A x = x( cosh (x + A sih (x. (! ( +! =0 =0 The geeral solutio is the of the form (b Rewrite the equatio as y = c 0 x cosh (x + c x sih (x. x ( xy + xy y = 0. x The we ca see that R =, P =, Q =, ad all other P s, Q s ad R s are zeros. So f (s = s s, g (s = (s (s, g (s = s 3, ad g (s = 0 if 3. f (s = 0 has two roots: s = 0 ad s =. For s = 0, we have g (A + g (A 3 A = = f ( A A ( for. From this, it s easy to check by iductio that A =! if. So x y = + + A x = (e x x + A x = e x + (A x.! = Hece the geeral solutio is of the form (c Rewrite the equatio as x y = c 0 e + c x. 4x 4 y y + y = 0. x x The we ca see that Q 4 = 4, P 0 =, ad all other P s, Q s ad R s are zeros. So f (s = s s, g 4 (s = 4, ad g (s = 0 if = 4. f (s = 0 has two roots: s = 0 ad s =. 4 For s = 0, we have A = A 3 = 0, ad A = ( A 4 for all 4. From these, it s ( easy to check by iductio that A + = 0, A 4 = ( (!, ad A 4+ = (+! A for all 0. So ( ( 4+ y = x 4 + A x = cos (x + A si (x. (! ( +! =0 =0 A
9 9 The geeral solutio is the of the form y = c 0 cos (x + c si (x. (d Rewrite the equatio as ( ( + x y + x y + ( +! x ( +! x y = 0. =0 =0 The we ca see that Q 0 =, P = ( + ( (+!, R = (+! for all 0, ad all other P s, Q s ad R s are zeros. So f (s = (s (s, ad g (s = 0, g (s = ( (+!(s (s 4 3 for all. f (s = 0 has two roots: s = ad s =. For s =,usig the equatio f (s + A = g k (s + A k, k= ( it s easy to check by iductio that A = (+!, ad A + = ( (+! A for all 0. So y = x A x =0 ( ( = x x + A x x + ( +! ( +! =0 =0 + ( ( = x + A x + ( +! ( +! =0 =0 = si x + A ( cos x. The geeral solutio is the of the form y = c 0 si x + c ( cos x.
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