objectives - ira a. fulton college of engineering and ...vps/me505/iem/05 04.pdf · v.4.1....

Chapter V ODE V.4 Power Series Solution October 2, 2019

385

VV..44 PPoowweerr SSeerriieess SSoolluuttiioonn

Objectives: After the completion of this section the student

- should recall the power series solution of a linear ODE with variable coefficients

- should recall the definitions of ordinary and singular points

- should recall the Frobenius method of the power series solution about a regular singular point

Contents: V.4.1. Definitions V.4.2. Power Series Solution V.4.3. The Method of Frobenius V.4.4. Taylor Series Solution V.4.5. Power Series Solution with Maple

V.4.6. Review Questions and Exercises

Chapter V ODE V.4 Power Series Solution October 2, 2019 386

5.4 POWER SERIES SOLUTION V.4.1. DEFINITIONS In the previous section we studied how to solve a linear ODE with constant

coefficients, and we discovered that for them we always can find n linearly independent solutions (fundamental set). For one special form of a linear ODE with variable coefficients, namely the Euler-Cauchy equation, we can find the solution by reduction to an equation with constant coefficients. In general, for linear equations with variable coefficients, the fundamental set cannot be determined by some universal anzats – their solutions have a wide variety of functional forms. But if the coefficients of the differential equation are well enough behaved functions, then the solutions of the linear ODE are analytical functions (recall Section II.4) and they can be represented by the power series expansion (analytic function) about some point 0x x= :

( ) ( ) ( ) ( )n n0 1 0 0 n 0

n 0y x c c x x ... x x ... c x x

∞

=

= + − + + − + = −∑ (1)

where coefficients nc ∈ have to be determined.

power series ( ) ( )nn 0

n 0y x c x x

∞

=

= −∑ power series about 0x x=

( ) nn

n 0y x c x

∞

=

= ∑ power series about x 0=

convergence Series converges at x if the sequence of partial sums converges

( )N n

n 0n 0

c x x S=

− →∑ ⇒ ( )nn 0

n 0c x x S

∞

=

− =∑

absolute convergence if the series of absolute values ( )nn 0

n 0c x x

∞

=

−∑ converges.

radius of convergence n

nn 1

cR lim

c→∞+

=

( )nn 0

n 0c x x

∞

=

−∑ absolutely converges for all 0 0x R x x R− < < +

Convergence at boundary points 0x x R= ± has to be investigated separately.

analytic function Function is called analytic at 0x if it has a Taylor series expansion about 0x x= .

Power series defines an analytic function in its interval of convergence:

( ) ( )nn 0

n 0f x c x x

∞

=

= −∑ , ( )0 0x x R,x R∈ − +

shift of index 0 0

n k mn m k

n n m n ka x a x

∞ ∞+

−= = +

=∑ ∑ = += −

m n kn m k

“dummy index”

identity theorem If ( )nn 0

n 0c x x 0

∞

=

− =∑ for all x , then all coefficients nc 0=

0x0x R+0x R−

diverges divergesconverges

interval of convergence


387

operations with series Let ( ) ( )nn 0

n 0f x a x x

∞

=

= −∑ and ( ) ( )nn 0

n 0g x b x x

∞

=

= −∑ converge in ( )0 0x R,x R− + :

summation ( ) ( ) ( )( )nn n 0

n 0f x g x a b x x

∞

=

± = ± −∑

multiplication ( ) ( )f x g x⋅ ( ) ( )n nn 0 n 0

n 0 n 0a x x b x x

∞ ∞

= =

= − ⋅ −

∑ ∑ ( )n

n 0n 0

c x x∞

=

= −∑ , n

n k n kk 0

c a b −=

= ∑

differentiation ( ) ( )n 1n 0

n 1f x na x x

∞−

=

′ = −∑

( ) ( ) ( )n 2n 0

n 1f x n n 1 a x x

∞−

=

′′ = − −∑

integration ( ) ( ) ( )n n 1nn 0 0

n 0 n 0

af x dx a x x dx x xn 1

∞ ∞+

= =

= − = −+∑ ∑∫ ∫

Taylor series ( )f x ( ) ( ) ( )k

k00

k 0

f xx x

k !

∞

=

= −∑

( ) ( ) ( )( ) ( ) ( )k

k0 00 0 0

f x f xf x x x ... x x ...

1! k !′

= + − + + − +

Maclaurin series ( )f x ( ) ( )k

k

k 0

f 0x

k !

∞

=

= ∑ ( ) ( ) ( ) ( )kkf 0 f 0

f 0 x ... x ...1! k !′

= + + + +

Table of Taylor series 1. 11 x−

2 k1 x x ... x ...= + + + + + k

k 0x

∞

=

= ∑ 1 x 1− < <

2. 11 x+

( )k k

k 01 x

∞

=

= −∑ 1 x 1− < <

3. 1x

( ) ( )k k

k 01 x 1

∞

=

= − −∑ 0 x 2< <

4. sin x ( )( )

k2k 1

k 0

1x

2k 1 !

∞+

=

−=

+∑ x−∞ < < ∞

5. cos x ( )( )

k2k

k 0

1x

2k !

∞

=

−= ∑ x−∞ < < ∞

6. sinh x ( )

2k 1

k 0

x2k 1 !

+∞

=

=+∑ x−∞ < < ∞

7. cosh x ( )

2k

k 0

x2k !

∞

=

= ∑ x−∞ < < ∞

8. ( )ln 1 x+( )k

k 1

k 0

1x

k 1

∞+

=

−=

+∑ 1 x 1− < <

9. xe k

k 0

xk !

∞

=

= ∑ x−∞ < < ∞

For more details about the sequences and series see Section II.4, p.120


2nd order ODE In our analysis, without significant loss of generality, we restrict ourselves to solution of the 2nd order homogeneous linear ODE for x D∈ ⊂ :

( ) ( ) ( )0 1 2a x y a x y a x y 0′′ ′+ + = general form (2)

In which coefficients ( )ja x are analytic functions in D. Dividing equation (2)

by the leading coefficient ( )0a x we can rewrite it in the normal form

( ) ( )y p x y q x y 0′′ ′+ + = standard form (3)

( )p x and ( )q x where coefficients are ( ) ( )( )

1

0

a xp x

a x= and ( ) ( )

( )2

0

a xq x

a x= . However, we do not

require ( )0a x 0≠ for all x D.∈ We will distinguish the points of the domain by the following criteria:

ordinary point The point 0x x= is called the ordinary point of the ODE (2), if ( )0 0a x 0≠ . singular point The point 1x x= is called the singular point of the ODE (2), if ( )0 1a x 0= . regular singular point The singular point 1x x= is called the regular singular point of ODE (3), if

( ) ( ) ( ) ( )2 k0 1 1 2 1 k 1

k 0xp x p p x x p x x ... p x x

∞

=

= + − + − + = −∑ (4)

( ) ( ) ( ) ( )2 k20 1 1 2 1 k 1

k 0x q x q q x x q x x ... q x x

∞

=

= + − + − + = −∑ (5)

are analytic at 1x x= .

If the point 0x x= is an ordinary point then we will be able to find two linearly independent solutions of ODE (2) in the form of the power series (1) about the ordinary point (analytic solution). If the point 1x x= is a singular point, then for the special case of the regular singular point we will be able to find one solution in the form of the power series expansion about the singular point, and the second solution can be constructed in some special way by the Frobenius method.

Examples 1) ( )2y x y sin x y 0′′ ′+ + = all points are ordinary

2) ( ) 2x 2 y x y y 0′′ ′− + + = 1x 2= is the only singular point

3) ( )2 2x 1 y xy x y 0′′ ′+ + + = 1x i= − , 2x i= are the singular points

4) ( )xy x 1 y y 0′′ ′+ − − = ⇒ x 1 1y y y 0x x− ′′ ′+ − =

( )xp x 1 x= − + is analytic, 0p 1= −

( )2x q x x= − is analytic, 0q 0=

Therefore, x 0= is a regular singular point.


389

V.4.2. POWER SERIES SOLUTION Theorem 1 (power series solution about the ordinary point)

Let all coefficients ( )na x of ODE (2) be analytic in some open interval D ⊂ , and let 0x D∈ be an ordinary point of ODE (2). Then two linearly independent solutions of ODE (2) can be found in the form of the power series

( ) ( )nn 0

n 0y x c x x

∞

=

= −∑

with the radius of convergence at least (can be bigger) 0 1R x x= − where 1x ∈ is a singular point (can be complex) nearest to 0x .

In other words, if coefficients of equation (2) are analytical, then analytical solution of (2) can be found.

The procedure of finding the solution consists of the following steps:

1. Choose an ordinary point 0x (usually, 0x 0= , if not, then by the change of variable 0x xξ = − , the ordinary point can be translated to 0x 0= ).

2. Assume the solution in the form ( ) nn

n 0y x c x

∞

=

= ∑ and substitute it into ODE (2)

(recall that absolutely convergent power series can be differentiated term by term).

3. Combine the obtained equation into a singular power series and use the Identity Theorem to establish the recurrence relation for coefficients nc . Application of the recurrence relation will yield two sets of coefficients containing the arbitrary constants 0c and 1c .

4. Each set of coefficients will produce a linearly independent solution in the form of the power

series (1). Then both solutions can be tested for the radius of convergence.

Example 1 (power series solution about the ordinary point)

Find the general solution of the differential equation y xy 2y 0′′ ′+ + = (6)

Solution: We will look for the solution of this linear ODE with variable coefficients in the form of power series according to Theorem 1.

All coefficients of the given equation are analytic for all x . There are no singular points of this equation. 1. Choose an ordinary point 0x 0= . Assume the solution in the form

( ) nn

n 0y x c x

∞

=

= ∑

2. Find the derivatives of the assumed solution (differentiate term by term):

( ) n 1n

n 1y x nc x

∞−

=

′ = ∑ (summation effectively starts with n 1= )

( ) ( ) n 2n

n 2y x n n 1 c x

∞−

=

′′ = −∑ (summation effectively starts with n 2= )

Substitute into equation (5):

( ) n 2 n 1 nn n n

n 2 n 1 n 0n n 1 c x x nc x 2 c x 0

∞ ∞ ∞− −

= = =

− + + =∑ ∑ ∑

( ) n 2 n nn n n

n 2 n 1 n 0n n 1 c x nc x 2c x 0

∞ ∞ ∞−

= = =

− + + =∑ ∑ ∑

interval ofconvergence

singularpoint

0x 0x R+

1xR

0x R−


3. To combine all sigma-terms in a single summation we need all of them to have the same index in the powers of mx and the same starting index of summation 0m m= . First, rename the indices in each sigma-term (recall that index of summation is “dummy” and that the same result can be achieved with any other letter):

( )

m n 2 n m n mn m 2

n 2 n nn n n

n 2 n 1 n 0n n 1 c x nc x 2c x 0

= − = == +

∞ ∞ ∞−

= = =

− + + =∑ ∑ ∑

( )( ) m m mm 2 m m

m 0 m 1 m 0m 2 m 1 c x mc x 2c x 0

∞ ∞ ∞

+= = =

+ + + + =∑ ∑ ∑

To have the common starting index of summation m 1= , write the first term in the first and the third sums explicitly:

( )( ) m m m2 m 2 m 0 m

m 1 m 1 m 12 1 c m 2 m 1 c x mc x 2 c 2c x 0

∞ ∞ ∞

+= = =

⋅ ⋅ + + + + + ⋅ + =∑ ∑ ∑

Now we can organize the equation in one power series

( ) ( )( ) ( ) m0 2 m 2 m

m 12 c c m 2 m 1 c m 2 c x 0

∞

+=

+ + + + + + = ∑

We have that the power series is equal to zero. Then according to the Identity Theorem all coefficients should be equal to zero:

( )0 22 c c 0+ =

( )( ) ( )m 2 mm 2 m 1 c m 2 c 0++ + + + = m 1,2,...= From which we have

2 0c c= − (7)

mm 2

cc

m 1+

−=

+ m 1,2,...= (the recurrence relation) (8)

Coefficients 0c and 1c are not defined by equations (7-8) and can be

chosen arbitrarily. Then varying the index in equations (7-8), we have: 0c arbitrary= 1c arbitrary= 2 0c c= −

m 1= 13

cc2

−=

m 2= 024

ccc3 1 3

−= =

⋅

m 3= 3 15

c cc4 2 4

−= =

⋅

m 4= 4 06

c cc

5 1 3 5− −

= =⋅ ⋅

m 5= 5 17

c cc6 2 4 6

− −= =

⋅ ⋅


391

m 2k 2= − ( )( )

k

2k 0

1c c

1 3 2k 1−

=⋅ ⋅⋅ ⋅ +

m 2k 1= − ( )k

2k 1 1

1c c

2 4 2k+

−=

⋅ ⋅⋅ ⋅

4. Substitute the obtained coefficients into the power series

( ) nn

n 0y x c x

∞

=

= ∑ ( )

( )( )k k

2k 2k 10 1

k 0

1 1c x c x

1 3 2k 1 2 4 2k

∞+

=

− −= +

⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ ∑

( )( )

( )k k2k 2k 1

0 1k 0 k 0

1 1c x c x

1 3 2k 1 2 4 2k

∞ ∞+

= =

− −= +

⋅ ⋅⋅ ⋅ + ⋅ ⋅ ⋅ ⋅∑ ∑

( )

( )( )

( )

( )21

k kk2k 2k 1

0 1 kk 0 k 0

y xy x

1 2 k ! 1c x c x

2k ! 2 k !

∞ ∞+

= =

− −= +∑ ∑

5. General solution in power series form:

( )y x ( )( )

( )k kk2k 2k 1

0 1 0 1kk 0 k 0

1 2 k ! 1 c x c x , c ,c

2k ! 2 k !

∞ ∞+

= =

− −= + ∈∑ ∑

The last expression is obtained by manipulation with the indices in the coefficients in order to obtain the simpler form of the result (this transformation is not always obvious and should not necessarily be performed). The obtained solution in the form of two power series with the arbitrary coefficients 0c and 1c is the general solution of the ODE (6). There are no singular points of the equation (6), therefore, both power series have an infinite radius of convergence (it also can be confirmed by the Ratio Test).

Use Maple to sketch the solution curves. Note that in the Maple solution

(see Example), the recurrence equation (8) is used directly for evaluation of the coefficients.

1 2 3 n ⋅ ⋅ ⋅ ⋅ ⋅

2 4 6 2n ⋅ ⋅ ⋅ ⋅ ⋅

( )1 3 5 2n 1⋅ ⋅ ⋅ ⋅ ⋅ +

Tricks with factorials

( )n n2 1 2 3 n 2 n!= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅

n!=

( ) ( )( )

( ) ( )

( )n

1 3 5 2n 1 2 4 6 2n 2 4 6 2n

1 2 3 4 5 2n 2n 12 4 6 2n

2n 1 !

2 n!

⋅ ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅=

⋅ ⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ +=

⋅ ⋅ ⋅ ⋅ ⋅+

=⋅


V.4.3. THE METHOD OF FROBENIUS

If the linear ODE with variable coefficients has singular points then the radius of convergence of the power series solution about the ordinary point can be reduced by the presence of singular points.

Sometimes an expansion about a singular point can have some advantage because such solutions can have a bigger radius of convergence. The following theorem provides the basis for the power series solution about the regular singular point. For simplicity, we formulate this theorem for the regular singular point 1x 0= .

Frobenius’ Theorem Theorem 2 (power series solution about the regular singular point 1x 0= )

Let the linear equation ( ) ( )y p x y q x y 0′′ ′+ + = (3)

has a regular singular point 1x 0= , and let

( ) 2 k0 1 2 k

k 0xp x p p x p x ... p x

∞

=

= + + + = ∑ for 1x R< (4a)

( )2 2 k0 1 2 k

k 0x q x q q x q x ... q x

∞

=

= + + + = ∑ for 2x R< (5a)

Let 1 2r and r be the roots of the indicial equation

( )20 0r p 1 r q 0+ − + = (10)

indexed such that ( ) ( )1 2Re r Re r≥ . Then the linear ODE (3) has a general solution

( ) ( ) ( )1 2y x ay x by x= +

for 0 x R< < , where { }1 2R min R ,R≥ , and linearly independent functions 1y and 2y have the form: 1. If 1 2r r− ∉ (do not differ by an integer) then

( ) 1n r1 n

n 0y x c x

∞+

=

= ∑ 0c 0≠

( ) 2n r2 n

n 0y x d x

∞+

=

= ∑ 0d 0≠

2. If 1 2r r− ∈ (differ by a positive integer) then

( ) 1n r1 n

n 0y x c x

∞+

=

= ∑ 0c 0≠

( ) ( )2n r2 n 1

n 0y x d x cy x ln x

∞+

=

= +∑ 0d 0≠ , c can be zero

3. If 1 2r r 0− = (equal roots 1 2r r r= = ) then

( ) n r1 n

n 0y x c x

∞+

=

= ∑ 0c 0≠

( ) ( )n r2 n 1

n 1y x d x y x ln x

∞+

=

= +∑

interval ofconvergence

ordinary point

singularpoint

singularpoint

converges converges

( )Ferdinand Georg Frobenius 1849-1917


393

Example 2 Use the Frobenius method to find a series solution for the ODE

( )xy x 1 y y 0′′ ′+ − − = (11)

around the point x 0= . Sketch the solution curves.

1. Point x 0= is the only singular point of the ODE (11).

Check if it is a regular singular point.

Rewrite the equation in the normal form ( ) ( )y p x y q x y 0′′ ′+ + =

x 1 1y y y 0x x− ′′ ′+ − =

( )xp x 1 x= − + is analytic 0p 1= −

( )2x q x x= − is analytic 0q 0=

Therefore, x 0= is a regular singular point.

Indicial equation (10): ( )20 0r p 1 r q 0+ − + =

( )2r 1 1 r 0 0+ − − + =

( )r r 2 0− =

roots 1r 2= , 2r 0= , 2 1r r 2− = ∈ differ by an integer

That corresponds to Case 2 of Frobenius’ Theorem:

2. The first solution ( )1y x can be found in the form:

1n r n 2n n

n 0 n 0y c x c x

∞ ∞+ +

= =

= =∑ ∑

( ) n 1n

n 0y n 2 c x

∞+

=

′ = +∑ differentiate and substitute into Equation (11)

( )( ) nn

n 0y n 2 n 1 c x

∞

=

′′ = + +∑

( )( ) ( ) ( )n n 1 n 1 n 2n n n n

n 0 n 0 n 0 n 0x n 2 n 1 c x x n 2 c x n 2 c x c x 0

∞ ∞ ∞ ∞+ + +

= = = =

+ + + + − + − =∑ ∑ ∑ ∑

( )( ) ( ) ( )n 1 n 2 n 1 n 2n n n n

n 0 n 0 n 0 n 0n 2 n 1 c x n 2 c x n 2 c x c x 0

∞ ∞ ∞ ∞+ + + +

= = = =

+ + + + − + − =∑ ∑ ∑ ∑

n m= n 2 m 1m n 1n m 1

+ = += += −

n m= n 2 m 1m n 1n m 1

+ = += += −

change of the indices

( )( ) ( ) ( )m 1 m 1 m 1 m 1m m 1 m m 1

m 0 m 1 m 0 m 1m 2 m 1 c x m 1 c x m 2 c x c x 0

∞ ∞ ∞ ∞+ + + +

− −= = = =

+ + + + − + − =∑ ∑ ∑ ∑


write the zero terms explicitly

( )( ) ( ) ( )

m 0

m 1 m 1 m 1 m 10 0 m m 1 m m 1

m 1 m 1 m 1 m 12c 2c m 2 m 1 c x m 1 c x m 2 c x c x 0

=

∞ ∞ ∞ ∞+ + + +

− −= = = =

− + + + + + − + − =∑ ∑ ∑ ∑

( )( ) ( ) ( ){ } m 1m m 1

m 1m 2 m 1 m 2 c m 1 1 c x 0

∞+

−=

+ + − + + + − = ∑

( ){ } m 1m m 1

m 1m 2 m c mc x 0

∞+

−=

+ + = ∑ use the Identity Theorem

( ) m m 1m 2 m c mc 0− + + =

( ) m m 1m 2 c c 0−+ + = ⇒ m 1m

cc

m 2−−

=+

m 1≥ recurrence relation

0c any=

m 1= 01

cc

3−

=

m 2= 012

ccc4 3 4

−= =

⋅

m 3= 023

ccc5 3 4 5

−−= =

⋅ ⋅

m k= ( )( )

( )( )

( )( )

k k k0 0 0

k

1 c 2 1 c 1 cc

3 4 5 k 2 2 3 4 5 k 2 k 2 !− − −

= = =⋅ ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ + +

( ) ( )( )

nn 2

1 0n 0

1y x c x

n 2 !

∞+

=

−=

+∑ 1st solution

3. The second solution (case 2 of Frobenius’ Theorem)

Substitution of the trial form (case 2) into equation: multiply by and add terms

n2 n 1

n 0y d x cy ln x

∞

=

= +∑ 1−

n 12 n 1 1

n 1

1y nd x cy ln x cyx

∞−

=

′ ′= + +∑ x 1−

( ) n 22 n 1 1 1 1 2

n 2

1 1 1y n n 1 d x cy ln x cy cy cyx x x

∞−

=

′′ ′′ ′ ′= − + + + −∑ x

( ) ( )n 2 n 1 nn 1 1 1 n 1 1 1 1 n 12

n 2 n 1 n 0

2 1 1 1x n n 1 d x xcy ln x xcy xcy x 1 nd x xcy ln x xcy cy ln x cy d x cy ln x 0x x xx

∞ ∞ ∞− −

= = =

′′ ′ ′ ′− + + − + − + + − − − − =∑ ∑ ∑

2 3 4 50 0 0 01

c c c cy x x x x ...

2! 3! 4! 5!= − + − +

02 is absorbed by c


395

( ) n 1 n n 1 nn n n n 1 1 1 1 1 1 1 1

n 2 n 1 n 1 n 0

1 1n n 1 d x nd x nd x d x xcy ln x xcy ln x 2cy cy cy cy ln x cy cy ln x 0xx

∞ ∞ ∞ ∞− −

= = = =

′′ ′ ′ ′− + − − + + + − + − − − =∑ ∑ ∑ ∑

( )1

n 1 n n 1 nn n n n 1 1 1 1 1 1 1

n 2 n 1 n 1 n 0 0 because y is a solution

2n n 1 d x nd x nd x d x 2cy cy cy c xy xy y y ln x 0x

∞ ∞ ∞ ∞− −

= = = =

′ ′′ ′ ′− + − − + − + + + − − =

∑ ∑ ∑ ∑

( ) n 1 n n 1 nn n n n 1 1 1

n 2 n 1 n 1 n 0

2cn n 1 d x nd x nd x d x 2cy y cy 0x

∞ ∞ ∞ ∞− −

= = = =

′− + − − + − + =∑ ∑ ∑ ∑

( )( )

nn 2

1n 0

1y x

n 2 !

∞+

=

−=

+∑ recall the 1st solution

( )( )

nn 1

1n 0

1y x

n 1 !

∞+

=

−′ =

+∑ differentiate and substitute into the equation

( ) ( )( )

( )( )

( )( )

n n nn 1 n n 1 n n 1 n 2 n 2

n n n nn 2 n 1 n 1 n 0 n 0 n 0 n 0

1 1 12cn n 1 d x nd x nd x d x 2c x x c x 0n 1 ! n 2 ! n 2 !x

∞ ∞ ∞ ∞ ∞ ∞ ∞− − + + +

= = = = = = =

− − −− + − − + − + =

+ + +∑ ∑ ∑ ∑ ∑ ∑ ∑

( ) ( )( )

( )( )

( )( )

m n 1 m n 1 m n 2m n 1 m n m n 1 m n n m 1 n m 1 n m 2n m 1 n m 1 n n nn 1 n n 1 n n 1 n 1 n 2

n n n nn 2 n 1 n 1 n 0 n 0 n 0 n 0

1 1 1n n 1 d x nd x nd x d x 2c x 2c x c x 0

n 1 ! n 2 ! n 2 !

= + = + = += − = = − = = − = − = −= + = +∞ ∞ ∞ ∞ ∞ ∞ ∞

− − + + +

= = = = = = =

− − −− + − − + − + =

+ + +∑ ∑ ∑ ∑ ∑ ∑ ∑

( ) ( ) ( )( )

( )( )

( )( )

m 1 m 1 m 2m m m m m m m

m 1 m m 1 mm 1 m 1 m 0 m 0 m 1 m 1 n 2

1 1 1m 1 md x md x m 1 d x d x 2c x 2c x c x 0

m ! m 1 ! m !

− − −∞ ∞ ∞ ∞ ∞ ∞ ∞

+ += = = = = = =

− − −+ + − + − + − + =

+∑ ∑ ∑ ∑ ∑ ∑ ∑

write terms for m=0 and m=1 explicitly:

m 0

1 0d d=

− −

m 1

2 1 2 11 2d x d x 2d x d x 2cx 2c x +2

=

+ + − − + −

( ) ( ) ( )( )

( )( )

( )( )

m 1 m 1 m 2m m m m m m m

m 1 m m 1 mm 2 m 2 m 2 m 2 m 2 m 2 n 2

1 1 1m 1 md x md x m 1 d x d x 2c x 2c x c x 0

m ! m 1 ! m !

− − −∞ ∞ ∞ ∞ ∞ ∞ ∞

+ += = = = = = =

− − −+ + + − + − + − + =

+∑ ∑ ∑ ∑ ∑ ∑ ∑

for m=0 1 0d d= −

for m=1 2 1 2 11 2 d x d x 2 d x d x 2cx 2c x = cx =02

+ − − + − ⇒ c 0=

then the equation becomes:

( ) ( ) ( )m m m m0 1 m 1 m m 1 m

m 2 m 2 m 2 m 2d d m 1 md x md x m 1 d x d x 0

∞ ∞ ∞ ∞

+ += = = =

− + + + + − + − =∑ ∑ ∑ ∑ combine in one summation


( ) ( ) ( ){ } mm 1 m

m 2m 1 m m 1 d m 1 d x 0

∞

+=

+ − + + − = ∑

( )( ) ( ){ } mm 1 m

m 2m 1 m 1 d m 1 d x 0

∞

+=

+ − + − = ∑

( )( ) ( )m 1 mm 1 m 1 d m 1 d 0+ + − + − = use the Identity Theorem

( ) m 1 mm 1 d d 0++ + =

mm 1

dd

m 1+ = −+

recurrence formula for m 2≥

0d any= (but cannot be chosen as 0d 0= )

1 0d d= −

2d any= choose 2d 0= then

mm 1

dd 0

m 1+ = − =+

m 2≥ then the second solution becomes

n2 n 1

n 0y d x cy ln x

∞

=

= +∑ ( )1

nn 0 1 0 0 0

n 0d x d d x d d x d 1 x

=

= = + = − = −∑

or it can be simply written as

( )2y x 1 x= − The second solution of (11)

4. Alternative solution for ( )1y x .

Because the second solution is in a simple closed form

( )2y x 1 x= −

we can use the reduction formula (Section 5.2, Equation (13)) to obtain the other linear independent

solution as:

1

0

adx

a

1 2 22

ey y dxy

−

=∫

∫ ( )( )

x 1dxx

2

e1 x dx1 x

−−

= −−

∫∫

( )( )

11 dxx

2

e1 x dx1 x

− −

= −−

∫∫

( )( )

x ln x

2

e1 x dx1 x

− +

= −−

∫

( )( )

x

2

xe1 x dx1 x

−

= −−

∫ (table integral)

xe−=


397

Let us see if we can retrieve this solution from the power-series

solution obtained in Part 2. We derived

( )1y x ( )

( )

nn 2

n 0

1x

n 2 !

∞+

=

−=

+∑

2 3 41 1 1x x x ...2! 3! 4!

= − + −

add and subtract term x-1 ( ) 2 3 41 1 11 x 1 x x x x ...2! 3! 4!

= − − + − + − + −

( )

xe

2 3 41 1 11 x 1 x x x x ...2! 3! 4!

−

= − − + − + − + −

( ) x1 x e−= − − +

xx 1 e−= − +

This solution can be verified by direct substitution into ODE (11).

So, the general solution of the ODE (11) can be written as

( )y x 1 1 2 2c y c y= +

( ) ( )x1 2c x 1 e c x 1−= − + + −

( )( ) x1 2 1c c x 1 c e−= + − + rename arbitrary constants

( ) xa x 1 be−= − +

Then the General Solution of equation (11) can be written in closed form:

( ) ( ) x1 2y x c x 1 c e−= − + for all x ∈

5. Solution curves for different values of 1c and 2c


V.4.4 Taylor series solution

( ) ( )y p x y q x y 0′′ ′+ + = ( )0 0y x c= ( )0 1y x c′ =

Look for solution in the form of the Taylor series about 0x x= :

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )iv

2 3 40 0 00 0 0 0 0 0

y x y x y xy x y x y x x x x x x x x x ...

2! 3! 4!′′ ′′′

′= + ⋅ − + ⋅ − + ⋅ − + ⋅ − +

The first two coefficients are from the initial conditions:

( )0 0y x c=

( )0 1y x c′ =

The third coefficient can be found from the given differential equation rewritten as

( ) ( ) ( )y x p x y q x y′′ ′= − − , then evaluate ( ) ( ) ( ) ( ) ( )

1 0c c

0 0 0 0 0y x p x y x q x y x′′ ′= − −

To find the next coefficients, differentiate the equation and evaluate it at 0x :

( ) ( ) ( )y x p x y q x y ′′′′ ′= − − ⇒ ( )0y x ...′′′ =

( ) ( )ivy x y x ′′′′= ⇒ ( )iv0y x ...=

and so on …

Then with the found ( ) ( )n0y x construct the Taylor series solution

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )iv

2 3 40 0 00 0 0 0 0 0

y x y x y xy x y x y x x x x x x x x x ...

2! 3! 4!′′ ′′′

′= + ⋅ − + ⋅ − + ⋅ − + ⋅ − +


399

Example 3 Find the first six terms of the power series solution ( ) nn

n 0y x c x

∞

=

= ∑

of the differential equation ( )cos x y xy 2y 0′′ ′+ − =

subject to initial conditions: ( ) ( )y 0 2, y 0 3′= = .


V.4.5. POWER SERIES SOLUTION WITH MAPLE 1. Find the solution of the differential equation y xy 2y 0′′ ′+ + = (see Example 1)

> restart; > M:=4;

:= M 4

> c[0]; c0

> c[1]; c1

> c[2]:=-c[0]; := c2 −c0

> for m from 1 to M do c[m+2]:=-c[m]/(m+1) od;

:= c3 −12 c1

:= c413 c0

:= c518 c1

:= c6 −115 c0

> m:='m': > y(x):=sum(c[m]*x^m,m=0..M+2);

:= ( )y x + − − + + − c0 c1 x c0 x2 12 c1 x3 1

3 c0 x4 18 c1 x5 1

15 c0 x6

> y(x):=collect(y(x),{c[0],c[1]});

:= ( )y x +

− + − 1 x2 1

3 x4 115 x6 c0

− + x 1

2 x3 18 x5 c1

> f:=subs({c[0]=i,c[1]=j},y(x));

:= f +

− + − 1 x2 1

3 x4 115 x6 i

− + x 1

2 x3 18 x5 j

> p:={seq(seq(f,i=-4..4),j=-4..4)}:

> plot(p,x=-1.5..1.5,color=black);


401

2. Solution of IVP: y(1)=1, y'(1)=2

> E1:=subs(x=1,y(x));

:= E1 + 58 c1

415 c0

> yp(x):=diff(y(x),x);

:= ( )yp x +

− + 1 3

2 x2 58 x4 c1

− + − 2 x 4

3 x3 25 x5 c0

> E2:=subs(x=1,yp(x));

:= E2 − 18 c1

1615 c0

> solve({E1=1,E2=2},{c[0],c[1]});

{ }, = c0-4528 = c1

167

> y(x):=subs({c[0]=-45/28,c[1]=16/7},y(x));

:= ( )y x − + − + + − + 4528

4528 x2 15

28 x4 328 x6 16

7 x 87 x3 2

7 x5

> plot(y(x),x=-2..2);


3. Power Series Solution with the POWERSERIES package:

Find the solution of the differential equation 3y x y 0′′ − = about the ordinary point 0x 0=

> restart;

> with (powseries):

> ODE:=diff(y(x),x$2)-x^3*y(x)=0;

:= ODE = −

d

d2

x2 ( )y x x3 ( )y x 0

Find the power series solution with error of order O(x^15):

> y(x):=powsolve(ODE);

:= ( )y x proc ( ) ... end procpowparm

> y(x):=tpsform(y(x),x,15);

:= ( )y x + + + + + + C0 C1 x C020 x5 C1

30 x6 C01800 x10 C1

3300 x11 ( )O x15

Convert to polynomial:

> y(x):=convert(y(x),polynom);

:= ( )y x + + + + + C0 C1 x 120 C0 x5 1

30 C1 x6 11800 C0 x10 1

3300 C1 x11

Plot solution curves:

> f:=subs({C0=i,C1=j},y(x));

:= f + + + + + i j x 120 i x5 1

30 j x6 11800 i x10 1

3300 j x11

> p:={seq(seq(f,i=-2..2),j=-2..2)}:

> plot(p,x=-3..2,y=-5..5,color=black);


403

V.4.6. REVIEW QUESTIONS 1. What is an analytical function?

2. What equations are solved in the form of a power series?

3. What is an ordinary point?

4. What is a singular point?

5. What is a regular singular point?

6. What is the radius of convergence of a power series solution about an

ordinary point?

7. What are the main steps in finding a power series solution about an ordinary

point?

8. What is the Method of Frobenius?

9. What is the indicial equation?

10. What cases for the roots of the indicial equation are considered in the

Frobenius Theorem?

11. What is the form of one solution which can be found for all three cases?

12. Why can it be advantageous to find a solution about a singular point?

EXERCISES 1) Give the power series expansions of xx 1−

and 1x 1−

in x and determine

their radius of convergence. 2) Find the interval of convergence of the power series:

a) ( )

n

2n 0

xn 3

∞

= +∑ b) ( )

2n

nn 0

n x 12

∞

=

−∑ c) n

nn 0

n! xn

∞

=∑ d) ( ) ( )

kk

2 kk 1

1x 3

k 4

∞

=

−+∑

3) Find the singular points of the equations

a) y 2xy y 0′′ ′+ + = b) ( )21 x y y y 0′′ ′− + − =

c) 2 22x y 3 y 0λ′′ − = d) 2x y y xy 0′′ ′− + =

4) Show that ( )( )

23

3xx q xx 1

=−

is analytic at 0x 0= .

5) Using the power series method or the method of Frobenius, find the general solution of the following differential equations:

a) y y 0λ′ − = f) 3y y 2y 0x

′′ ′+ − =

b) 2y y 0λ′′ − = g) ( )22x y x x 1 y y 0′′ ′− − − = c) y xy y 0′′ ′− + = h) xy y xy 0′′ ′+ − =

d) 3y x y 0′′ − = i) 2 2 1x y xy x y 04

′′ ′+ + − =

e) y xy 0′′ − = ( ) ( )y 0 1; y 0 0′= = k) ( )2 22x y 2x x y y 0′′ ′+ + − =

6) Consider the differential equation ( )x 1 y xy y 0′′ ′− − + =

a) find the general solution of the given ODE in the form of a power series about the point 0x 0= ;

b) What is the radius of convergence of the obtained power series solution?

c) Sketch the solution curves.

d) Find the solution subject to the initial conditions: ( ) ( )y 0 2; y 0 6′= − = .

objectives - ira a. fulton college of engineering and ...vps/me505/iem/05 04.pdf · v.4.1....

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