Chapter 1

1.Find the general solution ofd2y

dx2− y = 2 + x2 + ex.

Solution:m2 = 1⇒ m = ±1. Thus the complementary solution isY = Aex +Be−x.We also have

1D2 − 1

2 = −2 ,1

D2 − 1x2 = −(1 +D2 + · · ·)x2 = −x2 − 2.

1D2 − 1

ex = ex1

(D + 1)2 − 11 = ex

1D2 + 2D

1

= ex1

D(D + 2)1 = ex

1D(2)

1 =xex

2.

Therefore, the general solution of the differential equation is

y = Aex +Be−x − 4− x2 +xex

2.

2. Find the general solution of the ordinary differential equationd4y

dx4+ 4y = sin 2x+ 2x+ ex.

Solution:m4 = −4 =⇒ m = 1 + i,−1 + i,−1− i, 1− i.Thus the complementary solution isY (x) = ex(a cosx+ b sinx) + e−x(c cosx+ d sinx).A particular solution is

yp(x) =1

D4 + 4[sin 2x+ 2x+ ex] =

sin 2x20

+x

2+ex

5.

The general solution is

y(x) = ex(a cosx+ b sinx) + e−x(c cosx+ d sinx) +sin 2x

20+x

2+ex

5.

3. Find the general solution of the ordinary differential equationd4y

dx4− y = 3e2x + ex + 2x.

Solution:m4 − 1 = 0⇒ m = ±1,±i.ThusY = aex + be−x + c cosx+ d sinx.We also haveyP =

1D4 − 1

(3e2x + ex + 2x)

=e2x

5+

1(D − 1)(D + 1)(D2 + 1)

ex − (1 +D4 + · · ·)2x

1

=e2x

5+

1(D − 1)

14ex − 2x

=e2x

5+ ex

1D

14− 2x =

e2x

5+exx

4− 2x.

Therefore,

y = aex + be−x + c cosx+ d sinx+e2x

5+xex

4− 2x.

4. Find the general solution ford3y

dx3− y = 2 + cosx.

Solution:

The complementary solution satisfiesd3Y

dx3− Y = 0. Thus

Y = emx,m3 = 1⇒ mn = e2inπ/3.The three roots are: m0 = 1, m1 = (−1 + i

√3)/2, m2 = (−1 + i

√3)/2.

We getY = Aex + e−x/2[B cos(x

√3/2) + C sin(x

√3/2)].

The particular solution is

yp =1

D3 − 1(2 + cosx) = −2− 1

D + 1cosx

= −2− D − 1D2 − 1

cosx = −2− − sinx− cosx−1− 1

.

We havey = Aex + e−x[B cos(x

√3/2) + C sin(x

√3/2)]− 2− sinx+ cosx

2.

5. Consider the equation of motion

··x+ a

·x+ x = b cos t,

·x ≡ dx

dt, a > 0.

which describes a harmonic oscillator under a linear damping force as wellas an external harmonic force of the same frequency. Find the general solutionof this equation.

Solution:m2+am+ 1 = 0⇒ m1 = −a/2 +

√a2/4− 1, m2 = −a/2−

√a2/4− 1.

If a < 2, the two roots above are −a/2± iω, where ω =√

1− a2/4, and thecomplementary solution X is given by

X = e−at/2(c cosωt+ d sinωt), where c and d are constants.We note the motion is a damped oscillation with the angular fequency ω,

which is smaller than unity–the angular frequency when a = 0.

If a > 2, the two roots are −a/2 ±√a2/4− 1. The complementry solution

isX = e−at/2(ce

√a2/4−1t + de−

√a2/4−1t).

2

Both solutions above describe motions vanishing as t → ∞ without oscilla-tions.

If a = 2,X = e−t(c+ dt).Both solutions vanish as t→∞ without oscillations.

The particular solution is

xp =1

D2 + aD + 1b cos t =

1aD

b cos t =b sin ta

.

The general solution is

x = X +b sin ta

.

6. Show that the particular solution of a linear differential equation withconstant coefficients can always be expressed by an integral.

Solution:Let the equation bedny

dxn+ an−1

dn−1y

dxn−1+ · · ·+ a1

dy

dx+ a0y = f(x),

where am, m = 0, 1 · ·· n, are constants.We haveY = c1e

m1x + c2em2x + · · ·+ cne

mnx,where mj , j = 1, 2 · · · n are the roots of the algebric equationmn + an−1m

n−1 + · · ·+ a1m+ a0 = 0.We assume here that they are distinct. If the algebraic equation has multiple

roots, only slight modifications are needed.The particular solution is given by

yp =1

Dn + an−1Dn−1 + · · ·+ a1D + a0f(x).

We shall write1

Dn + an−1Dn−1 + · · ·+ a1D + a0=

1(D −m1)(D −m2) · · · (D −mn)

.

By partial fraction, we get1

Dn + an−1Dn−1 + · · ·+ a1D + a0

=1

(m1 −m2) · · · (m1 −mn)1

D −m1+

1(m2 −m1) · · · (m2 −mn)

1D −m2

+ · · ·+ 1(mn −m1) · · · (mn −mn−1)

1D −mn

.

Since1

D −mf(x) = emx

1D

[e−mxf(x)] =emx∫ x

e−mx′f(x′)dx′,

we have

3

yp(x) =∫ x[

em1(x−x′)

(m1 −m2)(m1 −m3) · · · (m1 −mn)+

em2(x−x′)

(m2 −m1)(m2 −m3) · · · (m2 −mn)+·

· ·+ emn(x−x′)

(mn −m1)(mn −m2) · · · (mn −mn−1)] f(x′)dx′.

Chapter 2

1.As we know, the sum of two complex numbers can be represented as that oftwo vectors in a two dimensional space. How about the product of two complexnumbers?

Solution:Let α ≡ a1 + ia2, β ≡ b1 + ib2.These two complex numbers can be reprsented by the vectors→v ≡ a1

→e1 + a2

→e2,→u ≡ b1

→e1 + b2

→e2.

where→e1 and

→e2 are unit vectors in the direction of the x-axis and the y-axis,

respectively.We haveα∗β = (a1 − ia2)(b1 + ib2) = (a1b1 + a2b2) + i(a1b2 − a2b1).The real part of α∗β is the scalar product of

→v and

→u ; the imaginary part

of α∗β is the cross product of→v and

→u .

2. Find the real part and the imaginary part of eiz as well as those of cos z.For what values of z is the real part of eiz equal to cos z?

Solution:eiz = eix−y = e−y(cosx+ i sinx) = u+ iv,e−iz = e−ix+y = ey(cosx− i sinx).Thus

cosz =eiz + e−iz

2= cosx cosh y − i sinx sinh y.

If u is equal to cos z, thene−y cosx = cosx cosh y − i sinx sinh y.The real part and the imaginary part of the equation above are:(e−y − cosh y) cosx = 0 and sinx sinh y = 0.If y 6= 0, these two equalities give sinx = cosx = 0, which is not possible.Both equalities are satisfied ify = 0.

Thus the real part of eiz is equal to cos z only if z is real.

3. For what constant value (or values ) of a is the function u = x + ax3

the real part of an analytic function? Find the imaginary part of this analyticfunction. Express this analytic function as a function of z.

4

Solution:

uxx = 6ax , uyy = 0 .Thus ∇2u = 0 only if a = 0, in which caseu = x.The conjugate of x, denoted as v, satisfiesvy = ux = 1 and vx = −uy = 0.We getv = y + c,andu+ iv = z + ic,where c is a constant.

4. Find the roots of (z + 1)8 = (z2 − 1)8 .

Solution:The equation above gives(z2 − 1) = ei2πn/8(z + 1), n = 0, 1, 2 · ··, 7 ,or(z + 1)(z − 1− ei2πn/8) = 0, n = 0, 1, 2 · ··, 7.Thus the roots arez = −1andz = 1 + ei2πn/8, n = 0, 1, 2 · ··, 7 .

5. Find the Laurent series expanded around z = 0 for

f(z) =1

z(z − 2).

Specify the region in which each of these series is valid.

Solution:In the region |z| < 2 , we have

f(z) = − 12z(1− z/2)

= − 12z

(1 +z

2+ · · ·+ zn

2n+ · · ·) .

And in the region |z| > 2 , we have

f(z) =1z2

11− 2/z

=1z2

(1 +2z

+ · · ·+ 2n

zn+ · · ·).

6.Evaluate the following integrals:

a.∫∞−∞

(1 + cosπx)dx(1− x2)

,

b.∫∞−∞

sin2 x

1 + x2dx,

5

c.∫∞−∞

1 + 3x2

(1 + x2)(4 + x2)dx.

Solutions:a. We get

J ≡∫∞−∞

(1 + eiπx)dx(1− x2)

= 0 by closing contour upstairs. ( Note that J is a

convergent integral.) ThusI = Re J = 0.

b. I =∫∞−∞

sin2 x

1 + x2dx = −1

4∫∞−∞

e2ix − 2 + e−2ix

1 + x2dx.

Let

J ≡∫∞−∞

e2ix − 11 + x2

dx.

By clsoing contour upstairs, we get

J = 2πie−2 − 1

2i= −π(1− e−2).

ThusI = −Re J/2 = π(1− e−2)/2.

c. By closing contour upstairs, we get∫∞−∞

1 + 3x2

(1 + x2)(4 + x2)dx = 2πi[

−2(2i)(3)

+−11

(−3)(4i)] = 7π/6.

7. Evaluate the following integrals:

a.∫ 2π

0

13− 2 sin θ

dθ.

b.∫∞−∞(

1x2 + 1

)3dx.

c.∫∞−∞

cos(πx/2)x2 − 1

dx.

a. Let z = eiθ. We haveI = −

∮c

dz

z2 − 3iz − 1,

where c is a unit circle with its center at the origin. The integrand abovehas two simple poles:

z1 = i(3 +√

5)/2, z2 = i(3−√

5)/2.Since z2 is the pole inside c,

I = 2πi Residue(z2) = − 2π(3−

√5)− 3

=2π√

5.

b. Closing the contour upstairs, we getI = 2πiRes(i).Since the integrand has a third-order pole at i, we get

6

Res(i) =12

limz→i

d2

dz2

1(z + i)3

=6

(2i)5=

316i

.

We have I =3π8.

c. Since both the numerator and the denominator vanish at x = ±1, theintegrand is entire. By deforming the contour, we get

I =12

∫c

eizπ/2 + e−izπ/2

z2 − 1dz,

where c passes around the points z = ±1 from above. By closing the contourupstairs, we find the first term in the integrand equal to zero. We get

I =12

∫c

e−izπ/2

z2 − 1dz = −πi(− i

2+

i

−2) = −π.

8. Evaluate the following integrals with the use of the Cauchy residue theo-rem.

a.∫∞0

dx

(9 + x2)(4 + x2),

b.∫∞−∞

sin2 x

x2dx,

Solution:

a. I =12

∫∞−∞

dx

(9 + x2)(4 + x2).

Closing the contour upstairs, we get

I = πi[1

(6i)(−5)+

1(5)(4i)

] = π/60.

b. Since the integrand is an entire function, we deform the contour and get

I = −14

∫c

e2iz − 2 + e−2iz

z2dz,

where c is a contour in the upper half-plane joining −∞ and ∞. The firsttwo terms in the integrand give zero. We get

I =2πi4

limz→0

de−2iz

dz= π.

9. Evaluate the following integrals:

a.∫∞−∞

sin cx(x+ a)2 + b2

dx,

where a is a real number while b and c are both real and positive.

b.∫∞−∞

x2

x4 + 1dx.

c.∫C

e1/z

zdz, where c is a circle around the origin in the counterclockwise

direction.

7

Solution:

a. I = Im∫∞−∞

eicx

(x+ a)2 + b2dx = Im[2πiRes(−a+ ib)]

= Im[2πie−iace−bc

2ib] = −πe

−bc sin acb

.

b. Closing the contour upstairs, we get I = (2πi)[Res(eiπ/4)+Res(−e−iπ/4)].Since

Res(eiπ/4)=z2

4z3]z=exp(iπ/4) , Res(−e−iπ/4)=

z2

4z3]z=− exp(−iπ/4) ,

I = (2πi)[1

4(eiπ/4)+

14(−e−iπ/4)

] = (2πi)−2i sin(π/4)

4= π√

2/2.

c. We have

I =∫C

e1/z

zdz =

∫C

1z

(1 + z + z2/2 + · · ·)dz.By the Cauchy residue theorem, we getI = 2πi,

10. Evaluate the integral I ≡∫∞0

x

1 + xndx.

Solution:Let J ≡

∫C

z

1 + zndz,

where C is the contour of a pie with two of its boundaries the infinite ray ofθ = 0 and that of θ = 2π/n. Then

J = (1− e4iπ/n)I .By the Cauchy residue theorem, we have

J =2πi

nei(n−2)π/n.

ThusI =

π

n sin(2π/n).

11. Find the branch points of the function f(z) = (z4−1)1/4 . Draw a set ofbranch cuts for this function to make this function single-valued in the complexplane.

Solution:The branch points of this function are atz = 1, i,−1,−i .As z →∞, f(z)→ z(1− z−4/4 + · · ·).Thus z∞ is not a branch point of the

function. We draw straight lines joining 1 to i , i to −1 , and −1 to −i , andthe function is single-valued in the cut-plane.

12. (a) Locate the singularities of the function (1 + z

1− z)1/3

11 + z2

in the ex-

tended z-plane. Draw branch cut (or cuts) to make this function single-valued.

8

(b) Evaluate the integral∫ 1

−1(1 + x

1− x)1/3

dx

1 + x2.

Solution:(a) The function (

1 + z

1− z)1/3

11 + z2

has simple poles at z = ±i and branch

points at ±1.Let z = 1/t. Then the function is

(t+ 1t− 1

)1/3t2

t2 + 1.

This shows that t = 0, or z = ∞, is not a branch point of the function.Therefore, drawing a finite branch cut from −1 to 1 makes the function single-valued.

(b) Let

J =∮c

(1 + z

1− z)1/3

11 + z2

dz,

where c is the “toilet seat” contour wrapping around the branch cut.

We choose (1 + z

1− z)1/3 to be real and positive when z is infinitesimally above

the branch cut. The value of (1 + z

1− z)1/3 at a point infinitesimally below the

branch cut can be obtained as follows. We start at a point infinitesimally abovethe branch cut and go around z = −1 to reach the point infinitesimally below thestarting point. Since the path does not enclose the point z = 1, the argument of

(1−z) does not change. The argument of (1+z) changes by 2π. Thus (1 + z

1− z)1/3

changes by a phase factor e2πi/3.From this consideration we getJ = (1− e2πi/3)I.Note that the infinitesimal circles around z = ±1 contribute nothing to J ,

as ε−1/3ε→ 0 when ε→ 0.Since the integrand vanishes like 1/z2 at infinity, we may close the contour

by adding an infinite circle to the toilet seat contour. By the Cauchy residuetheorem, we get

J = 2πi{[(1 + z

1− z)1/3]z=i

12i

+ [(1 + z

1− z)1/3]z=−i

1−2i}.

To evaluate (1 + z

1− z)1/3]z=i, we note that this factor is real and positive at

the origin infinitesimally above the branch cut. As we move from z = 0 to z = i,the argument of (z+ 1) increases by π/4, while that of (z−1) decreases by π/4.Thus we have

(1 + z

1− z)1/3]z=i = (

eiπ/4

e−iπ/4) = eiπ/6.

Similarly, since (1 + z

1− z)1/3 at the origin infinitesimally below the branch cut

is equal to e2πi/3, we have

[(1 + z

1− z)1/3]z=−i = e2πi/3(

e−iπ/4

eiπ/4)1/3 = e2πi/3e−iπ/6.

9

Therefore,J = π(eiπ/6 − e2πi/3e−iπ/6).We get

I = πeiπ/6 − e2πi/3e−iπ/6

1− e2πi/3= π

e−iπ/3eiπ/6 − eiπ/3e−iπ/6

e−iπ/3 − eiπ/3

= πsin(π/6)sin(π/3)

=π√3.

13. (a) Locate and classify the singularities of the function (z + 2)−1(1 −z2)−1/2 in the extended z-plane. Draw branch cut (or cuts) to make the functionsingle-valued.

(b) Evalaute the integral

I =∫ 1

−1

dx

(x+ 2)(1− x2)1/2.

Solution:(a) The function (z + 2)−1(1− z2)−1/2 has a simple pole at −2 and branch

points at ±1. This function is single-valued in the plane cut from −1 to 1.(b) Let

J =∮c

dz

(z + 2)(1− z2)1/2,

where c is the closed contour wrapping around the branch cut in the clockwisedirection. We note that

J = 2I.We may view this contour as the one enclosing the cut-plane in the counter-

clockwise direction. Since the integrand vanishes like z−2 as z → ∞, the onlysingularity enclosed by the contour is at z = −2. Therefore, the Cauchy residuetheorem gives

J = 2πi Res(−2) =2πi

(1− 4)1/2=

2πii√

3=

2π√3.

ThusI =

π√3.

14. Let f(z) be analytic at z0 and f(z0) 6= 0. Prove log f(z) does not havea branch point at z0.

Solution:If f(z0) 6= 0, f(z) has no zero in a sufficiently small neighborhood of z0.

Therefore, the argument of f(z) does not change as we go around a closedcontour in this neighborhood. As a result, the value of log f(z) does not changeas we go around this closed contour. Therefore, z0 is not a branch point of f(z).

15. Evaluate the integral∫∞0

dx

xa(1 + x), where 0 < Re a < 1.

. .

10

Solution:The function f(z) =

1za(1 + z)

has a branch point at z = 0. We choose the

branch cut of this function to be the positive x-axis.LetJ =

∫c

dz

za(1 + z),

where c wraps around the positive x-axis in the clockwise direction. Thecontour goes around the singularity at the origin via a circle of radius ε.

The integral of f(z) over a circle of radius R is of the order ofR

(Ra)(R),

which vanishes as R→∞. ( This is because a > 0.) Therefore, we may closethe contour by adding an infinite circle to c. By the Cauchy residue theorem,

J = 2πiRes(-1)= 2πi1eiπa

.

Now the contour c had three sub-contours:1. Sub-contour c1, the positive x-axis. The integral over c1 is I.2. Sub-contour c2, the circle of radius ε around the origin. The integral over

c2 is proportional toε

εa,

which vanishes as ε→ 0. (This is because a < 1.)3. Sub-contour c3, which is the x-axis below the branch cut. The integrand

on c3 is equal to1

(e2iπx)a(1 + x)= e−2πia 1

xa(1 + x).

Thus the integral over c3 is equal to−e−2πiaI.The minus sign above us due to the direction of the contour eing clockwise.

Thus we haveJ = (1− e−2πia)I.We getI =

π

sinπa.

16. Find the value ofI =

∫ ∞0

lnx1 + x5

dx. Reminder:∫ ∞

0

11 + x5

dx =π

5 sin(π/5).

Solution: LetJ =

∮c

ln z1 + z5

dz,

where c is the closed pie-shape contour with angle 2π/5 which encloses thesimple pole at eiπ/5. By the Cauchy residue theorem, we get

J = 2πiiπ/5

5e4iπ/5= −2π2e−4iπ/5

25=

2π2eiπ/5

25.

To extract I from J , we have

11

J = I − e2πi/5∫ ∞

0

ln r + 2iπ/51 + r5

dr = (1− e2πi/5)I − 2iπ2e2πi/5

25 sin(π/5).

Thus we have

(1−e2πi/5)I =2π2

25eiπ/5(1+

eiπ/5i

sin(π/5)) =

2π2

25eiπ/5(1+

i cos(π/5)− sinπ/5sin(π/5)

) =

2π2

25eiπ/5

i cosπ/5sin(π/5)

,

or

I = − π2 cos(π/5)

25 sin2(π/5).

17. Evaluate I =∫∞−∞

x

sinhxdx.

Solution:We note sinh z = (ez−e−z)/2 is finite if the real part of z is finite, regardless

of the vaue of the imaginary part of z. As a result,z

sinh zdoes not vanish at

all points of the upper infinite half circle. Consequently, we cannot close thecontour upstairs. For the same reason, we cannot close the contour downstairs.

Let’s close the contour in a differently way. We define

J =∮c

z

sinh zdz,

where c is the boundary of the ( infinite) rectangle the horizontal sides ofwhich are the x-axis and the line z = x+ iπ.

On the line z = x+ iπ, the integrand is equal tox+ iπ

sinh(x+ iπ).

Sincesinh(x+ iπ) = − sinhx,the integrand on this line is equal to

−x+ iπ

sinhx,

which blows up at x = 0.To avoid the singularity at z = iπ, we make the indentation of an infinitesi-

mal half circle below iπ.We shall denote the x-axis as c1, the line z = x+ iπ with |x| > ε as c2, and

the indented half circle as c3.Since c encloses no singularity of the integrand,J = 0.Now the integral over c1 is I.The integral over c2 is

P∫∞−∞

x+ iπ

sinhxdx = P

∫∞−∞

x

sinhxdx = I,

asP

∫∞−∞

iπdx

sinhx= 0.

The residue ofz

sinh zat iπ is

12

iπ

−1.

Thus the integral over c3 is equal to−iπ(−iπ) = −π2.Finally, the integral over the two vertical boundary lines vanishes.Thus we haveJ = 2I − π2.We getI = π2/2.

18. Find the Fourier transform of the function f(x) = e−x2,−∞ < x <∞.

Solution:The Fourier transform of e−x

2is∫∞

−∞ dx e−ikx−x2

= e−k2/4

∫∞−∞ dxe−(x+ik/2)2 .

By deforming the contour of integration to the linez = −ik/2 + t,where t is real, we getI = e−k

2/4∫∞−∞ dte−t

2=√πe−k

2/4.

19. Find the Fourier transform of the function f(x) = (x2 +1)−1,−∞ <x <∞.

Solution:f(x) = (x2 + 1)−1,∼f (k) =

∫∞−∞ dx

e−ikx

1 + x2.

If k > 0, we close contour downstairs and get∼f (k) = −2πi

e−k

−2i= πe−k, k > 0.

If k < 0, we close contour upstairs and get

∼f (k) = 2πi

ek

2i= πek, k < 0.

Thus∼f (k) = πe−|k|, all k.

20. (a) LetF (x) = 1, −a < x < a,

= 0, |x|> a,

where a is a positive constant. Find˜

F (k), the Fourier transform of F (x).(b) Show that the inversion formula holds in this example.

Solution:

13

(a)˜

F (k) =∫ a−a e

−ikxdx =e−ika − eika

−ik=

2 sin(ka)k

.

(b) We need to prove that∫∞−∞ eikx

e−ika − eika

−ikdk

2π= 0, |x| > a,

= 1, |x| < a.

Sincee−ika − eika

−ikis an even function of k, the integral above is an even

function of x. Thus we only need to show that the above is true when x ispositive, or∫∞

−∞ eikxe−ika − eika

−ikdk

2π= 0, x > a, (I)

= 1, 0 < x < a. (II)We deform the contour to the upper half plane, avoiding the origin k = 0.If x > a, both (x + a) and (x − a) are positive. Thus we may close the

contour upstairs and get∫eikx

e−ika − eika

−ikdk

2π= 0,

and (I) is proven.If 0 < x < a, (x+a) is positive while (x−a) is negative. Closing the contour

upstairs, we get∫eikx

eika

−ikdk

2π= 0.

We also get, by closing the contour downstairs,∫eikx

e−ika

−ikdk

2π= (−2πi)(

1−i

)(1

2π) = 1.

Thus (II) is proven.

21. Find the Fourier transform off(x) =

11 + x4

, −∞ < x <∞ .

Solution:The Fourier transform of

11 + x4

is∫∞−∞

e−ikx

1 + x4dx

The poles are located ate±iπ/4, e±3iπ/4 .For k > 0 , we close the contour downstairs; and for k < 0 , we close the

contour upstairs. We then obtain the Fourier transform of the function asπ

2e−|k|(1+i)/

√2+iπ/4 + c.c. = πe−|k|/

√2 cos(k/

√2− π/4) .

22. Find the Fourier coefficient cn of the function f(θ) = eθ, −π << π. Byevauating the Fourier series for f(θ) at θ = π and at θ = −π, prove

14

∞∑n=1

11 + n2

=π

2eπ + e−π

eπ − e−π− 1

2.

Solution:

cn =∫ π−π e

−inθeθdθ

2π=

e−inθeθ

2π(1− in)|π−π= (−1)n

eπ − e−π

(1− in).

Thus

eθ =eπ − e−π

2π

∞∑n=−∞

(−1)neinθ

1− in.

At θ = π and θ = −π, the Fourier seris has the same valueeπ − e−π

2π

∞∑n=−∞

11− in

=eπ − e−π

2π(1 +

21 + 1

+2

1 + 22+ · · ·+ 2

1 + n2+ · · ·)

=eπ − e−π

2π(1 + 2

∞∑n=1

11 + n2

).

Since f(π) = eπ, f(−π) = e−π, the Fourier seris at θ = ±π is equal to theaverage of f(π) and f(−π). Thus

eπ + e−π

2=eπ − e−π

2π(1 + 2

∞∑n=1

11 + n2

),

or∞∑n=1

11 + n2

=π

2eπ + e−π

eπ − e−π− 1

2.

23. Find Fourier coefficient an for

f(θ) =1

a+ b cos θ, with a and b positive and b < a.

Solution:We have

a−n =1

2π∫ π−π

einθdθ

a+ b cos θ.

Note thatan = a∗−n.Let z ≡ eiθ, then we have

a−n =1πi

∮c

zn−1dz

[2a+ b(z + z−1)]=

1πbi

∮c

zndz

[2az/b+ (z2 + 1)].

where c is the unit circle and n is non-negative. The integrand has simplepoles at

z = −a/b±√a2/b2 − 1.

The pole inside the unit circle is located at−a/b+√a2/b2 − 1.Using Cauchy’s

residue theorem, we get, for n > 0,

a−n =(−1)n(a−

√a2 − b2)n

bn√a2 − b2

.

Since a−n is real, we have

an = a−n =(−1)n(a−

√a2 − b2)n

bn√a2 − b2

, n ≥ 0.

15

24. Use the Laplace transform to find the solution ofd2y

dt2+ y = δ(t− 1) + 2δ(t− 2)

satisfying the initial conditions y(0) =·y(0) = 0, where

·y ≡ dy

dt.

Solution:Let L(s) be the Lapalce transform of y(s), then(s2 + 1)L(s) = e−s + 2e−2s,or L(s) = (e−s + 2e−2s)/(s2 + 1).Thus

y(x) =∫ i∞−i∞

es(t−1) + 2es(t−2)

s2 + 1ds

2πi.

By carrying out the integration, we gety(x) = θ(t− 1) sin(t− 1) + 2θ(t− 2) sin(t− 2),whereθ(τ) = 1, τ > 0,

= 0, τ < 0.

25. Solve with the method of Laplace transform the initial-value problemd2y

dx2+ y =

√x, x > 0,

with the initial conditiony(0) = 1, y′(0) = 0.

Solution:We have−s+ (s2 + 1)Y (s) = Γ(3/2)/s3/2,where Y (s) is the Laplace transform of y(x). Thus

Y (s) =s

s2 + 1+

Γ(3/2)s3/2(s2 + 1)

.

By performing an inverse Laplace transform, we gety(x) = cosx+

∫ x0x′1/2 sin(x− x′)dx′.

The last integral is obtained by using the convolution theorem.

26. Let the Laplace equation ∇2u(r, θ) = 0 be satisfied inside the disk r < 5,where r and θ are the polar coordinates,and let the boundary condition be

u(5, θ) = sin2 θ.Find u(r, θ).

Solution:

sin2 θ =(eiθ − e−iθ)2

−4=

1− cos 2θ2

.

Thus

u(r, θ) =1− r2 cos 2θ/25

2.

16

Chapter 3

1. Find the solution u(x, y) of

ux + (x2 + y)uy = 0, u(0, y) =1

1 + y2.

Solution:

The equation for the characteristic curves is given bydx

1=

dy

x2 + y,

ordy

dx= y + x2.

The complementary solution of the equation above isY = cex,and a particular solution is

yp = − 11−D

x2 = −(1 +D +D2 + · · ·)x2 = −(x2 + 2x+ 2).

Thus the characteristic curves are given byy = cex − (x2 + 2x+ 2),or(y + x2 + 2x+ 2)e−x = c.Therefore, the general solution of the PDE isu(x, y) = f((y + x2 + 2x+ 2)e−x).The initial condition gives

11 + y2

= f(y + 2),orf(y) =

11 + (y − 2)2

.

Therefore,

u(x, y) =1

1 + [(y + x2 + 2x+ 2)e−x − 2)]2.

2. Find the solution u(x, y) ofux + (e2x + y)uy = 0, u(0, y) = ey

Solution:The equation for the characteristic curves isdx

1=

dy

e2x + y,

or y′ = y + e2x ⇒ y = cex + e2x ⇒ (y − e2x)e−x = c.Therefore, the general solution for thie PDE isu(x, y) = f(ye−x − ex).From the initial condition, we get

17

ey = f(y − 1)⇒ f(y) = ey+1.Therefore we haveu(x, y) = eye

−x−ex+1.

3. Solveut + xux = 0with the initial condition u(x, 0) = e−x

2.

Solution:

The equation for the characteristic curves aredt

1=dx

x.

Thus the characteristic curves are given byt = lnx+ c′,orxe−t = c.Therefore, the general solution of the equation isu(x, t) = f(xe−t).To determine f , we set t = 0 and find thatf(x) = e−x

2.

Thusu(x, t) = e−x

2e−2t.

Chapters 4 and 5

1. The equation∂2u

∂x2+∂2u

∂y2= u holds in the rectangle with vertices (0, 0),

(a, 0), (a, b), (0, b). At the right vertical side, u(a, y) = sin(πy

b), 0 < y < b. The

function u vanishes at the other three sides of the rectangle. Find u.

Solution:

Since u vanishes at the two horizontal sides, we express u as a Fourier sineseries of y :

u(x, y) =∞∑n=1

An(x) sin(nπy

b).

Substituting u into the partial differential equation, we getd2An(x)dx2

− n2π2 + b2

b2An(x) = 0.

An(x) = cn cosh(x√n2π2 + b2/b) + dn sinh(x

√n2π2 + b2/b).

Thus

18

u(x, y) =∞∑n=1

[cn cosh(x√n2π2 + b2/b) + dn sinh(x

√n2π2 + b2/b)] sin(

nπy

b).

The boundary condition u(0, y) = 0 gives

0 =∞∑n=1

cn sin(nπy

b),

or cn = 0.That u(a, y) = sin(

πy

b) gives

∞∑n=1

dn sinh(a√n2π2 + b2/b)] sin(

nπy

b) = sin(

πy

b),

ordn = 0, n 6= 1,d1 = 1/(sinh(a

√π2 + b2/b).

We haveu(x, y) = sinh(x

√π2 + b2/b)] sin(

πy

b)/(sinh(a

√π2 + b2/b).

2. Consider the wave equation∂2u

∂x2=∂2u

∂t2, −∞ < x < ∞, t > 0, with the

initial conditions u(x, 0) = f(x), ut(x, 0) = 0.a. Find the equation satisfied by L(x, s), whereL(x, s) ≡

∫∞0dt e−stu(x, t).

b. Assuming that both f(x) and L(x, s) have Fourier transforms, find L(x, s)in the form of a Fourier integral. (You are allowed to differentiate a Fourierintegral by differentiating its integrand.)

c. Find u(x, t).Note: the Laplace transform of u′′(t) is equal to s2L(s)−u′(0)−su(0), where

L(s) is the Laplace trandform of u(t).

Solution:a. Let the Laplace transform of u(x, t) be L(s, x).We have, by multiplying

the equation above with e−st and integrate with respect to t from 0 to ∞,d2L(x, s)dx2

− s2L(x, s) = −sf(x).b. LetL(x, s) =

∫∞−∞

dk

2πeikxl(k, s)

and substitute it into the equation for L(x, s) above, we have(−k2 − s2)l(k, s) = −sF (k),where F (k) is the Fourier transform of f(x).Therefore,

l(k, s) =sF (k)k2 + s2

⇒ L(x, s) =∫∞−∞

dk

2πeikx

sF (k)k2 + s2

.

c. We have

u(x, t) =∫c

ds

2πiest

∫∞−∞

dk

2πeikx

sF (k)k2 + s2

,

where c is a vertical contour to the right of all singularities of L(x, s). Byclosing contour in the s-plane to the left, we get

u(x, t) =12

∫∞−∞

dk

2πeikx(eikt + e−ikt)F (k) =

f(x+ t) + f(x− t)2

, t > 0.

19

Note: It is easier to find the solution of this problem by making the Fouriertransform with respect to x first, which we have demonstrated in the textbook.

3. Consider the Laplace equation∇2u(r, θ) = 0which holds inside a disk of radius 3 with the center at the origin. The

boundary condition is

u(3, θ) =∞∑n=1

sinnθn!

.

Find u(r, θ) in a closed form.

Hint: Make use of ex = 1 + x+ x2/2! + · · ·+ xn/n! + · · ·.Solution:

u(r, θ) =∞∑n=1

sinnθn!

(r

3)n =

12i

∞∑n=0

einθ − e−inθ

n!(r

3)n

=12i

∞∑n=0

(eiθr

3)n − (e−iθ

r

3)n

n!=

12i

[ereiθ/3 − ere−iθ/3]

=12ier cos θ/3[eir sin θ/3 − e−ir sin θ/3] = er cos θ/3 sin(r sin θ/3).

Note: u(x, y) is the imaginary part of the analytic function ez/3.

4. Consider the Sch··0dinger equation

i∂Ψ∂t

= −∂2Ψ∂x2

, Ψ = Ψ(x, t),which holds for all values of x and all positive values of t. Let the initial

value of Ψ(x, t) beΨ(x, 0) = f(x)and let the boundary conditions for Ψ beΨ(±∞, t) = 0.

(a) Make a Fourier transform of the Sch··0dinger equation with respect to x

and solve the resulting ordinary differential equation.(b) Express the solution Ψ(x, t) satisfying the initial condition (B) in the

form of a Fourier integral.

Solution:a. After making the Fourier transform, we have

i∂∼Ψ(k, t)∂t

= k2∼Ψ(k, t).

The solution of the equation above is∼Ψ(k, t) = a(k)e−ik

2t.ThusΨ(x, t) =

∫∞−∞

dk

2πeikx−ik

2ta(k).b. To satisfy the initial condition, we have

20

a(k) =∼f (k).

ThereforeΨ(x, t) =

∫∞−∞

dk

2πeikx−ik

2t∼f (k).

The answer above can be expressed as an integral over x′:

Ψ(x, t) =∫∞−∞

dk

2πeikx−ik

2t∫∞−∞ dx′e−ikx

′f(x′)

=∫∞−∞ dx′

√1

4iπtexp[

(x− x′)2i4t

] f(x′).

5. Find the eigenvalues and the eigenfunctions of the Sturm-LIouviile prob-lem

d2u

dx2= −λu, 0 < x < L; u′(0) = 0 and u(L) = 0.

Solution:The general solution of the differential equation above isu(x) = a cos(

√λx) +b sin(

√λx).

Taking the derivative of the expression above, we getu′(x) =

√λ[−a sin(

√λx) +b cos(

√λx)].

The boundary condition u′(0) = 0 givesb√λ = 0⇒ either λ = 0 or b = 0.

If λ = 0, the solution of the differential equation is u = c1+c2x. This solutionvanishes identically as the boundary conditions u′(0) = u(L) = 0 are imposed.

If λ 6= 0, we have b = 0, oru = a cos(

√λx).

The boundary condition u(L) = 0 givesa cos(

√λL) = 0.

Since a 6= 0 for a non-trivial solution, the eigenvalues of the non-trivialsolutions are√

λnL = (n+ 1/2)π, n = 0, 1, 2, · · ·or

λn =(n+ 1/2)2π2

L2, n = 0, 1, 2, · · ·.

The eigenfunction corresponding to the eigenvalue above is

un(x) =√

2L

cos[(n+ 1/2)πx

L].

These eigenfunctions are orthonormal:∫ L0un(x)um(x)dx = δnm.

6. Find the solution of uxx + uyy = 0, ∞ > y > 0 and 0 < x < 1,with ux(0, y) = u(L, y) = 0, u(x, 0) = δ(x− 1/2) and u(x,∞) = 0.

Solution:We express u by the Fourier cosine series

u(x, y) =∞∑n=0

An(y) cos[(n+ 1/2)πx],

which satisfies the boundary condition ux(0, y) = u(L, y) = 0.

21

Substituting this expression of u into the PDE, we getd2An(y)dy2

= [(n+ 1/2)π]2An(y).

The solution of the differential equation above isAn(y) = ane

(n+1/2)πy + bne−(n+1/2)πy.

The boundary condition at y =∞ gives an = 0. Therefore

u(x, y) =∞∑n=0

bne−(n+1/2)πy cos[(n+ 1/2)πx].

The boundary condition at y = 0 requires

δ(x− 1/2 ) =∞∑n=0

bn cos[(n+ 1/2]πx].

Thus we have12bn =

∫ 1

0δ(x− 1/2) cos[(n+ 1/2)πx]dx = cos[(n+ 1/2)π/2],

orbn = 2 cos[(n+ 1/2)π/2].Therefore we have

u(x, y) = 2∞∑n=0

e−(n+1/2)πy cos[(n+ 1/2)π

2] cos[(n+ 1/2)πx].

7. Consider the vibration of the surface of a drum of unit radious. Theequation of motion is

∂2u

∂t2=

1r

∂

∂r(r∂u

∂r) +

1r2∂2u

∂θ2, r < 1, 0 ≤ θ ≤ 2π,

where u(r, θ, t) is the displacement of the drum surface, and r and θ are thepolar coordinates. The boundary condition is

ur(1, θ, t) = 0, where ur ≡∂u

∂r.

The initial conditions areu(r, θ, 0) = cos θ, ut(r, θ, 0) = 0.Find u(r, θ, t).

Solution:

Substituting u = T (t)R(r)Θ(θ) into the PDE, we getT”T

=1rR

d

dr(rR′) +

Θ”r2Θ

= −c.

Thus T” = −cT, Θ” = −n2Θ,d

dr(rdR

dr)− n2

rR = −crR.

Since Θ is required to satisfy Θ(θ + 2π) = Θ(θ), n is an integer.The solution of the equation for R isR = Jn(

√cr),

which satisfies the boundary condition of R(0) being finite, where Jn is theBessel function of order n. Let rm,n be the mth root of J ′n, i.e., J ′n(rm,n) = 0,m = 1, 2, · · ·

22

Then∑n,m

cos(rm,nt)Jn(rm,nr)(Am,n cosnθ+Bm,n sinnθ) satifies the boudary

condition at r = 1 as well as the the initial condition ut(r, θ, 0) = 0.To satisfy the initial condition u(r, θ, 0) = cos θ we require∑n,m

Jn(rm,nr)(Am,n cosnθ +Bm,n sinnθ) = cos θ.

Thus all Fourier sine coefficients Bm.n = 0, while the only non-vanishingFourier cosine coefficients are Am,1. These coefficients are determined from∑

mJ1(rm,1r)Am,1 = 1.

We get

Am,1 =

∫ 1

0J1(rm,1r)rdr∫ 1

0[J1(rm,1r)]2rdr

andu(r, θ, t) =

∞∑m=1

Am,1 cos(rm,1t)Jn(rm,1r) cos θ.

8. Consider the partial differential equation

[1r2

∂

∂rr2∂

∂r+

1r2 sin2 θ

∂2

∂φ2+

1r2 sin θ

∂

∂θsin θ

∂

∂θ]u = V (r, θ, φ)u,

where

V (r, θ, φ) = V1(r) +V2(φ)r2 sin2 θ

+V3(θ)r2

.

Perform separation of variables and reduce the equation into three ordinarydifferential equations.

Solution:

Substitutingu = R(r)Θ(θ)Φ(φ)into the PDE and multiplying the resulting equation by r2, we get

[1R

d

dr(r2

dR

dr) − r2V1(r)] +

1sin2 θ

[1Φd2Φdφ2

− V2(φ)] + [1Θ

1sin θ

(d

dθsin θ

dΘdθ

) −

V3(θ)] = 0.Note that the first bracket above is a function of r only, while the second

and the third brackets are functions of θ and φ only. Thus we have1R

d

dr(r2

dR

dr) − r2V1(r) = − 1

sin2 θ[1Φd2Φdφ2− V2(φ)] − [

1Θ

1sin θ

d

dθ(sin θ

dΘdθ

) −

V3(θ)] = a,where a is a constant. The equation above contains two ordinary differential

equations:d

dr(r2

dR

dr)− [r2V1(r) + a]R = 0,

and1

sin2 θ[1Φd2Φdφ2− V2(φ)] + [

1Θ

1sin θ

d

dθ(sin θ

dΘdθ

)− V3(θ)] + a = 0.

23

Multiplying the equation above by sin2 θ, we get

[1Φd2Φdφ2− V2(φ)] + [

1Θ

sin θd

dθ(sin θ

dΘdθ

)− sin2 θV3(θ) + a sin2 θ] = 0.

We see that the first bracket above is a function of φ only, while the secondbracket above is a function of θ only. Thus we have

1Φd2Φdφ2− V2(φ) = − 1

Θsin θ

d

dθ(sin θ

dΘdθ

) + sin2 θV3(θ)− a sin2 θ = b,

where b is a constant. Therefore we have the ordinary differential equationsd2Φdφ2− [b+ V2(φ)]Φ = 0

andsin θ

d

dθ[sin θ

dΘdθ

] + [b+ a sin2 θ − sin2 θV3(θ)]Θ = 0.

(Indeed, it is not difficult to prove that V = V1(r) +V2(φ)r2 sin2 θ

+V3(θ)r2

is the

most general form of V for the equation in this problem to be separable.)

6.d2u

dx2= −λu, 0 < x < L; u′(0) = 0 and u(L) = 0.

The general solution of the differential equation isu(x) = a cos(

√λx) +b sin(

√λx).

Taking the derivative of the expression above, we getu′(x) =

√λ[−a sin(

√λx) +b cos(

√λx)].

The boundary condition u′(0) = 0 givesb√λ = 0⇒ either λ = 0 or b = 0.

If λ = 0, the solution of the differential equation is u = c1+c2x. This solutionvanishes identically as the boundary conditions are imposed.

If λ 6= 0, we have b = 0, oru = a cos(

√λx).

The boundary condition u(L) = 0 givesa cos(

√λL) = 0.

Since a 6= 0, the eigenvalues of the non-trivial solutions are√λnL = (n+ 1/2)π, n = 0, 1, 2, · · ·

or

λn =(n+ 1/2)2π2

L2, n = 0, 1, 2, · · ·.

The eigenfunction corresponding to the eigenvalue above is

un(x) =√

2L

cos[(n+ 1/2)πx

L].

These eigenfunctions are orthonormal:∫ L0un(x)um(x)dx = δnm.

3. uxx + uyy = 0, ∞ > y > 0 and 0 < x < 1;ux(0, y) = 0, u(L, y) = 0andu(x, 0) = δ(x− 1/2) and u(x,∞) = 0.Let us express u by the Fourier cosine series

24

u(x, y) =∞∑n=0

An(y) cos[(n+ 1/2)πx],

which satisfies the boundary condition at x = 0 and x = 1. The coefficientAn(y) satisfies the differential equation

d2An(y)dy2

= [(n+ 1/2)π]2An(y).

The solution of the differential equation above isAn(y) = ane

(n+1/2)πy + bne−(n+1/2)πy.

The boundary condition at y =∞ gives an = 0. Therefore

u(x, y) =∞∑n=0

bne−(n+1/2)πy cos[(n+ 1/2)πx].

The boundary condition at y = 0 requires

δ(x− 1/2 ) =∞∑n=0

bn cos[(n+ 1/2]πx].

Thus we have12bn =

∫ 1

0δ(x− 1/2) cos[(n+ 1/2)πx]dx = cos[(n+ 1/2)π/2],

orbn = 2 cos[(n+ 1/2)π/2].Therefore,

u(x, y) = 2∞∑n=0

e−(n+1/2)πy cos[(n+ 1/2)π

2] cos[(n+ 1/2)πx].

Chapter 6

1. Find the two independent solutions ofd2y

dx2+ x

dy

dx+ y = 0

in the form of Maclaurin series. One of the series can be summed into aclosed form. Find this closed form.

Solution:

Since x = 0 is an ordinary point of the equation, we put y =∑Anx

n. Thusd2y

dx2=

∑Ann(n− 1)xn−2,

xdy

dx+ y =

∑An(n+ 1)xn =

∑An−2(n− 1)xn−2.

We haveAn = −An−2

n.

25

The recurrence formula above for n = 2m gives

A2m = −A2(m−1)

2m=⇒ A2m = (−1)m

A0

2mm!.

Thus one of the solutions is

y1(x) =∞∑m=0

(−x2)m

2mm!= e−x

2/2,

which is in a closed form. The recurrence formula above for n = 2m + 1gives

A2m+1 = −A2(m−1)+1

2(m+ 1/2)= (−1)m

A1Γ(3/2)2mΓ(m+ 3/2)

.

Thus the second solution is

y2(x) = x∞∑m=0

(−x2/2)m

Γ(m+ 3/2).

The general solution of the equation is

y(x) = ae−x2/2 + bx

∞∑m=0

(−x2/2)m

Γ(m+ 3/2).

Note: We may take advantage of the fact that the Wronskian between y1and y2 is a known function, and express y2 as an integral.

2. Find the Frobenius solutions of

x(1− x)d2y

dx2+dy

dx+

34y = 0,

the truncated series of which are good approximations of the solutions nearx = 0.

Solution:

a. The point x = 0 is a regular singular point. Thus we try y =∑Anx

n+s.We have

xd2y

dx2+dy

dx=

∑An[(n+s)(n+s−1)+(n+s)]xn+s−1 =

∑An(n+s)2xn+s−1,

−x2 d2y

dx2+

34y = −

∑An[(n+ s)(n+ s− 1)− 3/4]xn+s

= −∑An(n+ s− 3/2)(n+ s+ 1/2)xn+s = −

∑An−1(n+ s− 5/2)(n+ s−

1/2)xn+s−1.Thus we have(n+ s)2An = (n+ s− 5/2)(n+ s− 1/2)An−1.The indicial equation is obtained by setting n in the equation above to zero.

We gets2 = 0 =⇒ s1 = s2 = 0.

An =(n+ s− 5/2)(n+ s− 1/2)

(n+ s)2An−1

=⇒ An =[(n+ s− 5/2) · · · (s− 3/2)][(n+ s− 1/2) · · · (s+ 1/2)

(n+ s)2 · · · (1 + s)2

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=Γ(n+ s− 3/2)Γ(n+ s+ 1/2)

[Γ(n+ s+ 1)]2[Γ(s+ 1)]2

Γ(s− 3/2)Γ(s+ 1/2)

′

,

where we have set A0 to unity. Thus one of the solutions is obtained bysetting s = 0 :

y1(x) =1

Γ(−3/2)Γ(1/2)∑ Γ(n− 3/2)Γ(n+ 1/2)

[n!]2xn.

The second solution is

y2(x) =∂

∂s

∞∑n=0

[(n+ s− 5/2) · · · (s− 3/2)][(n+ s− 1/2) · · · (s+ 1/2)(n+ s)2 · · · (1 + s)2

xn+s]s=0

= (lnx)y1(x) +∞∑n=0

[(n− 5/2) · · · (−3/2)][(n− 1/2) · · · (1/2)](n!)2

·[( 1n− 5/2

+ · · ·+ 1−3/2

) + (1

n− 1/2+ · · ·+ 1

1/2)− 2(

1n

+ · · ·+ 11

)]xn

3.Consider the equation

xd2y

dx2+ (

12− x)

dy

dx− by = 0 ,

where b is a constant.a.Locate and classify the singular points of this equation in the extended

plane.b.Find by the Frobenius method the two independent solutions of this equa-

tion, the point of expansion being the origin.

Solution:

1.(a)The point x = 0 is a regular singular point.Let x = 1/t ,then we have

td

dtt2dy

dt− (

12t2 − t)dy

dt− by = 0 ,

ord2y

dt2+

3t/2 + 1t2

dy

dt− b

t3y = 0 .

Thus the point ∞ is an irregular singualr point of rank 1 .

(b) Lety =

∑anx

n+s ,wherea0 6= 0 , and a−1 = a−2 = · · · = 0 .Then−by = −

∑banx

n+s = −∑ban−1x

n+s−1 ,12y′ =

∑ 12an(n+ s)xn+s−1 ,

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−xy′ = −∑an−1(n+ s− 1)xn+s−1

andxy′′ =

∑an(n+ s)(n+ s− 1)xn+s−1

We then have

an(n+ s)(n+ s− 1/2) = an−1(n+ s− 1 + b)Setting n = 0 in the equation above, we gets = 0 or 1/2 .For n 6= 0, we have

an =(n+ s− 1 + b)

(n+ s)(n+ s− 1/2)an−1

=Γ(n+ s+ b)Γ(s+ 1)Γ(s+ 1/2)

Γ(s+ b)Γ(n+ s+ 1)Γ(n+ s+ 1/2)a0 .

Setting s = 0 , we find that one of the solutions is

y1 =∞∑n=0

Γ(n+ b)Γ(n+ 1/2)n!

xn .

Setting s = 1/2 , we find the other solution is

y2 =∞∑n=0

Γ(n+ 1/2 + b)Γ(n+ 3/2)n!

xn+1/2 .

4. Consider the differential equationy” + x2y = 0 .

a.Find and classify the singular points of this equation in the extended com-plex plane.

b.Find one (any non-trivial one) of its solutions in the form of a series ex-panded around the origin.

Solution:a.There are no singular points in the finite plane.

Let x =1t

,then

y” = t2d

dtt2dy

dt= t4

d2y

dt2+ 2t3

dy

dt,

and henced2y

dt2+

2t

dy

dt+

1t6y = 0 .

Thus the infinity is an irregular singular point of the differential equation.

b. Lety =

∑anx

n

with the understanding thata−1 = a−2 = · · · = 0 .

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We havey” =

∑ann(n− 1)xn−2

andx2y =

∑anx

n+2 =∑an−4x

n−2 .Thus we haveann(n− 1) = −an−4 ,oran = − an−4

n(n− 1).

Thus

a4m = −a4(m−1)

16m(m− 1/4)= (− 1

16)m

a0Γ(3/4)m!Γ(m+ 3/4)

.

Thus a solution is∞∑m=0

(− 116

)mx4m

m!Γ(m+ 3/4).

5. Find the general solution ofxy”− y = 0in two steps.a. Find a solution of this ODE with the Frobenius method.b. Find another independent solution of this ODE in the form of an infinite

series.

Solution:

6. Lety”− (x4 − 3x−2/16)y = 0.

a. Locate and classify all the singular points of this equation, finite orinfinite. Give the rank of each of the irregular singular points, if any.

b. Find the two independent solutions of this equation which are useful whenx is small.

7.a. Find with the Frobenius method one of the independent solutions inthe form of a series (expanded around the origin) for

x(1− x)d2y

dx2+dy

dx+

34y = 0.

b. Find the second independence solution of the equation.

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