Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

3.4E: Exercises

This page is a draft and is under active development. 

( \newcommand{\kernel}{\mathrm{null}\,}\)

In Exercises (3.4E.1) to (3.4E.18), find the general solution of the given Euler equation on (0,).

Exercise 3.4E.1

x2y+7xy+8y=0

Exercise 3.4E.2

x2y7xy+7y=0

Answer

p(r)=r(r1)7r+7=(r7)(r1) ; y=c1x+c2x7 .

Exercise 3.4E.3

x2yxy+y=0

Exercise 3.4E.4

x2y+5xy+4y=0

Answer

p(r)=r(r1)+5r+4=(r+2)2 ; y=x2(c1+c2lnx)

Exercise 3.4E.5

x2y+xy+y=0

Exercise 3.4E.6

x2y3xy+13y=0

Answer

p(r)=r(r1)3r+13=(r2)2+9 ;
  y=x2[c1cos(3lnx)+c2sin(3lnx)] .

Exercise 3.4E.7

x2y+3xy3y=0

Exercise 3.4E.8

12x2y5xy+6y=0

Answer

p(r)=12r(r1)5r+6=(3r2)(4r3) ;
y=c1x2/3+c2x3/4 .

Exercise 3.4E.9

4x2y+8xy+y=0

Exercise 3.4E.10

3x2yxy+y=0

Answer

p(r)=3r(r1)r+1=(r1)(3r1) ; y=c1x+c2x1/3 .

Exercise 3.4E.11

2x2y3xy+2y=0

Exercise 3.4E.12

x2y+3xy+5y=0

Answer

p(r)=r(r1)+3r+5=(r+1)2+4 ;
y=1x[c1cos(2lnx)+c2sin(2lnx]

Exercise 3.4E.13

9x2y+15xy+y=0

Exercise 3.4E.14

x2yxy+10y=0

Answer

p(r)=r(r1)r+10=(r1)2+9 ;
  y=x[c1cos(3lnx)+c2sin(3lnx)] .

Exercise 3.4E.15

x2y6y=0

Exercise 3.4E.16

2x2y+3xyy=0

Answer

p(r)=2r(r1)+3r1=(r+1)(2r1) ;

y=c1x+c2x1/2.
 

Exercise 3.4E.17

x2y3xy+4y=0

Exercise 3.4E.18

2x2y+10xy+9y=0

Answer

p(r)=2r(r1)+10r+9=2(r+2)2+1 ;
y=1x2[c1cos(12lnx)+c2sin(12lnx)] .

Exercise 3.4E.19

(a) Adapt the proof of Theorem (3.4.3) to show that y=y(x) satisfies the Euler equation

(3.4E.1)ax2y+bxy+cy=0

on (,0) if and only if Y(t)=y(et)

ad2Ydt2+(ba)dYdt+cY=0.

on (,).

(b) Use part (a) to show that the general solution of (3.4E.1) on (,0) is

y=c1|x|r1+c2|x|r2 if r1 and r2 are distinct real numbers; y=|x|r1(c1+c2ln|x|) if r1=r2y=|x|λ[c1cos(ωln|x|)+c2sin(ωln|x|)] if r1,r2=λ±iω with ω>0.

Exercise 3.4E.20

Use reduction of order to show that if

ar(r1)+br+c=0

has a repeated root r1 then y=xr1(c1+c2lnx) is the general solution of

ax2y+bxy+cy=0

on (0,).

Answer

If p(r)=ar(r1)+br+c=a(rr1)2 , then (A) p(r1)=p(r1)=0 .
If y=uxr1 , then y=uxr1+r1uxr11
and y=uxr1+2r1ux1r11+r1(r11)x1r12 , so
ax2y+bxy+cy=axr1+2u+(2ar1+b)xr1+1u+(ar1(r11)+br1+c)xr1u=axr1+2u+p(r1)x1r1+1u+p(r)xr1u=axr1+2u,
from (A). Therefore,\) u''=0\) , so \) u=c_1+c_2x\) and
y=xr1(c1+c2x) .

Exercise 3.4E.21

A nontrivial solution of

P0(x)y+P1(x)y+P2(x)y=0

is said to be oscillatory on an interval (a,b) if it has infinitely many zeros on (a,b). Otherwise y is said to be nonoscillatory on (a,b). Show that the equation

x2y+ky=0(k=constant)

has oscillatory solutions on (0,) if and only if k>1/4.

Exercise 3.4E.22

In Example (3.4.2) we saw that x0=1 and x0=1 are regular singular points of Legendre's equation

(3.4E.2)(1x2)y2xy+α(α+1)y=0.

(a) Introduce the new variables t=x1 and Y(t)=y(t+1), and show that y is a solution of (3.4E.2) if and only if Y is a solution of

t(2+t)d2Ydt2+2(1+t)dYdtα(α+1)Y=0,

which has a regular singular point at t0=0.

(b) Introduce the new variables t=x+1 and Y(t)=y(t1), and show that y is a solution of (3.4E.2) if and only if \(Y\( is a solution of

t(2t)d2Ydt2+2(1t)dYdt+α(α+1)Y=0,

which has a regular singular point at t0=0.

Answer

a. If t=x1 and y(t)=y(t+1)=y(x) , then
(1x2)y2xy+α(α+1)y=t(2+t)d2Ydt22(1+t)dYdt+α(α+1)Y=0 ,
so y satisfies Legendre's equation if and only if y satisfies
(A) \) \displaystyle t(2+t){d^2Y\over dt^2}+2(1+t){dY\over
dt}-\alpha(\alpha+1)Y=0\) . Since (A) can be rewritten as
\) \displaystyle t^2(2+t){d^2Y\over dt^2}+2t(1+t){dY\over
dt}-\alpha(\alpha+1)tY=0\) , (A) has a regular singular point at
\) t=0_0\) .


b. If t=x+1 and y(t)=y(t1)=y(x) , then
(1x2)y2xy+α(α+1)y=t(2t)d2Ydt2+2(1t)dYdt+α(α+1)Y ,
so y satisfies Legendre's equation if and only if y satisfies
(B) \) \displaystyle t(2-t){d^2Y\over dt^2}+2(1-t){dY\over
dt}+\alpha(\alpha+1)Y\) ,
 Since (B) can be rewritten as
(B) t2(2t)d2Ydt2+2t(1t)dYdt+α(α+1)tY ,
(B) has a regular singular point at
t=00 .

Exercise 3.4E.23

Let P0,P1, and P2 be polynomials with no common factor, and suppose x00 is a singular point of

(3.4E.3)P0(x)y+P1(x)y+P2(x)y=0.

Let t=xx0 and Y(t)=y(t+x0).

(a) Show that y is a solution of (3.4E.3) if and only if Y is a solution of

(3.4E.4)R0(t)d2Ydt2+R1(t)dYdt+R2(t)Y=0.

where

Ri(t)=Pi(t+x0),i=0,1,2.

(b) Show that R0, R1, and R2 are polynomials in t with no common factors, and R0(0)=0; thus, t0=0 is a singular point of (3.4E.4).


This page titled 3.4E: Exercises is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by William F. Trench.

Support Center

How can we help?