7.5E: Regular Singular Points Euler Equations (Exercises)
( \newcommand{\kernel}{\mathrm{null}\,}\)
Q7.4.1
In Exercises 7.4.1-7.4.18 find the general solution of the given Euler equation on (0,∞).
1. x2y″+7xy′+8y=0
2. x2y″−7xy′+7y=0
3. x2y″−xy′+y=0
4. x2y″+5xy′+4y=0
5. x2y″+xy′+y=0
6. x2y″−3xy′+13y=0
7. x2y″+3xy′−3y=0
8. 12x2y″−5xy″+6y=0
9. 4x2y″+8xy′+y=0
10. 3x2y″−xy′+y=0
11. 2x2y″−3xy′+2y=0
12. x2y″+3xy′+5y=0
13. 9x2y″+15xy′+y=0
14. x2y″−xy′+10y=0
15. x2y″−6y=0
16. 2x2y″+3xy′−y=0
17. x2y″−3xy′+4y=0
18. 2x2y″+10xy′+9y=0
Q7.4.2
19.
- Adapt the proof of Theorem 7.4.3 to show that y=y(x) satisfies the Euler equation ax2y″+bxy′+cy=0 on (−∞,0) if and only if Y(t)=y(−et) ad2Ydt2+(b−a)dYdt+cY=0. on (−∞,∞).
- Use (a) to show that the general solution of Equation A 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=r2; y=|x|λ[c1cos(ωln|x|)+c2sin(ωln|x|)] if r1,r2=λ±iω with ω>0.
20. Use reduction of order to show that if
ar(r−1)+br+c=0
has a repeated root r1 then y=xr1(c1+c2lnx) is the general solution of
ax2y″+bxy′+cy=0
on (0,∞).
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.
22. In Example 7.4.2 we saw that x0=1 and x0=−1 are regular singular points of Legendre’s equation
(1−x2)y″−2xy′+α(α+1)y=0.
- Introduce the new variables t=x−1 and Y(t)=y(t+1), and show that y is a solution of (A) if and only if Y is a solution of t(2+t)d2Ydt2+2(1+t)dYdt−α(α+1)Y=0, which has a regular singular point at t0=0.
- Introduce the new variables t=x+1 and Y(t)=y(t−1), and show that y is a solution of (A) if and only if Y is a solution of t(2−t)d2Ydt2+2(1−t)dYdt+α(α+1)Y=0, which has a regular singular point at t0=0.
23. Let P0,P1, and P2 be polynomials with no common factor, and suppose x0≠0 is a singular point of
P0(x)y″+P1(x)y′+P2(x)y=0.
Let t=x−x0 and Y(t)=y(t+x0).- Show that y is a solution of (A) if and only if Y is a solution of R0(t)d2Ydt2+R1(t)dYdt+R2(t)Y=0. where Ri(t)=Pi(t+x0),i=0,1,2.
- 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 (B).