7.5E: Regular Singular Points Euler Equations (Exercises)

Q7.4.1

In Exercises 7.4.1-7.4.18 find the general solution of the given Euler equation on (0,∞).

1. 𝑥2⁢𝑦″ +7⁢𝑥⁢𝑦′ +8⁢𝑦 =0

2. 𝑥2⁢𝑦″ −7⁢𝑥⁢𝑦′ +7⁢𝑦 =0

3. 𝑥2⁢𝑦″ −𝑥⁢𝑦′ +𝑦 =0

4. 𝑥2⁢𝑦″ +5⁢𝑥⁢𝑦′ +4⁢𝑦 =0

5. 𝑥2⁢𝑦″ +𝑥⁢𝑦′ +𝑦 =0

6. 𝑥2⁢𝑦″ −3⁢𝑥⁢𝑦′ +13⁢𝑦 =0

7. 𝑥2⁢𝑦″ +3⁢𝑥⁢𝑦′ −3⁢𝑦 =0

8. 12⁢𝑥2⁢𝑦″ −5⁢𝑥⁢𝑦″ +6⁢𝑦 =0

9. 4⁢𝑥2⁢𝑦″ +8⁢𝑥⁢𝑦′ +𝑦 =0

10. 3⁢𝑥2⁢𝑦″ −𝑥⁢𝑦′ +𝑦 =0

11. 2⁢𝑥2⁢𝑦″ −3⁢𝑥⁢𝑦′ +2⁢𝑦 =0

12. 𝑥2⁢𝑦″ +3⁢𝑥⁢𝑦′ +5⁢𝑦 =0

13. 9⁢𝑥2⁢𝑦″ +15⁢𝑥⁢𝑦′ +𝑦 =0

14. 𝑥2⁢𝑦″ −𝑥⁢𝑦′ +10⁢𝑦 =0

15. 𝑥2⁢𝑦″ −6⁢𝑦 =0

16. 2⁢𝑥2⁢𝑦″ +3⁢𝑥⁢𝑦′ −𝑦 =0

17. 𝑥2⁢𝑦″ −3⁢𝑥⁢𝑦′ +4⁢𝑦 =0

18. 2⁢𝑥2⁢𝑦″ +10⁢𝑥⁢𝑦′ +9⁢𝑦 =0

Q7.4.2

19.

1. Adapt the proof of Theorem 7.4.3 to show that 𝑦 =𝑦⁡(𝑥) satisfies the Euler equation 𝑎⁢𝑥2⁢𝑦″+𝑏⁢𝑥⁢𝑦′+𝑐⁢𝑦=0(A) on (−∞,0) if and only if 𝑌⁡(𝑡) =𝑦⁡(−𝑒𝑡) 𝑎⁢𝑑2⁢𝑌𝑑⁢𝑡2+(𝑏−𝑎)⁢𝑑⁢𝑌𝑑⁢𝑡+𝑐⁢𝑌=0. on (−∞,∞).
2. 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+𝑐2⁢ln⁡|𝑥|)⁢ if 𝑟1=𝑟2; 𝑦=|𝑥|𝜆⁢[𝑐1⁢cos⁡(𝜔⁢ln⁡|𝑥|)+𝑐2⁢sin⁡(𝜔⁢ln⁡|𝑥|)]⁢ if 𝑟1,𝑟2=𝜆±𝑖⁢𝜔 with 𝜔>0.

20. Use reduction of order to show that if

𝑎⁢𝑟⁢(𝑟−1)+𝑏⁢𝑟+𝑐=0

has a repeated root 𝑟1 then 𝑦 =𝑥𝑟1⁢(𝑐1 +𝑐2⁢ln⁡𝑥) is the general solution of

𝑎⁢𝑥2⁢𝑦″+𝑏⁢𝑥⁢𝑦′+𝑐⁢𝑦=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

𝑥2⁢𝑦″+𝑘⁢𝑦=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″−2⁢𝑥⁢𝑦′+𝛼⁢(𝛼+1)⁢𝑦=0.(A)

1. 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)⁢𝑑2⁢𝑌𝑑⁢𝑡2+2⁢(1+𝑡)⁢𝑑⁢𝑌𝑑⁢𝑡−𝛼⁢(𝛼+1)⁢𝑌=0, which has a regular singular point at 𝑡0 =0.
2. 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)⁢𝑑2⁢𝑌𝑑⁢𝑡2+2⁢(1−𝑡)⁢𝑑⁢𝑌𝑑⁢𝑡+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.(A)

Let 𝑡 =𝑥 −𝑥0 and 𝑌⁡(𝑡) =𝑦⁢(𝑡 +𝑥0).

1. Show that 𝑦 is a solution of (A) if and only if 𝑌 is a solution of 𝑅0⁡(𝑡)⁢𝑑2⁢𝑌𝑑⁢𝑡2+𝑅1⁡(𝑡)⁢𝑑⁢𝑌𝑑⁢𝑡+𝑅2⁡(𝑡)⁢𝑌=0.(B) where 𝑅𝑖⁡(𝑡)=𝑃𝑖⁢(𝑡+𝑥0),𝑖=0,1,2.
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).