7.4.1: 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 \((0,\infty)\).

1. \(x^2y''+7xy'+8y=0\)

2. \(x^2y''-7xy'+7y=0\)

3. \(x^2y''-xy'+y=0\)

4. \(x^2y''+5xy'+4y=0\)

5. \(x^2y''+xy'+y=0\)

6. \(x^2y''-3xy'+13y=0\)

7. \(x^2y''+3xy'-3y=0\)

8. \(12x^2y''-5xy''+6y=0\)

9. \(4x^2y''+8xy'+y=0\)

10. \(3x^2y''-xy'+y=0\)

11. \(2x^2y''-3xy'+2y=0\)

12. \(x^2y''+3xy'+5y=0\)

13. \(9x^2y''+15xy'+y=0\)

14. \(x^2y''-xy'+10y=0\)

15. \(x^2y''-6y=0\)

16. \(2x^2y''+3xy'-y=0\)

17. \(x^2y''-3xy'+4y=0\)

18. \(2x^2y''+10xy'+9y=0\)

Q7.4.2

19.

1. Adapt the proof of Theorem 7.4.3 to show that \(y=y(x)\) satisfies the Euler equation \[ax^2y''+bxy'+cy=0\tag{A} \] on \((-\infty,0)\) if and only if \(Y(t)=y(-e^t)\) \[a {d^2Y\over dt^2}+(b-a){dY\over dt}+cY=0.\nonumber \] on \((-\infty,\infty)\).
2. Use (a) to show that the general solution of Equation A on \((-\infty,0)\) is \[\begin{aligned} y&=c_1|x|^{r_1}+c_2|x|^{r_2}\mbox{ if $r_1$ and $r_2$ are distinct real numbers; } \\[4pt] y&=|x|^{r_1}(c_1+c_2\ln|x|)\mbox{ if $r_1=r_2$; } \\[4pt] y&=|x|^{\lambda}\left[c_1\cos\left(\omega\ln|x|\right)+ c_2\sin\left(\omega\ln|x| \right)\right]\mbox{ if $r_1,r_2=\lambda\pm i\omega$ with $\omega>0$}.\end{aligned}\nonumber \]

20. Use reduction of order to show that if

\[ar(r-1)+br+c=0\nonumber \]

has a repeated root \(r_1\) then \(y=x^{r_1}(c_1+c_2\ln x)\) is the general solution of

\[ax^2y''+bxy'+cy=0\nonumber \]

on \((0,\infty)\).

21. A nontrivial solution of

\[P_0(x)y''+P_1(x)y'+P_2(x)y=0\nonumber \]

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

\[x^2y''+ky=0 \quad (k=\; \mbox{constant})\nonumber \]

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

22. In Example 7.4.2 we saw that \(x_0=1\) and \(x_0=-1\) are regular singular points of Legendre’s equation

\[(1-x^2)y''-2xy'+\alpha(\alpha+1)y=0. \tag{A} \]

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 \[t(2+t){d^2Y\over dt^2}+2(1+t){dY\over dt}-\alpha(\alpha+1)Y=0,\nonumber \] which has a regular singular point at \(t_0=0\).
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 \[t(2-t){d^2Y\over dt^2}+2(1-t){dY\over dt}+\alpha(\alpha+1)Y=0,\nonumber \] which has a regular singular point at \(t_0=0\).

23. Let \(P_0,P_1\), and \(P_2\) be polynomials with no common factor, and suppose \(x_0\ne0\) is a singular point of

\[P_0(x)y''+P_1(x)y'+P_2(x)y=0. \tag{A} \]

1. Show that \(y\) is a solution of (A) if and only if \(Y\) is a solution of \[R_0(t){d^2Y\over dt^2}+R_1(t){dY\over dt}+R_2(t)Y=0. \tag{B} \] where \[R_i(t)=P_i(t+x_0),\quad i=0,1,2.\nonumber \]
2. Show that \(R_0\), \(R_1\), and \(R_2\) are polynomials in \(t\) with no common factors, and \(R_0(0)=0\); thus, \(t_0=0\) is a singular point of (B).