7.5E: Regular Singular Points Euler Equations (Exercises)

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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.

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 $(-\infty,\infty)$.
Use (a) to show that the general solution of Equation A on $(-\infty,0)$ is \[\begin{aligned} y&=c_1|x|^{r_1}+c_2|x|^{r_2}\mbox{ if $r_1$ and $r_2$ are distinct real numbers; } \\ y&=|x|^{r_1}(c_1+c_2\ln|x|)\mbox{ if $r_1=r_2$; } \\ y&=|x|^{\lambda}\left[c_1\cos\left(\omega\ln|x|\right)+ c_2\sin\left(\omega\ln|x| \right)\right]\mbox{ if $r_1,r_2=\lambda\pm i\omega$ with $\omega>0$}.\end{aligned}\nonumber\]

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

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

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 $t_0=0$.
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

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

Let $t=x-x_0$ and $Y(t)=y(t+x_0)$.

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 \[R_i(t)=P_i(t+x_0),\quad i=0,1,2.\nonumber\]
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).