
# 7.5.1: 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; } \\ 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

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

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

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

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