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).