3.4E: Exercises

In Exercises \((3.4E.1)\) to \((3.4E.18)\), find the general solution of the given Euler equation on \((0,\infty)\).

Exercise \(\PageIndex{1}\)

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

Exercise \(\PageIndex{2}\)

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

Answer

\(p(r)=r(r-1)-7r+7=(r-7)(r-1)\) ; \(y=c_1x+c_2x^7\) .

Exercise \(\PageIndex{3}\)

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

Exercise \(\PageIndex{4}\)

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

Answer

\(p(r)=r(r-1)+5r+4=(r+2)^2\) ;\(y=x^{-2}(c_1+c_2 \ln x)\)

Exercise \(\PageIndex{5}\)

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

Exercise \(\PageIndex{6}\)

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

Answer

\(p(r)=r(r-1)-3r+13=(r-2)^2+9\) ;
 \(y=x^2[c_1 \cos (3 \ln x)+c_2 \sin (3 \ln x)]\) .

Exercise \(\PageIndex{7}\)

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

Exercise \(\PageIndex{8}\)

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

Answer

\(p(r)=12r(r-1)-5r+6=(3r-2)(4r-3)\) ;
\(y=c_1x^{2/3}+c_2 x^{3/4}\) .

Exercise \(\PageIndex{9}\)

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

Exercise \(\PageIndex{10}\)

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

Answer

\(p(r)=3r(r-1)-r+1=(r-1)(3r-1)\) ; \(y=c_1x+c_2x^{1/3}\) .

Exercise \(\PageIndex{11}\)

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

Exercise \(\PageIndex{12}\)

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

Answer

\(p(r)=r(r-1)+3r+5=(r+1)^2+4\) ;
\(y=\displaystyle{
 {1\over x}\left[c_1\cos(2 \ln x)+c_2\sin(2 \ln x\right]}\)

Exercise \(\PageIndex{13}\)

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

Exercise \(\PageIndex{14}\)

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

Answer

\(p(r)=r(r-1)-r+10=(r-1)^2+9\) ;
 \(y=x\left[c_1\cos(3 \ln x)+c_2\sin(3 \ln x)\right]\) .

Exercise \(\PageIndex{15}\)

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

Exercise \(\PageIndex{16}\)

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

Answer

\(p(r)=2r(r-1)+3r-1=(r+1)(2r-1)\) ;

\(y=\displaystyle {\frac{c_1}{x}+\frac{c_2}{x^{1/2}}}\).
 

Exercise \(\PageIndex{17}\)

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

Exercise \(\PageIndex{18}\)

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

Answer

\(p(r)=2r(r-1)+10r+9=2(r+2)^2+1\) ;
\(y=\displaystyle{{1\over x^2} \left[c_1\cos\left({1\over\sqrt{2}} \ln x\right)+c_2\sin\left({1\over\sqrt{2}} \ln x\right) \right]}\) .

Exercise \(\PageIndex{19}\)

(a) Adapt the proof of Theorem \((3.4.3)\) to show that \(y=y(x)\) satisfies the Euler equation

\begin{equation}\label{eq:3.4E.1}
ax^2y''+bxy'+cy=0
\end{equation}

on \((-\infty,0)\) if and only if \(Y(t)=y(-e^t)\)

\begin{eqnarray*}
a {d^2Y\over dt^2}+(b-a){dY\over dt}+cY=0.
\end{eqnarray*}

on \((-\infty,\infty)\).

(b) Use part (a) to show that the general solution of \eqref{eq:3.4E.1} on \((-\infty,0)\) is

\begin{eqnarray*}
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{eqnarray*}

Exercise \(\PageIndex{20}\)

Use reduction of order to show that if

\begin{eqnarray*}
ar(r-1)+br+c=0
\end{eqnarray*}

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

\begin{eqnarray*}
ax^2y''+bxy'+cy=0
\end{eqnarray*}

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

Answer

If \(p(r)=ar(r-1)+br+c=a(r-r_1)^2\) , then (A) \(p(r_1)=p'(r_1)=0\) .
If \(y=ux^{r_1}\) , then \(y'=u'x^{r_1}+r_1ux^{r_1-1}\)
and \(y''=u''x^{r_1}+2r_1u'x_1^{r_1-1}+r_1(r_1-1)x_1^{r_1-2}\) , so
\begin{eqnarray*}
ax^2y''+bxy'+cy&=&ax^{r_1+2}u''+(2ar_1+b)x^{r_1+1}u'
+\left(ar_1(r_1-1)+br_1+c\right)x^{r_1}u\\
&=&ax^{r_1+2}u''+p'(r_1)x_1^{r_1+1}u'+p(r)x^{r_1}u=ax^{r_1+2}u'',
\end{eqnarray*}
from (A). Therefore,\) u''=0\) , so \) u=c_1+c_2x\) and
\(y=x^{r_1}(c_1+c_2x)\) .

Exercise \(\PageIndex{21}\)

A nontrivial solution of

\begin{eqnarray*}
P_0(x)y''+P_1(x)y'+P_2(x)y=0
\end{eqnarray*}

is said to be \( \textcolor{blue}{\mbox{oscillatory}} \) on an interval \((a,b)\) if it has infinitely many zeros on \((a,b)\). Otherwise \(y\) is said to be \( \textcolor{blue}{\mbox{nonoscillatory}} \) on \((a,b)\). Show that the equation

\begin{eqnarray*}
x^2y''+ky=0 \quad (k=\; \mbox{constant})
\end{eqnarray*}

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

Exercise \(\PageIndex{22}\)

In Example \((3.4.2)\) we saw that \(x_0=1\) and \(x_0=-1\) are regular singular points of Legendre's equation

\begin{equation}\label{eq:3.4E.2}
(1-x^2)y''-2xy'+\alpha(\alpha+1)y=0.
\end{equation}

(a) Introduce the new variables \(t=x-1\) and \(Y(t)=y(t+1)\), and show that \(y\) is a solution of \eqref{eq:3.4E.2} if and only if \(Y\) is a solution of

\begin{eqnarray*}
t(2+t){d^2Y\over dt^2}+2(1+t){dY\over dt}-\alpha(\alpha+1)Y=0,
\end{eqnarray*}

which has a regular singular point at \(t_0=0\).

(b) Introduce the new variables \(t=x+1\) and \(Y(t)=y(t-1)\), and show that \(y\) is a solution of \eqref{eq:3.4E.2} if and only if \(Y\( is a solution of

\begin{eqnarray*}
t(2-t){d^2Y\over dt^2}+2(1-t){dY\over dt}+\alpha(\alpha+1)Y=0,
\end{eqnarray*}

which has a regular singular point at \(t_0=0\).

Answer

a. If \( t=x-1\) and \(y(t)=y(t+1)=y(x)\) , then
\( \displaystyle(1-x^2)y''-2xy'+\alpha(\alpha+1)y=
-t(2+t){d^2Y\over dt^2}-2(1+t){dY\over dt}+\alpha(\alpha+1)Y=0\) ,
so \(y\) satisfies Legendre's equation if and only if \(y\) satisfies
(A) \) \displaystyle t(2+t){d^2Y\over dt^2}+2(1+t){dY\over
dt}-\alpha(\alpha+1)Y=0\) . Since (A) can be rewritten as
\) \displaystyle t^2(2+t){d^2Y\over dt^2}+2t(1+t){dY\over
dt}-\alpha(\alpha+1)tY=0\) , (A) has a regular singular point at
\) t=0_0\) .

b. If \( t=x+1\) and \(y(t)=y(t-1)=y(x)\) , then
\( \displaystyle(1-x^2)y''-2xy'+\alpha(\alpha+1)y=
t(2-t){d^2Y\over dt^2}+2(1-t){dY\over dt}+\alpha(\alpha+1)Y\) ,
so \(y\) satisfies Legendre's equation if and only if \(y\) satisfies
(B) \) \displaystyle t(2-t){d^2Y\over dt^2}+2(1-t){dY\over
dt}+\alpha(\alpha+1)Y\) ,
 Since (B) can be rewritten as
(B) \( \displaystyle t^2(2-t){d^2Y\over dt^2}+2t(1-t){dY\over
dt}+\alpha(\alpha+1)tY\) ,
(B) has a regular singular point at
\( t=0_0\) .

Exercise \(\PageIndex{23}\)

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

\begin{equation}\label{eq:3.4E.3}
P_0(x)y''+P_1(x)y'+P_2(x)y=0.
\end{equation}

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

(a) Show that \(y\) is a solution of \eqref{eq:3.4E.3} if and only if \(Y\) is a solution of

\begin{equation}\label{eq:3.4E.4}
R_0(t){d^2Y\over dt^2}+R_1(t){dY\over dt}+R_2(t)Y=0.
\end{equation}

where

\begin{eqnarray*}
R_i(t)=P_i(t+x_0),\quad i=0,1,2.
\end{eqnarray*}

(b) 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 \eqref{eq:3.4E.4}.