3.3E: Exercises
- Page ID
- 17625
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In Exercises \((3.3E.1)\) to \((3.3E.12)\), find the coefficients \(a_0\), \(\dots\), \(a_N\) for \(N\) at least \(7\) in the series solution \(y=\sum_{n=0}^\infty a_nx^n\) of the initial value problem.
Exercise \(\PageIndex{1}\)
\((1+3x)y''+xy'+2y=0,\quad y(0)=2,\quad y'(0)=-3\)
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Exercise \(\PageIndex{2}\)
\((1+x+2x^2)y''+(2+8x)y'+4y=0,\quad y(0)=-1,\quad y'(0)=2\)
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Exercise \(\PageIndex{3}\)
\((1-2x^2)y''+(2-6x)y'-2y=0,\quad y(0)=1,\quad y'(0)=0\)
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Exercise \(\PageIndex{4}\)
\((1+x+3x^2)y''+(2+15x)y'+12y=0,\quad y(0)=0,\quad y'(0)=1\)
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Exercise \(\PageIndex{5}\)
\((2+x)y''+(1+x)y'+3y=0,\quad y(0)=4,\quad y'(0)=3\)
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Exercise \(\PageIndex{6}\)
\((3+3x+x^2)y''+(6+4x)y'+2y=0,\quad y(0)=7,\quad y'(0)=3\)
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Exercise \(\PageIndex{7}\)
\((4+x)y''+(2+x)y'+2y=0,\quad y(0)=2,\quad y'(0)=5\)
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Exercise \(\PageIndex{8}\)
\((2-3x+2x^2)y''-(4-6x)y'+2y=0,\quad y(1)=1,\quad y'(1)=-1\)
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Exercise \(\PageIndex{9}\)
\((3x+2x^2)y''+10(1+x)y'+8y=0,\quad y(-1)=1,\quad y'(-1)=-1\)
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Exercise \(\PageIndex{10}\)
\((1-x+x^2)y''-(1-4x)y'+2y=0,\quad y(1)=2,\quad y'(1)=-1\)
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Exercise \(\PageIndex{11}\)
\((2+x)y''+(2+x)y'+y=0,\quad y(-1)=-2,\quad y'(-1)=3\)
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Exercise \(\PageIndex{12}\)
\(x^2y''-(6-7x)y'+8y=0,\quad y(1)=1,\quad y'(1)=-2\)
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Exercise \(\PageIndex{13}\)
Do the following experiment for various choices of real numbers \(a_0\), \(a_1\), and \(r\), with \(0<r<1/\sqrt2\).
(a) Use differential equations software to solve the initial value problem
\begin{equation}\label{eq:3.3E.1}
(1+x+2x^2)y''+(1+7x)y'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1,
\end{equation}
numerically on \((-r,r)\). (See Example \((3.3.1)\).)
(b) For \(N=2\), \(3\), \(4\), \(\dots\), compute \(a_2\), \(\dots\), \(a_N\) in the power series solution \(y=\sum_{n=0}^\infty a_nx^n\) of \eqref{eq:3.3E.1}, and graph
\begin{eqnarray*}
T_N(x)=\sum_{n=0}^N a_nx^n
\end{eqnarray*}
and the solution obtained in (a) on \((-r,r)\). Continue increasing \(N\) until there's no perceptible difference between the two graphs.
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Exercise \(\PageIndex{14}\)
Do the following experiment for various choices of real numbers \(a_0\), \(a_1\), and \(r\), with \(0<r<2\).
(a) Use differential equations software to solve the initial value problem
\begin{equation}\label{eq:3.3E.2}
(3+x)y''+(1+2x)y'-(2-x)y=0,\quad y(-1)=a_0,\quad y'(-1)=a_1,
\end{equation}
numerically on \((-1-r,-1+r)\). (See Example \((3.3.2)\). Why this interval?)
(b) For \(N=2\), \(3\), \(4\), \(\dots\), compute \(a_2,\dots,a_N\) in the power series solution
\begin{eqnarray*}
y=\sum_{n=0}^\infty a_n(x+1)^n
\end{eqnarray*}
of \eqref{eq:3.3E.2}, and graph
\begin{eqnarray*}
T_N(x)=\sum_{n=0}^N a_n(x+1)^n
\end{eqnarray*}
and the solution obtained in (a) on \((-1-r,-1+r)\). Continue increasing \(N\) until there's no perceptible difference between the two graphs.
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Exercise \(\PageIndex{15}\)
Do the following experiment for several choices of \(a_0\), \(a_1\), and \(r\), with \(r>0\).
(a) Use differential equations software to solve the initial value problem
\begin{equation}\label{eq:3.3E.3}
y''+3xy'+(4+2x^2)y=0,\quad y(0)=a_0,\quad y'(0)=a_1,
\end{equation}
numerically on \((-r,r)\). (See Example \((3.3.3)\).)
(b) Find the coefficients \(a_0\), \(a_1\), \(\dots\), \(a_N\) in the power series solution \(y=\sum_{n=0}^\infty a_nx^n\) of \eqref{eq:3.3E.3}, and graph
\begin{eqnarray*}
T_N(x)=\sum_{n=0}^N a_nx^n
\end{eqnarray*}
and the solution obtained in (a) on \((-r,r)\). Continue increasing \(N\) until there's no perceptible difference between the two graphs.
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Exercise \(\PageIndex{16}\)
Do the following experiment for several choices of \(a_0\) and \(a_1\).
(a) Use differential equations software to solve the initial value problem
\begin{equation}\label{eq:3.3E.4}
(1-x)y''-(2-x)y'+y=0,\quad y(0)=a_0,\quad y'(0)=a_1,
\end{equation}
numerically on \((-r,r)\).
(b) Find the coefficients \(a_0\), \(a_1\), \(\dots\), \(a_N\) in the power series solution \(y=\sum_{n=0}^Na_nx^n\) of \eqref{eq:3.3E.4}, and graph
\begin{eqnarray*}
T_N(x)=\sum_{n=0}^N a_nx^n
\end{eqnarray*}
and the solution obtained in (a) on \((-r,r)\). Continue increasing \(N\) until there's no perceptible difference between the two graphs. What happens as you let \(r\to1\)?
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Exercise \(\PageIndex{17}\)
Follow the directions of Exercise \((3.3E.16)\) for the initial value problem
\begin{eqnarray*}
(1+x)y''+3y'+32y=0,\quad y(0)=a_0,\quad y'(0)=a_1.
\end{eqnarray*}
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Exercise \(\PageIndex{18}\)
Follow the directions of Exercise \((3.3E.16)\) for the initial value problem
\begin{eqnarray*}
(1+x^2)y''+y'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1.
\end{eqnarray*}
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In Exercises \((3.3E.19)\) to \((3.3E.28)\), find the coefficients \(a_0\), \(\dots\), \(a_N\) for \(N\) at least \(7\) in the series solution
\begin{eqnarray*}
y=\sum_{n=0}^\infty a_n(x-x_0)^n
\end{eqnarray*}
of the initial value problem. Take \(x_0\) to be the point where the initial conditions are imposed.
Exercise \(\PageIndex{19}\)
\((2+4x)y''-4y'-(6+4x)y=0,\quad y(0)=2,\quad y'(0)=-7\)
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Exercise \(\PageIndex{20}\)
\((1+2x)y''-(1-2x)y'-(3-2x)y=0,\quad y(1)=1,\quad y'(1)=-2\)
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Exercise \(\PageIndex{21}\)
\((5+2x)y''-y'+(5+x)y=0,\quad y(-2)=2,\quad y'(-2)=-1\)
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Exercise \(\PageIndex{22}\)
\((4+x)y''-(4+2x)y'+(6+x)y=0,\quad y(-3)=2,\quad y'(-3)=-2\)
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Exercise \(\PageIndex{23}\)
\((2+3x)y''-xy'+2xy=0,\quad y(0)=-1,\quad y'(0)=2\)
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Exercise \(\PageIndex{24}\)
\((3+2x)y''+3y'-xy=0,\quad y(-1)=2,\quad y'(-1)=-3\)
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Exercise \(\PageIndex{25}\)
\((3+2x)y''-3y'-(2+x)y=0,\quad y(-2)=-2,\quad y'(-2)=3\)
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Exercise \(\PageIndex{26}\)
\((10-2x)y''+(1+x)y=0,\quad y(2)=2,\quad y'(2)=-4\)
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Exercise \(\PageIndex{27}\)
\((7+x)y''+(8+2x)y'+(5+x)y=0,\quad y(-4)=1,\quad y'(-4)=2\)
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Exercise \(\PageIndex{28}\)
\((6+4x)y''+(1+2x)y=0,\quad y(-1)=-1,\quad y'(-1)=2\)
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Exercise \(\PageIndex{29}\)
Show that the coefficients in the power series in \(x\) for the general solution of
\begin{eqnarray*}
(1+\alpha x+\beta x^2)y''+(\gamma+\delta x)y'+\epsilon y=0
\end{eqnarray*}
satisfy the recurrence relation
\begin{eqnarray*}
a_{n+2}=-{\gamma+\alpha n\over n+2}\,a_{n+1}-{\beta n(n-1)+\delta n+\epsilon\over(n+2)(n+1)}\, a_n.
\end{eqnarray*}
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Exercise \(\PageIndex{30}\)
(a) Let \(\alpha\) and \(\beta\) be constants, with \(\beta\ne0\). Show that \(y=\sum_{n=0}^\infty a_nx^n\) is a solution of
\begin{equation}\label{eq:3.3E.5}
(1+\alpha x+\beta x^2)y''+(2\alpha+4\beta x)y'+2\beta y=0
\end{equation}
if and only if
\begin{equation}\label{eq:3.3E.6}
a_{n+2}+\alpha a_{n+1}+\beta a_n=0,\quad n\ge0.
\end{equation}
An equation of this form is called a \( \textcolor{blue}{\mbox{second order homogeneous linear difference equation}} \). The polynomial \(p(r)=r^2+\alpha r+\beta\) is called the \( \textcolor{blue}{\mbox{characteristic polynomial}} \) of \eqref{eq:3.3E.6}. If \(r_1\) and \(r_2\) are the zeros of \(p\), then \(1/r_1\) and \(1/r_2\) are the zeros of
\begin{eqnarray*}
P_0(x)=1+\alpha x+\beta x^2.
\end{eqnarray*}
(b) Suppose \(p(r)=(r-r_1)(r-r_2)\) where \(r_1\) and \(r_2\) are real and distinct, and let \(\rho\) be the smaller of the two numbers \(\{1/|r_1|,1/|r_2|\}\). Show that if \(c_1\) and \(c_2\) are constants then the sequence
\begin{eqnarray*}
a_n=c_1r_1^n+c_2r_2^n,\quad n\ge0
\end{eqnarray*}
satisfies \eqref{eq:3.3E.6}. Conclude from this that any function of the form
\begin{eqnarray*}
y=\sum_{n=0}^\infty (c_1r_1^n+c_2r_2^n)x^n
\end{eqnarray*}
is a solution of \eqref{eq:3.3E.5} on \((-\rho,\rho)\).
(c) Use (b) and the formula for the sum of a geometric series to show that the functions
\begin{eqnarray*}
y_1={1\over1-r_1x}\quad\mbox{ and }\quad y_2={1\over1-r_2x}
\end{eqnarray*}
form a fundamental set of solutions of \eqref{eq:3.3E.5} on \((-\rho,\rho)\).
(d) Show that \(\{y_1,y_2\}\) is a fundamental set of solutions of \eqref{eq:3.3E.5} on any interval that doesn't contain either \(1/r_1\) or \(1/r_2\).
(e) Suppose \(p(r)=(r-r_1)^2\), and let \(\rho=1/|r_1|\). Show that if \(c_1\) and \(c_2\) are constants then the sequence
\begin{eqnarray*}
a_n=(c_1+c_2n)r_1^n,\quad n\ge0
\end{eqnarray*}
satisfies \eqref{eq:3.3E.6}. Conclude from this that any function of the form
\begin{eqnarray*}
y=\sum_{n=0}^\infty (c_1+c_2n)r_1^nx^n
\end{eqnarray*}
is a solution of \eqref{eq:3.3E.5} on \((-\rho,\rho)\).
(f) Use (e) and the formula for the sum of a geometric series to show that the functions
\begin{eqnarray*}
y_1={1\over1-r_1x}\quad\mbox{ and }\quad y_2={x\over(1-r_1x)^2}
\end{eqnarray*}
form a fundamental set of solutions of \eqref{eq:3.3E.5} on \((-\rho,\rho)\).
(g) Show that \(\{y_1,y_2\}\) is a fundamental set of solutions of \eqref{eq:3.3E.5} on any interval that does not contain \(1/r_1\).
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Exercise \(\PageIndex{31}\)
Use the results of Exercise \((3.3E.30)\) to find the general solution of the given equation on any interval on which polynomial multiplying \(y''\) has no zeros.
(a) \((1+3x+2x^2)y''+(6+8x)y'+4y=0\)
(b) \((1-5x+6x^2)y''-(10-24x)y'+12y=0\)
(c) \((1-4x+4x^2)y''-(8-16x)y'+8y=0\)
(d) \((4+4x+x^2)y''+(8+4x)y'+2y=0\)
(e) \((4+8x+3x^2)y''+(16+12x)y'+6y=0\)
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In Exercises \((3.3E.32)\) to \((3.3E.38)\), find the coefficients \(a_0\), \(\dots\), \(a_N\) for \(N\) at least \(7\) in the series solution \(y=\sum_{n=0}^\infty a_nx^n\) of the initial value problem.
Exercise \(\PageIndex{32}\)
\(y''+2xy'+(3+2x^2)y=0,\quad y(0)=1,\quad y'(0)=-2\)
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Exercise \(\PageIndex{33}\)
\(y''-3xy'+(5+2x^2)y=0,\quad y(0)=1,\quad y'(0)=-2\)
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Exercise \(\PageIndex{34}\)
\(y''+5xy'-(3-x^2)y=0,\quad y(0)=6,\quad y'(0)=-2\)
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Exercise \(\PageIndex{35}\)
\(y''-2xy'-(2+3x^2)y=0,\quad y(0)=2,\quad y'(0)=-5\)
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Exercise \(\PageIndex{36}\)
\(y''-3xy'+(2+4x^2)y=0,\quad y(0)=3,\quad y'(0)=6\)
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Exercise \(\PageIndex{37}\)
\(2y''+5xy'+(4+2x^2)y=0,\quad y(0)=3,\quad y'(0)=-2\)
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Exercise \(\PageIndex{38}\)
\(3y''+2xy'+(4-x^2)y=0,\quad y(0)=-2,\quad y'(0)=3\)
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Exercise \(\PageIndex{39}\)
Find power series in \(x\) for the solutions \(y_1\) and \(y_2\) of
\begin{eqnarray*}
y''+4xy'+(2+4x^2)y=0
\end{eqnarray*}
such that \(y_1(0)=1\), \(y'_1(0)=0\), \(y_2(0)=0\), \(y'_2(0)=1\), and identify \(y_1\) and \(y_2\) in terms of familiar elementary functions.
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In Exercises \((3.3E.40)\) tp \((3.3E.49)\), find the coefficients \(a_0\), \(\dots\), \(a_N\) for \(N\) at least \(7\) in the series solution
\begin{eqnarray*}
y=\sum_{n=0}^\infty a_n(x-x_0)^n
\end{eqnarray*}
of the initial value problem. Take \(x_0\) to be the point where the initial conditions are imposed.
Exercise \(\PageIndex{40}\)
\((1+x)y''+x^2y'+(1+2x)y=0,\quad y(0)-2,\quad y'(0)=3\)
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Exercise \(\PageIndex{41}\)
\(y''+(1+2x+x^2)y'+2y=0,\quad y(0)=2,\quad y'(0)=3\)
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Exercise \(\PageIndex{42}\)
\((1+x^2)y''+(2+x^2)y'+xy=0,\quad y(0)=-3,\quad y'(0)=5\)
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Exercise \(\PageIndex{43}\)
\((1+x)y''+(1-3x+2x^2)y'-(x-4)y=0,\quad y(1)=-2,\quad y'(1)=3\)
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Exercise \(\PageIndex{44}\)
\(y''+(13+12x+3x^2)y'+(5+2x),\quad y(-2)=2,\quad y'(-2)=-3\)
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Exercise \(\PageIndex{45}\)
\((1+2x+3x^2)y''+(2-x^2)y'+(1+x)y=0,\quad y(0)=1,\quad y'(0)=-2\)
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Exercise \(\PageIndex{46}\)
\((3+4x+x^2)y''-(5+4x-x^2)y'-(2+x)y=0,\quad y(-2)=2,\quad y'(-2)=-1\)
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Exercise \(\PageIndex{47}\)
\((1+2x+x^2)y''+(1-x)y=0,\quad y(0)=2,\quad y'(0)=-1\)
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Exercise \(\PageIndex{48}\)
\((x-2x^2)y''+(1+3x-x^2)y'+(2+x)y=0,\quad y(1)=1,\quad y'(1)=0\)
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Exercise \(\PageIndex{49}\)
\((16-11x+2x^2)y''+(10-6x+x^2)y'-(2-x)y,\quad y(3)=1,\quad y'(3)=-2\)
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