# 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 \(\PageIndex{1}\)

\((1+3x)y''+xy'+2y=0,\quad y(0)=2,\quad y'(0)=-3\)

**Answer**-
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## Exercise \(\PageIndex{2}\)

\((1+x+2x^2)y''+(2+8x)y'+4y=0,\quad y(0)=-1,\quad y'(0)=2\)

**Answer**-
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## Exercise \(\PageIndex{3}\)

\((1-2x^2)y''+(2-6x)y'-2y=0,\quad y(0)=1,\quad y'(0)=0\)

**Answer**-
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## Exercise \(\PageIndex{4}\)

\((1+x+3x^2)y''+(2+15x)y'+12y=0,\quad y(0)=0,\quad y'(0)=1\)

**Answer**-
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## Exercise \(\PageIndex{5}\)

\((2+x)y''+(1+x)y'+3y=0,\quad y(0)=4,\quad y'(0)=3\)

**Answer**-
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## Exercise \(\PageIndex{6}\)

\((3+3x+x^2)y''+(6+4x)y'+2y=0,\quad y(0)=7,\quad y'(0)=3\)

**Answer**-
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## Exercise \(\PageIndex{7}\)

\((4+x)y''+(2+x)y'+2y=0,\quad y(0)=2,\quad y'(0)=5\)

**Answer**-
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## Exercise \(\PageIndex{8}\)

\((2-3x+2x^2)y''-(4-6x)y'+2y=0,\quad y(1)=1,\quad y'(1)=-1\)

**Answer**-
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## Exercise \(\PageIndex{9}\)

\((3x+2x^2)y''+10(1+x)y'+8y=0,\quad y(-1)=1,\quad y'(-1)=-1\)

**Answer**-
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## Exercise \(\PageIndex{10}\)

\((1-x+x^2)y''-(1-4x)y'+2y=0,\quad y(1)=2,\quad y'(1)=-1\)

**Answer**-
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## Exercise \(\PageIndex{11}\)

\((2+x)y''+(2+x)y'+y=0,\quad y(-1)=-2,\quad y'(-1)=3\)

**Answer**-
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## Exercise \(\PageIndex{12}\)

\(x^2y''-(6-7x)y'+8y=0,\quad y(1)=1,\quad y'(1)=-2\)

**Answer**-
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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.

**Answer**-
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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.

**Answer**-
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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.

**Answer**-
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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 \(r\to1\)?

**Answer**-
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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*}

**Answer**-
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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*}

**Answer**-
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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 \(\PageIndex{19}\)

\((2+4x)y''-4y'-(6+4x)y=0,\quad y(0)=2,\quad y'(0)=-7\)

**Answer**-
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## Exercise \(\PageIndex{20}\)

\((1+2x)y''-(1-2x)y'-(3-2x)y=0,\quad y(1)=1,\quad y'(1)=-2\)

**Answer**-
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## Exercise \(\PageIndex{21}\)

\((5+2x)y''-y'+(5+x)y=0,\quad y(-2)=2,\quad y'(-2)=-1\)

**Answer**-
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## Exercise \(\PageIndex{22}\)

\((4+x)y''-(4+2x)y'+(6+x)y=0,\quad y(-3)=2,\quad y'(-3)=-2\)

**Answer**-
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## Exercise \(\PageIndex{23}\)

\((2+3x)y''-xy'+2xy=0,\quad y(0)=-1,\quad y'(0)=2\)

**Answer**-
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## Exercise \(\PageIndex{24}\)

\((3+2x)y''+3y'-xy=0,\quad y(-1)=2,\quad y'(-1)=-3\)

**Answer**-
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## Exercise \(\PageIndex{25}\)

\((3+2x)y''-3y'-(2+x)y=0,\quad y(-2)=-2,\quad y'(-2)=3\)

**Answer**-
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## Exercise \(\PageIndex{26}\)

\((10-2x)y''+(1+x)y=0,\quad y(2)=2,\quad y'(2)=-4\)

**Answer**-
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## Exercise \(\PageIndex{27}\)

\((7+x)y''+(8+2x)y'+(5+x)y=0,\quad y(-4)=1,\quad y'(-4)=2\)

**Answer**-
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## Exercise \(\PageIndex{28}\)

\((6+4x)y''+(1+2x)y=0,\quad y(-1)=-1,\quad y'(-1)=2\)

**Answer**-
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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*}

**Answer**-
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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 \(1/r_1\).

**Answer**-
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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) \((4+8x+3x^2)y''+(16+12x)y'+6y=0\)

**Answer**-
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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 \(\PageIndex{32}\)

\(y''+2xy'+(3+2x^2)y=0,\quad y(0)=1,\quad y'(0)=-2\)

**Answer**-
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## Exercise \(\PageIndex{33}\)

\(y''-3xy'+(5+2x^2)y=0,\quad y(0)=1,\quad y'(0)=-2\)

**Answer**-
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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 \(\PageIndex{38}\)

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

**Answer**-
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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.

**Answer**-
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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 \(\PageIndex{40}\)

\((1+x)y''+x^2y'+(1+2x)y=0,\quad y(0)-2,\quad y'(0)=3\)

**Answer**-
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## Exercise \(\PageIndex{41}\)

\(y''+(1+2x+x^2)y'+2y=0,\quad y(0)=2,\quad y'(0)=3\)

**Answer**-
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## Exercise \(\PageIndex{42}\)

\((1+x^2)y''+(2+x^2)y'+xy=0,\quad y(0)=-3,\quad y'(0)=5\)

**Answer**-
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## Exercise \(\PageIndex{43}\)

\((1+x)y''+(1-3x+2x^2)y'-(x-4)y=0,\quad y(1)=-2,\quad y'(1)=3\)

**Answer**-
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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 \(\PageIndex{45}\)

\((1+2x+3x^2)y''+(2-x^2)y'+(1+x)y=0,\quad y(0)=1,\quad y'(0)=-2\)

**Answer**-
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## Exercise \(\PageIndex{46}\)

\((3+4x+x^2)y''-(5+4x-x^2)y'-(2+x)y=0,\quad y(-2)=2,\quad y'(-2)=-1\)

**Answer**-
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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 \(\PageIndex{48}\)

\((x-2x^2)y''+(1+3x-x^2)y'+(2+x)y=0,\quad y(1)=1,\quad y'(1)=0\)

**Answer**-
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## Exercise \(\PageIndex{49}\)

\((16-11x+2x^2)y''+(10-6x+x^2)y'-(2-x)y,\quad y(3)=1,\quad y'(3)=-2\)

**Answer**-
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