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Mathematics LibreTexts

7.3E: Series Solutions Near an Ordinary Point II (Exercises)

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    18311
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    In Exercises [exer:7.3.1} –[exer:7.3.12} find the coefficients \(a_0\),…, \(a_N\) for \(N\) at least \(7\) in the series solution \(y=\sum_{n=0}^\infty a_nx^n\) of the initial value problem.

    [exer:7.3.1] \((1+3x)y''+xy'+2y=0,\quad y(0)=2,\quad y'(0)=-3\)

    [exer:7.3.2] \((1+x+2x^2)y''+(2+8x)y'+4y=0,\quad y(0)=-1,\quad y'(0)=2\)

    [exer:7.3.3] \((1-2x^2)y''+(2-6x)y'-2y=0,\quad y(0)=1,\quad y'(0)=0\)

    [exer:7.3.4] \((1+x+3x^2)y''+(2+15x)y'+12y=0,\quad y(0)=0,\quad y'(0)=1\)

    [exer:7.3.5] \((2+x)y''+(1+x)y'+3y=0,\quad y(0)=4,\quad y'(0)=3\)

    [exer:7.3.6] \((3+3x+x^2)y''+(6+4x)y'+2y=0,\quad y(0)=7,\quad y'(0)=3\)

    [exer:7.3.7] \((4+x)y''+(2+x)y'+2y=0,\quad y(0)=2,\quad y'(0)=5\)

    [exer:7.3.8] \((2-3x+2x^2)y''-(4-6x)y'+2y=0,\quad y(1)=1,\quad y'(1)=-1\)

    [exer:7.3.9] \((3x+2x^2)y''+10(1+x)y'+8y=0,\quad y(-1)=1,\quad y'(-1)=-1\)

    [exer:7.3.10] \((1-x+x^2)y''-(1-4x)y'+2y=0,\quad y(1)=2,\quad y'(1)=-1\)

    [exer:7.3.11] \((2+x)y''+(2+x)y'+y=0,\quad y(-1)=-2,\quad y'(-1)=3\)

    [exer:7.3.12] \(x^2y''-(6-7x)y'+8y=0,\quad y(1)=1,\quad y'(1)=-2\)

    [exer:7.3.13] Do the following experiment for various choices of real numbers \(a_0\), \(a_1\), and \(r\), with \(0<r<1/\sqrt2\).

    Use differential equations software to solve the initial value problem

    \[(1+x+2x^2)y''+(1+7x)y'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \eqno{\rm(A)}\nonumber \]

    numerically on \((-r,r)\). (See Example

    Example \(\PageIndex{1}\):

    Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at7.3.1}.)

    For \(N=2\), \(3\), \(4\), …, compute \(a_2\), …, \(a_N\) in the power series solution \(y=\sum_{n=0}^\infty a_nx^n\) of (A), and graph

    \[T_N(x)=\sum_{n=0}^N a_nx^n\nonumber \]

    and the solution obtained in (a) on \((-r,r)\). Continue increasing \(N\) until there’s no perceptible difference between the two graphs.

    [exer:7.3.14] Do the following experiment for various choices of real numbers \(a_0\), \(a_1\), and \(r\), with \(0<r<2\).

    Use differential equations software to solve the initial value problem \[(3+x)y''+(1+2x)y'-(2-x)y=0,\quad y(-1)=a_0,\quad y'(-1)=a_1, \eqno{\rm(A)}\nonumber \]

    numerically on \((-1-r,-1+r)\). (See Example 7.3.2}. Why this interval?)

    For \(N=2\), \(3\), \(4\), …, compute \(a_2,\dots,a_N\) in the power series solution

    \[y=\sum_{n=0}^\infty a_n(x+1)^n\nonumber \]

    of (A), and graph

    \[T_N(x)=\sum_{n=0}^N a_n(x+1)^n\nonumber \]

    and the solution obtained in (a) on \((-1-r,-1+r)\). Continue increasing \(N\) until there’s no perceptible difference between the two graphs.

    [exer:7.3.15] Do the following experiment for several choices of \(a_0\), \(a_1\), and \(r\), with \(r>0\).

    Use differential equations software to solve the initial value problem \[y''+3xy'+(4+2x^2)y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \eqno{\rm(A)}\nonumber \]

    numerically on \((-r,r)\). (See Example 7.3.3}.)

    Find the coefficients \(a_0\), \(a_1\), …, \(a_N\) in the power series solution \(y=\sum_{n=0}^\infty a_nx^n\) of (A), and graph

    \[T_N(x)=\sum_{n=0}^N a_nx^n\nonumber \]

    and the solution obtained in (a) on \((-r,r)\). Continue increasing \(N\) until there’s no perceptible difference between the two graphs.

    [exer:7.3.16] Do the following experiment for several choices of \(a_0\) and \(a_1\).

    Use differential equations software to solve the initial value problem

    \[(1-x)y''-(2-x)y'+y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \eqno{\rm(A)}\nonumber \]

    numerically on \((-r,r)\).

    Find the coefficients \(a_0\), \(a_1\), …, \(a_N\) in the power series solution \(y=\sum_{n=0}^Na_nx^n\) of (A), and graph

    \[T_N(x)=\sum_{n=0}^N a_nx^n\nonumber \]

    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 \(r\to1\)?

    [exer:7.3.17] Follow the directions of Exercise [exer:7.3.16} for the initial value problem \[(1+x)y''+3y'+32y=0,\quad y(0)=a_0,\quad y'(0)=a_1.\nonumber \]

    [exer:7.3.18] Follow the directions of Exercise [exer:7.3.16} for the initial value problem \[(1+x^2)y''+y'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1.\nonumber \]

    [exer:7.3.19] \((2+4x)y''-4y'-(6+4x)y=0,\quad y(0)=2,\quad y'(0)=-7\)

    [exer:7.3.20] \((1+2x)y''-(1-2x)y'-(3-2x)y=0,\quad y(1)=1,\quad y'(1)=-2\)

    [exer:7.3.21] \((5+2x)y''-y'+(5+x)y=0,\quad y(-2)=2,\quad y'(-2)=-1\)

    [exer:7.3.22] \((4+x)y''-(4+2x)y'+(6+x)y=0,\quad y(-3)=2,\quad y'(-3)=-2\)

    [exer:7.3.23] \((2+3x)y''-xy'+2xy=0,\quad y(0)=-1,\quad y'(0)=2\)

    [exer:7.3.24] \((3+2x)y''+3y'-xy=0,\quad y(-1)=2,\quad y'(-1)=-3\)

    [exer:7.3.25] \((3+2x)y''-3y'-(2+x)y=0,\quad y(-2)=-2,\quad y'(-2)=3\)

    [exer:7.3.26] \((10-2x)y''+(1+x)y=0,\quad y(2)=2,\quad y'(2)=-4\)

    [exer:7.3.27] \((7+x)y''+(8+2x)y'+(5+x)y=0,\quad y(-4)=1,\quad y'(-4)=2\)

    [exer:7.3.28] \((6+4x)y''+(1+2x)y=0,\quad y(-1)=-1,\quad y'(-1)=2\)

    [exer:7.3.29] Show that the coefficients in the power series in \(x\) for the general solution of \[(1+\alpha x+\beta x^2)y''+(\gamma+\delta x)y'+\epsilon y=0\nonumber \] satisfy the recurrrence relation \[a_{n+2}=-{\gamma+\alpha n\over n+2}\,a_{n+1}-{\beta n(n-1)+\delta n+\epsilon\over(n+2)(n+1)}\, a_n.\nonumber \]

    [exer:7.3.30] Let \(\alpha\) and \(\beta\) be constants, with \(\beta\ne0\). Show that \(y=\sum_{n=0}^\infty a_nx^n\) is a solution of \[(1+\alpha x+\beta x^2)y''+(2\alpha+4\beta x)y'+2\beta y=0 \eqno{\rm (A)}\nonumber \] if and only if \[a_{n+2}+\alpha a_{n+1}+\beta a_n=0,\quad n\ge0. \eqno{\rm (B)}\nonumber \]

    An equation of this form is called a second order homogeneous linear difference equation. The polynomial \(p(r)=r^2+\alpha r+\beta\) is called the characteristic polynomial of (B). If \(r_1\) and \(r_2\) are the zeros of \(p\), then \(1/r_1\) and \(1/r_2\) are the zeros of \[P_0(x)=1+\alpha x+\beta x^2.\nonumber \]

    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 \[a_n=c_1r_1^n+c_2r_2^n,\quad n\ge0\nonumber \] satisfies (B). Conclude from this that any function of the form \[y=\sum_{n=0}^\infty (c_1r_1^n+c_2r_2^n)x^n\nonumber \] is a solution of (A) on \((-\rho,\rho)\).

    Use (b) and the formula for the sum of a geometric series to show that the functions

    \[y_1={1\over1-r_1x}\quad\mbox{ and }\quad y_2={1\over1-r_2x}\nonumber \]

    form a fundamental set of solutions of (A) on \((-\rho,\rho)\).

    Show that \(\{y_1,y_2\}\) is a fundamental set of solutions of (A) on any interval that does’nt contain either \(1/r_1\) or \(1/r_2\).

    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

    \[a_n=(c_1+c_2n)r_1^n,\quad n\ge0\nonumber \]

    satisfies (B). Conclude from this that any function of the form

    \[y=\sum_{n=0}^\infty (c_1+c_2n)r_1^nx^n\nonumber \]

    is a solution of (A) on \((-\rho,\rho)\).

    Use (e) and the formula for the sum of a geometric series to show that the functions

    \[y_1={1\over1-r_1x}\quad\mbox{ and }\quad y_2={x\over(1-r_1x)^2}\nonumber \]

    form a fundamental set of solutions of (A) on \((-\rho,\rho)\).

    Show that \(\{y_1,y_2\}\) is a fundamental set of solutions of (A) on any interval that does not contain \(1/r_1\).

    [exer:7.3.31] Use the results of Exercise [exer:7.3.30} 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) \((4+8x+3x^2)y''+(16+12x)y'+6y=0\)

    [exer:7.3.32] \(y''+2xy'+(3+2x^2)y=0,\quad y(0)=1,\quad y'(0)=-2\)

    [exer:7.3.33] \(y''-3xy'+(5+2x^2)y=0,\quad y(0)=1,\quad y'(0)=-2\)

    [exer:7.3.34] \(y''+5xy'-(3-x^2)y=0,\quad y(0)=6,\quad y'(0)=-2\)

    [exer:7.3.35] \(y''-2xy'-(2+3x^2)y=0,\quad y(0)=2,\quad y'(0)=-5\)

    [exer:7.3.36] \(y''-3xy'+(2+4x^2)y=0,\quad y(0)=3,\quad y'(0)=6\)

    [exer:7.3.37] \(2y''+5xy'+(4+2x^2)y=0,\quad y(0)=3,\quad y'(0)=-2\)

    [exer:7.3.38] \(3y''+2xy'+(4-x^2)y=0,\quad y(0)=-2,\quad y'(0)=3\)

    [exer:7.3.39] Find power series in \(x\) for the solutions \(y_1\) and \(y_2\) of \[y''+4xy'+(2+4x^2)y=0\nonumber \] 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.

    [exer:7.3.40] \((1+x)y''+x^2y'+(1+2x)y=0,\quad y(0)-2,\quad y'(0)=3\)

    [exer:7.3.41] \(y''+(1+2x+x^2)y'+2y=0,\quad y(0)=2,\quad y'(0)=3\)

    [exer:7.3.42] \((1+x^2)y''+(2+x^2)y'+xy=0,\quad y(0)=-3,\quad y'(0)=5\)

    [exer:7.3.43] \((1+x)y''+(1-3x+2x^2)y'-(x-4)y=0,\quad y(1)=-2,\quad y'(1)=3\)

    [exer:7.3.44] \(y''+(13+12x+3x^2)y'+(5+2x),\quad y(-2)=2,\quad y'(-2)=-3\)

    [exer:7.3.45] \((1+2x+3x^2)y''+(2-x^2)y'+(1+x)y=0,\quad y(0)=1,\quad y'(0)=-2\)

    [exer:7.3.46] \((3+4x+x^2)y''-(5+4x-x^2)y'-(2+x)y=0,\quad y(-2)=2,\quad y'(-2)=-1\)

    [exer:7.3.47] \((1+2x+x^2)y''+(1-x)y=0,\quad y(0)=2,\quad y'(0)=-1\)

    [exer:7.3.48] \((x-2x^2)y''+(1+3x-x^2)y'+(2+x)y=0,\quad y(1)=1,\quad y'(1)=0\)

    [exer:7.3.49] \((16-11x+2x^2)y''+(10-6x+x^2)y'-(2-x)y,\quad y(3)=1,\quad y'(3)=-2\)