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Tutorial 6: Power Series Solution

Tutorial Question

  1. Find the radius of convergence and interval of cinvergence for the given power series.

    a. ∑n=1∞2nnxn\sum _{n=1}^{\infty }\frac{2^{n}}{n} x^{n}

    Solution
    lim⁡n→∞∣an+1an∣=lim⁡n→∞∣2n+1xn+1/(n+1)2nxn/n∣=lim⁡n→∞2nn+1∣x∣=2∣x∣\lim _{n\rightarrow \infty }\left\vert \frac{a_{n} +1}{a_{n}}\right\vert =\lim _{n\rightarrow \infty }\left\vert \frac{2^{n+1} x^{n+1} /( n+1)}{2^{n} x^{n} /n}\right\vert =\lim _{n\rightarrow \infty }\frac{2n}{n+1}\vert x\vert =2\vert x\vert

    The series is absolutely convergent for 2∣x∣<12\vert x\vert < 1 or ∣x∣<12\vert x\vert < \frac{1}{2}. At x=−12x=-\frac{1}{2}, the series ∑n=1∞(−1)nn\sum _{n=1}^{\infty }\frac{( -1)^{n}}{n} converges by the alternating series test. At x=12x=\frac{1}{2}, the series ∑n=1∞1n\sum _{n=1}^{\infty }\frac{1}{n} is the harmonic series which diverges. Thus, the given series converges on [−12,12]\left[ -\frac{1}{2} ,\frac{1}{2}\right].

    b. ∑n=1∞(−1)n4n(x+3)n\sum _{n=1}^{\infty }\frac{( -1)^{n}}{4^{n}}( x+3)^{n}

    Solution
    L=lim⁡n→∞∣(−1)n+1(n+1)(x+3)n+14n+14n(−1)n(n)(x+3)n∣=lim⁡n→∞∣−(n+1)(x+3)4n∣=∣x+3∣lim⁡n→∞n+14n=14∣x+3∣\begin{aligned} L & =\lim _{n\rightarrow \infty }\left\vert \frac{( -1)^{n+1}( n+1)( x+3)^{n+1}}{4^{n+1}}\frac{4^{n}}{( -1)^{n}( n)( x+3)^{n}}\right\vert \\ & =\lim _{n\rightarrow \infty }\left\vert \frac{-( n+1)( x+3)}{4n}\right\vert \\ & =\vert x+3\vert \lim _{n\rightarrow \infty }\frac{n+1}{4n}\\ & =\frac{1}{4}\vert x+3\vert \end{aligned}

    14∣x+3∣<1⟹∣x+3∣<4\frac{1}{4}\vert x+3\vert < 1\Longrightarrow \vert x+3\vert < 4 series converges 14∣x+3∣>1⟹∣x+3∣>4\frac{1}{4}\vert x+3\vert >1\Longrightarrow \vert x+3\vert >4 series diverges

    The radius of convergence for this power series is R=4R=4.

    −4<x+3<4−7<x<1\begin{aligned} -4 & < x+3< 4\\ -7 & < x< 1 \end{aligned}

    c. ∑n=0∞(100)nn!(x+7)n\sum _{n=0}^{\infty }\frac{( 100)^{n}}{n!}( x+7)^{n}

    Solution
    lim⁡n→∞∣an+1an∣=lim⁡n→∞∣100n+1(x+7)n+1/(n+1)!100n(x+7)n/n!∣=lim⁡n→∞100n+1∣x+7∣=0\lim _{n\rightarrow \infty }\left\vert \frac{a_{n+1}}{a_{n}}\right\vert =\lim _{n\rightarrow \infty }\left\vert \frac{100^{n+1}( x+7)^{n+1} /( n+1) !}{100^{n}( x+7)^{n} /n!}\right\vert =\lim _{n\rightarrow \infty }\frac{100}{n+1}\vert x+7\vert =0

    The series is absolutely convergent on (−∞,∞)( -\infty ,\infty ).

    d. ∑n=0∞n!(2x+1)n\sum _{n=0}^{\infty } n!( 2x+1)^{n}

    Solution
    L=lim⁡n→∞∣(n+1)!(2x+1)n+1n!(2x+1)n∣=lim⁡n→∞∣(n+1)n!(2x+1)n!∣=∣2x+1∣lim⁡n→∞(n+1)\begin{aligned} L & =\lim _{n\rightarrow \infty }\left\vert \frac{( n+1) !( 2x+1)^{n+1}}{n!( 2x+1)^{n}}\right\vert \\ & =\lim _{n\rightarrow \infty }\left\vert \frac{( n+1) n!( 2x+1)}{n!}\right\vert \\ & =\vert 2x+1\vert \lim _{n\rightarrow \infty }( n+1) \end{aligned}

    At this point we need to be careful. Yje limit is infinite, but there is that term with the xx's in front of the limit. We will have L=∞>1L=\infty >1 provided x≠−12x\neq -\frac{1}{2} will only converge if x=−12x=-\frac{1}{2}.

    The radius of convergence is R=0R=0 and the interval of convergence is x=−12x=-\frac{1}{2}.

  2. Rewrite the given power series by shifting the index, so that its general term involves xkx^{k}.

    a. ∑n=3∞(2n−1)cnxn−3\sum _{n=3}^{\infty }( 2n-1) c_{n} x^{n-3}

    Solution
    ∑n=3∞(2n−1)cnxn−3=∑k=0∞(2(k+3)−1)ck+3xk=∑k=0∞(2k+5)ck+3xk\sum _{n=3}^{\infty }( 2n-1) c_{n} x^{n-3} =\sum _{k=0}^{\infty }( 2( k+3) -1) c_{k+3} x^{k} =\mathbf{\sum _{k=0}^{\infty }( 2k+5) c_{k+3} x^{k}}

    b. ∑n=3∞3n(2n)!xn−3\sum _{n=3}^{\infty }\frac{3^{n}}{( 2n) !} x^{n-3}

    Solution
    ∑n=3∞3n(2n)!xn−2=∑k=1∞3k+2(2(k+2))!xk=∑k=1∞3k+2(2k+1)!xk\sum _{n=3}^{\infty }\frac{3^{n}}{( 2n) !} x^{n-2} =\sum _{k=1}^{\infty }\frac{3^{k+2}}{( 2( k+2)) !} x^{k} =\mathbf{\sum _{k=1}^{\infty }\frac{3^{k+2}}{( 2k+1) !} x^{k}}

    c. ∑n=3∞(−1)n(2n+1)!x2n+1\sum _{n=3}^{\infty }\frac{( -1)^{n}}{( 2n+1) !} x^{2n+1}

    Solution
    ∑n=3∞(−1)n(2n+1)!x2n+1=∑k=7∞(−1)k−12(2(k−12)+1)!xk=∑k=7∞(−1)k−12k!xk=∑k=7∞(−1)k−1k!xk\sum _{n=3}^{\infty }\frac{( -1)^{n}}{( 2n+1) !} x^{2n+1} =\sum _{k=7}^{\infty }\frac{( -1)^{\frac{k-1}{2}}}{\left( 2\left(\frac{k-1}{2}\right) +1\right) !} x^{k} =\sum _{k=7}^{\infty }\frac{( -1)^{\frac{k-1}{2}}}{k!} x^{k} =\mathbf{\sum _{k=7}^{\infty }\frac{\sqrt{( -1)^{k-1}}}{k!} x^{k}}

  3. Rewrite the given expression as a single power series whose general term involves xkx^k.

    a. ∑n=2∞n(n−1)Cnxn+2∑n=2∞n(n−1)cnxn−2+3∑n=1∞ncnxn\sum _{n=2}^{\infty } n( n-1) C_{n} x^{n} +2\sum _{n=2}^{\infty } n( n-1) c_{n} x^{n-2} +3\sum _{n=1}^{\infty } nc_{n} x^{n}

    Solution
    ∑n=2∞n(n−1)cnxn+2∑n=2∞n(n−1)cnxn−2+3∑n=1∞ncnxn=2⋅2⋅1c2x0+2⋅3⋅2c3x′+3⋅1⋅c1x′+∑n=2∞n(n−1)cn⏟k=nxk=nn+2∑n=4∞n(n−1)cnxn−2⏟k=n−2+3∑n=2∞ncnxn⏟k=n=4c2+(12c3+(12c3+3c1)x+∑n=2∞k(k−1)ckxk+2∑n=2∞(k+2)(k+1)xk+2xk+3∑n=2∞kckxk=4c2+(3c1+12c3)x+∑n=2∞([k(k−1)+3k]ck+2(k+2)(k+1)xk+2)xk=4c2+(3c1+12c3)x+∑n=2∞[k(k+2)ck+2(k+1)(k+2)ck+2]xk\begin{aligned} & \sum _{n=2}^{\infty } n( n-1) c_{n} x^{n} +2\sum _{n=2}^{\infty } n( n-1) c_{n} x^{n-2} +3\sum _{n=1}^{\infty } nc_{n} x_{n}\\ &= 2\cdot 2\cdot 1c_{2} x^{0} +2\cdot 3\cdot 2c_{3} x'+3\cdot 1\cdot c_{1} x'+\underbrace{ \sum _{n=2}^{\infty } n( n-1) c_{n}}_{k=n} x^{n}_{k=n} +2\underbrace{ \sum _{n=4}^{\infty } n( n-1) c_{n} x^{n-2}}_{k=n-2} +3\underbrace{ \sum _{n=2}^{\infty } nc_{n} x^{n}}_{k=n}\\ &= 4c_{2} +( 12c_{3} +( 12c_{3} +3c_{1}) x+\sum _{n=2}^{\infty } k( k-1) c_{k} x^{k} +2\sum _{n=2}^{\infty }( k+2)( k+1) x_{k+2} x^{k} +3\sum _{n=2}^{\infty } kc_{k} x^{k}\\ &= 4c_{2} +( 3c_{1} +12c_{3}) x+\sum _{n=2}^{\infty }([ k( k-1) +3k] c_{k} +2( k+2)( k+1) x_{k+2}) x^{k}\\ &= 4c_{2} +( 3c_{1} +12c_{3}) x+\sum _{n=2}^{\infty }[ k( k+2) c_{k} +2( k+1)( k+2) c_{k+2}] x^{k} \end{aligned}

    b. 3x2∑n=−2∞n(n−1)xn−2+x∑n=1∞nxn3x^{2}\sum _{n=-2}^{\infty } n( n-1) x^{n-2} +x\sum _{n=1}^{\infty } nx^{n}

    Solution
    3x2∑n=−2∞n(n−1)xn−2+x∑n=1∞nxn=3∑n=−2∞n(n−1)xn+∑n=1∞nxn+1=3∑k=−2∞k(k−1)xk+∑k=2∞(k−1)xk(1)3x^{2}\sum _{n=-2}^{\infty } n( n-1) x^{n-2} +x\sum _{n=1}^{\infty } nx^{n} =3\sum _{n=-2}^{\infty } n( n-1) x^{n} +\sum _{n=1}^{\infty } nx^{n+1} =3\sum _{k=-2}^{\infty } k( k-1) x^{k} +\sum _{k=2}^{\infty }( k-1) x^{k} \tag{1}

    Solving first part of Eq(1),

    ∑k=−2∞k(k−1)xk=−2(−2−1)x−2+(−1)(−1−1)x−1+0+1(1−1)x1+∑k=2∞k(k−1)xk=6x2+2x+∑k=2∞k(k−1)xk\begin{aligned} \sum _{k=-2}^{\infty } k( k-1) x^{k} & =-2( -2-1) x^{-2} +( -1)( -1-1) x^{-1} +0+1( 1-1) x^{1} +\sum _{k=2}^{\infty } k( k-1) x^{k}\\ & =\frac{6}{x^{2}} +\frac{2}{x} +\sum _{k=2}^{\infty } k( k-1) x^{k} \end{aligned}

    Substitute back to Eq (1),

    ∴3[6x2+2x+∑k=2∞k(k−1)xk]+∑k=2∞(k−1)xk=18x2+6x+∑k=2∞[3k(k−1)+(k−1)]xk=18x2+6x+∑k=2∞[(k−1)(3k+1)]xk\begin{aligned} \therefore 3\left[\frac{6}{x^{2}} +\frac{2}{x} +\sum _{k=2}^{\infty } k( k-1) x^{k}\right] +\sum _{k=2}^{\infty }( k-1) x^{k} & =\frac{18}{x^{2}} +\frac{6}{x} +\sum _{k=2}^{\infty }[ 3k( k-1) +( k-1)] x^{k}\\ & =\frac{18}{x^{2}} +\frac{6}{x} +\sum _{k=2}^{\infty }[( k-1)( 3k+1)] x^{k} \end{aligned}

    c. ∑n=1∞3n(2n)!xn−1+3x3∑n=−1∞(−1)n(2n+1)!x2n+1\sum _{n=1}^{\infty }\frac{3^{n}}{( 2n) !} x^{n-1} +3x^{3}\sum _{n=-1}^{\infty }\frac{( -1)^{n}}{( 2n+1) !} x^{2n+1}

    Solution
    ∑n=1∞3n(2n)!xn−1+2x3∑n=−1∞(−1)n(2n+1)!x2n+1=∑n=1∞3n(2n)!xn−1+2∑n=−1∞(−1)n(2n+1)!x2n+4=∑k=0∞3k+1(2(k+1))!xk+2∑k=2∞(−1)k−42(2(k−42)+1)!xk=∑k=0∞3k+1(2k+2)!xk+2∑k=2∞(−1)k−42(k−3)!xk(2)\begin{aligned} \sum _{n=1}^{\infty }\frac{3^{n}}{( 2n) !} x^{n-1} +2x^{3}\sum _{n=-1}^{\infty }\frac{( -1)^{n}}{( 2n+1) !} x^{2n+1} & =\sum _{n=1}^{\infty }\frac{3^{n}}{( 2n) !} x^{n-1} +2\sum _{n=-1}^{\infty }\frac{( -1)^{n}}{( 2n+1) !} x^{2n+4}\\ & =\sum _{k=0}^{\infty }\frac{3^{k+1}}{( 2( k+1)) !} x^{k} +2\sum _{k=2}^{\infty }\frac{( -1)^{\frac{k-4}{2}}}{\left( 2\left(\frac{k-4}{2}\right) +1\right) !} x^{k}\\ & =\sum _{k=0}^{\infty }\frac{3^{k+1}}{( 2k+2) !} x^{k} +2\sum _{k=2}^{\infty }\frac{( -1)^{\frac{k-4}{2}}}{( k-3) !} x^{k} \end{aligned} \tag{2}

    Solving first part of Eq (2),

    ∑k=0∞3k+1(2k+2)!xk=314!x0+324!x1+∑k=2∞3k+1(2k+2)!=32+38x+∑k=2∞3k+1(2k+2)!xk\begin{aligned} \sum _{k=0}^{\infty }\frac{3^{k+1}}{( 2k+2) !} x^{k} & =\frac{3^{1}}{4!} x^{0} +\frac{3^{2}}{4!} x^{1} +\sum _{k=2}^{\infty }\frac{3^{k+1}}{( 2k+2) !}\\ & =\frac{3}{2} +\frac{3}{8} x+\sum _{k=2}^{\infty }\frac{3^{k+1}}{( 2k+2) !} x^{k} \end{aligned}

    Subtitute back to Eq (2),

    ∴32+38x+∑k=2∞3k+1(2k+2)!xk+2∑k=2∞(−1)k−42(k−3)!xk=32+38x+∑k=2∞[3k+2(2k+2)!+2(−1)k−42(k−3)!]xk=32+38x+∑k=2∞[3k+2(2k+2)!+2(−1)k−4(k−3)!]xk\begin{aligned} \therefore \frac{3}{2} +\frac{3}{8} x+\sum _{k=2}^{\infty }\frac{3^{k+1}}{( 2k+2) !} x^{k} +2\sum _{k=2}^{\infty }\frac{( -1)^{\frac{k-4}{2}}}{( k-3) !} x^{k} & =\frac{3}{2} +\frac{3}{8} x+\sum _{k=2}^{\infty }\left[\frac{3^{k+2}}{( 2k+2) !} +\frac{2( -1)^{\frac{k-4}{2}}}{( k-3) !}\right] x^{k}\\ & =\frac{3}{2} +\frac{3}{8} x+\sum _{k=2}^{\infty }\left[\frac{3^{k+2}}{( 2k+2) !} +\frac{2\sqrt{( -1)^{k-4}}}{( k-3) !}\right] x^{k} \end{aligned}

  4. Find two power series solutions of given differential equation about the ordinary point x=0x=0.

    a. y′′+xy′+y=0y''+xy'+y=0

    Solution

    Let y(x)=∑n=0∞cnxn⟹ y′(x)=∑n=1∞ncnxn−1y( x) =\sum _{n=0}^{\infty } c_{n} x^{n} \Longrightarrow \ y'( x) =\sum _{n=1}^{\infty } nc_{n} x^{n-1} and y′′(x)=∑n=2∞(n+2)(n+1)cn+2xny''( x) =\sum _{n=2}^{\infty }( n+2)( n+1) c_{n+2} x^{n}.

    The differential equation becomes

    ∑n=0∞(n+2)(n+1)cn+2xn+x∑n=1∞ncnxn−1+∑n=0∞cnxn=0∑n=0∞[(n+2)(n+1)cn+2+ncn+cn]xn=0[△Since∑n=1∞ncnxn=∑n=0∞ncnxn]\begin{aligned} \sum _{n=0}^{\infty }( n+2)( n+1) c_{n+2} x^{n} +x\sum _{n=1}^{\infty } nc_{n} x^{n-1} +\sum _{n=0}^{\infty } c_{n} x^{n} & =0\\ \sum _{n=0}^{\infty }[( n+2)( n+1) c_{n+2} +nc_{n} +c_{n}] x^{n} & =0\\ \color{red} {\scriptsize { \text{[} \triangle \text{Since} \sum _{n=1}^{\infty }nc_{n}x^{n} = \sum _{n=0}^{\infty } nc_{n}x^{n} \text{]}}} & \end{aligned}

    Equating coefficients gives,

    (n+2)(n+1)cn+2+(n+1)cn=0( n+2)( n+1) c_{n+2} +( n+1) c_{n} =0

    Thus, the recursion relation is,

    cn+2=−(n+1)cn(n+2)(n+1)=−cnn+2,n=0,1,2,…c_{n+2} =\frac{-( n+1) c_{n}}{( n+2)( n+1)} =-\frac{c_{n}}{n+2} ,\quad n=0,1,2,\ldots

    Then, the even coefficients are given by c2=−c02,c4=−c24=c02⋅4,c6=−c46=−c02⋅4⋅6c_{2} =-\frac{c_{0}}{2} ,\quad c_{4} =-\frac{c_{2}}{4} =\frac{c_{0}}{2\cdot 4} ,\quad c_{6} =-\frac{c_{4}}{6} =-\frac{c_{0}}{2\cdot 4\cdot 6}. Hence,

    c2n=(−1)nc02⋅4⋅⋅⋅2n=(−1)nc02nn!c_{2n} =( -1)^{n}\frac{c_{0}}{2\cdot 4\cdot \cdot \cdot 2n} =\frac{( -1)^{n} c_{0}}{2^{n} n!}

    The odd coefficients are c3=−c13,c5=−c35=c13⋅5,c7=−c57=−c13⋅5⋅7c_{3} =-\frac{c_{1}}{3} ,\quad c_{5} =-\frac{c_{3}}{5} =\frac{c_{1}}{3\cdot 5} ,\quad c_{7} =-\frac{c_{5}}{7} =-\frac{c_{1}}{3\cdot 5\cdot 7}. Hence,

    c2n+1=(−1)nc13⋅5⋅7⋅⋅⋅(2n+1)=(−2)nn!c1(2n+1)!c_{2n+1} =( -1)^{n}\frac{c_{1}}{3\cdot 5\cdot 7\cdot \cdot \cdot ( 2n+1)} =\frac{( -2)^{n} n!c_{1}}{( 2n+1) !}

    The solution is,

    y(x)=c0∑n=0∞(−1)n2nn!x2n+c1∑n=0∞(−2)nn!(2n+1)!x2n+1\mathbf{y( x) =c_{0}\sum _{n=0}^{\infty }\frac{( -1)^{n}}{2^{n} n!} x^{2n} +c_{1}\sum _{n=0}^{\infty }\frac{( -2)^{n} n!}{( 2n+1) !} x^{2n+1}}

    b. (x−1)y′′+y′=0( x-1) y''+y'=0

    Solution

    Subtituting y=∑n=0∞cnxny=\sum _{n=0}^{\infty } c_{n} x^{n} into the differential equation,

    (x−1)y′′+y′=∑n=2∞n(n−1)cnxn−1⏟k=n−1−∑n=2∞n(n−1)cnxn−2⏟k=n−2+∑n=1∞ncnxn−1⏟k=n−1=∑k=1∞(k+1)kck+1xk−∑k=0∞(k+2)(k+1)ck+2xk+∑k=0∞(k+1)ck+1xk=−2c2+c1+∑k=1∞[(k+1)kck+1−(k+2)(k+1)ck+2+(k+1)ck+1]xk=0\begin{aligned} ( x-1) y''+y' & =\underbrace{ \sum _{n=2}^{\infty } n( n-1) c_{n} x^{n-1}}_{k=n-1} -\underbrace{ \sum _{n=2}^{\infty } n( n-1) c_{n} x^{n-2}}_{k=n-2} +\underbrace{ \sum _{n=1}^{\infty } nc_{n} x^{n-1}}_{k=n-1}\\ & ={ \sum _{k=1}^{\infty }( k+1) kc_{k+1} x^{k} -\sum _{k=0}^{\infty }( k+2)( k+1) c_{k+2} x^{k} +\sum _{k=0}^{\infty }( k+1) c_{k+1} x^{k}}\\ & =-2c_{2} +c_{1} +{ \sum _{k=1}^{\infty }[( k+1) kc_{k+1} -( k+2)( k+1) c_{k+2} +( k+1) c_{k+1}] x^{k} =0} \end{aligned}

    Thus,

    −2c2+c1=0(k+1)2ck+1−(k+2)(k+1)ck+2=0\begin{aligned} -2c_{2} +c_{1} & =0\\ ( k+1)^{2} c_{k+1} -( k+2)( k+1) c_{k+2} & =0 \end{aligned}

    and

    c2=12c1ck+2=k+1k+2ck+1,k=1,2,3,…\begin{aligned} c_{2} & =\frac{1}{2} c_{1}\\ c_{k+2} & =\frac{k+1}{k+2} c_{k+1} ,\quad k=1,2,3,\ldots \end{aligned}

    Choosing c0=1c_{0} =1 and c1=0c_{1} =0, we find c2=c3=c4=…=0c_{2} =c_{3} =c_{4} =\ldots=0. For c0=0c_{0} =0 and c1=1c_{1} =1, we obtain

    c2=12,c3=13,c4=14c_{2} =\frac{1}{2} ,\quad c_{3} =\frac{1}{3} ,\quad c_{4} =\frac{1}{4}

    Thus, the two solutions are

    y1=1y2=x+12x2+13x3+14x4+…\mathbf{y_{1} =1\quad \quad y_{2} =x+\frac{1}{2} x^{2} +\frac{1}{3} x^{3} +\frac{1}{4} x^{4} +\ldots}

    c. y′′+exy′−y=0y''+e^{x} y'-y=0

    Solution

    Substituting y=∑n=0∞cnxny=\sum _{n=0}^{\infty } c_{n} x^{n} into the differential equation we have,

    y′′−(x+1)y′−y=∑n=2∞n(n−1)cnxn−2⏟k=n−2−∑n=1∞ncnxn⏟k=n−∑n=1∞ncnxn−1⏟k=n−1−∑n=0∞cnxn⏟k=n=∑k=0∞(k+2)(k+1)ck+2xk−∑k=1∞kckxk−∑k=0∞(k+1)ck+1xk−∑k=0∞ckxk=2c2−c1−c0+∑k=1∞[(k+2)(k+1)ck+2−(k+1)ck+1−(k+1)ck]xk=0\begin{aligned} y''-( x+1) y'-y & =\underbrace{ \sum _{n=2}^{\infty } n( n-1) c_{n} x^{n-2}}_{k=n-2} -\underbrace{ \sum _{n=1}^{\infty } nc_{n} x^{n}}_{k=n} -\underbrace{ \sum _{n=1}^{\infty } nc_{n} x^{n-1}}_{k=n-1} -\underbrace{ \sum _{n=0}^{\infty } c_{n} x^{n}}_{k=n}\\ & =\sum _{k=0}^{\infty }( k+2)( k+1) c_{k+2} x^{k} -\sum _{k=1}^{\infty } kc_{k} x^{k} -\sum _{k=0}^{\infty }( k+1) c_{k+1} x^{k} -\sum _{k=0}^{\infty } c_{k} x^{k}\\ & =2c_{2} -c_{1} -c_{0} +\sum _{k=1}^{\infty }[( k+2)( k+1) c_{k+2} -( k+1) c_{k+1} -( k+1) c_{k}] x^{k} =0 \end{aligned}

    Thus,

    2c2−c1−c0=0(k+2)(k+1)ck+2−(k−1)(ck+1+ck)=0\begin{aligned} 2c_{2} -c_{1} -c_{0} & =0\\ ( k+2)( k+1) c_{k+2} -( k-1)( c_{k+1} +c_{k}) & =0 \end{aligned}

    and,

    c2=c1+c02ck+2=ck+1+ckk+2ck,k=2,3,4,…\begin{aligned} c_{2} & =\frac{c_{1} +c_{0}}{2}\\ c_{k+2} & =\frac{c_{k+1} +c_{k}}{k+2} c_{k} ,\quad k=2,3,4, \ldots \end{aligned}

    Choosing c0=1c_{0} =1 and c1=0c_{1} =0, we find,

    c2=12,c3=16,c4=16c_{2} =\frac{1}{2} ,\quad \quad c_{3} =\frac{1}{6} ,\quad \quad c_{4} =\frac{1}{6}

    and so on. For c0=0c_{0} =0 and c1=1c_{1} =1, we obtain,

    c2=12,c3=12,c4=14c_{2} =\frac{1}{2} ,\quad \quad c_{3} =\frac{1}{2} ,\quad \quad c_{4} =\frac{1}{4}

    and so on. Thus, the two solutions are,

    y1=1+12x2+16x3+16x4+…andy2=x+12x2+12x3+14x4+…y_{1} =1+\frac{1}{2} x^{2} +\frac{1}{6} x^{3} +\frac{1}{6} x^{4} +\ldots\quad \quad \text{and} \quad \quad y_{2} =x+\frac{1}{2} x^{2} +\frac{1}{2} x^{3} +\frac{1}{4} x^{4} +\ldots

    d. (x2+1)y′′+xy′−y=0\left( x^{2} +1\right) y''+xy'-y=0

    Solution

    Let y(x)=∑n=0∞cnxn⟹ y′′(x)=∑n=0∞n(n−1)cnxn−2y( x) =\sum _{n=0}^{\infty } c_{n} x^{n} \Longrightarrow \ y''( x) =\sum _{n=0}^{\infty } n( n-1) c_{n} x^{n-2} and xy′′=∑n=0∞ncnxnxy''=\sum _{n=0}^{\infty } nc_{n} x^{n}. Thus,

    (x2+1)y′′=∑n=0∞n(n−1cnxn+∑n=0∞(n+2)(n+1)cn+2xn\left( x^{2} +1\right) y''=\sum _{n=0}^{\infty } n( n-1c_{n} x^{n} +\sum _{n=0}^{\infty }( n+2)( n+1) c_{n+2} x^{n}

    The differential equation becomes,

    ∑n=0∞[(n+2)(n+1)cn+2+[n(n−1)+n−1]cn]xn=0\sum _{n=0}^{\infty }[( n+2)( n+1) c_{n+2} +[ n( n-1) +n-1] c_{n}] x^{n} =0

    The recursion relation is

    cn+2=−(n−1)cnn+2,n=0,1,2,…c_{n+2} =-\frac{( n-1) c_{n}}{n+2} ,\quad n=0,1,2,\ldots

    Given c0c_{0} and c1c_{1}, c2=c02c_{2} =\frac{c_{0}}{2}, c4=−c022⋅2!c_{4} =-\frac{c_{0}}{2^{2} \cdot 2!}, c6=−3c46=(−1)23c023⋅3!,…c_{6} =-\frac{3c_{4}}{6} =( -1)^{2}\frac{3c_{0}}{2^{3} \cdot 3!},\ldots

    c2n=(−1)n−11⋅3⋅⋅⋅(2n−3)c02nn!=(−1)n−1(2n−3)!c02n2n−2n!(n−2)!=(−1)n−1(2n−3)!c022n−2n!(n−2)! for n=2,3,…c_{2n} =( -1)^{n-1}\frac{1\cdot 3\cdot \cdot \cdot ( 2n-3) c_{0}}{2^{n} n!} =( -1)^{n-1}\frac{( 2n-3) !c_{0}}{2^{n} 2^{n-2} n!( n-2) !} =( -1)^{n-1}\frac{( 2n-3) !c_{0}}{2^{2n-2} n!( n-2) !} \ \text{for} \ n=2,3,\ldotsc3=0⋅c13=0⟹c2n+1=0 for n=1,2,…c_{3} =\frac{0\cdot c_{1}}{3} =0\Longrightarrow c_{2n+1} =0\ \text{for} \ n=1,2,\ldots

    Thus the solution is,

    y(x)=c0+c1x+c0x22+c0∑n=2∞(−1)n−1(2n−3)!22n−2n!(n−2)!x2n\mathbf{y( x) =c_{0} +c_{1} x+c_{0}\frac{x^{2}}{2} +c_{0}\sum _{n=2}^{\infty }\frac{( -1)^{n-1}( 2n-3) !}{2^{2n-2} n!( n-2) !} x^{2n}}

  5. Use the power series method to solve the given initial-value problem.

    a. y′′−xy′−y=0, y(0)=1, y′(0)=0y''-xy'-y=0, y(0)=1, y'(0)=0

    Solution

    Let y(x)=∑n=0∞cnxny( x) =\sum _{n=0}^{\infty } c_{n} x^{n}. Then −xy′(x)=−x∑n=1∞ncnxn−1=−∑n=1∞ncnxn=−∑n=0∞ncnxn-xy'( x) =-x\sum _{n=1}^{\infty } nc_{n} x^{n-1} =-\sum _{n=1}^{\infty } nc_{n} x^{n} =-\sum _{n=0}^{\infty } nc_{n} x^{n}.

    y′′(x)=∑n=0∞(n+2)(n+1)cn+2xny''( x) =\sum _{n=0}^{\infty }( n+2)( n+1) c_{n+2} x^{n}

    The equation y′′−xy′−y=0y''-xy'-y=0 becomes

    ∑n=0∞[(n+2)(n+1)cn+2−ncn−cn]xn=0\sum _{n=0}^{\infty }[( n+2)( n+1) c_{n+2} -nc_{n} -c_{n}] x^{n} =0

    Thus, the recursion relation is

    cn+2=ncn+cn(n+2)(n+1)=cn(n+1)(n+2)(n+1)=cnn+2 for n=0,1,2,…c_{n+2} =\frac{nc_{n} +c_{n}}{( n+2)( n+1)} =\frac{c_{n}( n+1)}{( n+2)( n+1)} =\frac{c_{n}}{n+2} \ \text{for} \ n=0,1,2,\ldots

    One of the condition is, y(0)=1y( 0) =1. But y(0)=∑n=0∞cn(0)n=c0+0+0+…=c0y( 0) =\sum _{n=0}^{\infty } c_{n}( 0)^{n} =c_{0} +0+0+\ldots=c_{0}, so c0=1c_{0} =1.

    Hence, c2=c02=12c_{2} =\frac{c_{0}}{2} =\frac{1}{2}, c4=c24=12⋅4c_{4} =\frac{c_{2}}{4} =\frac{1}{2\cdot 4}, c6=c46=12⋅4⋅6c_{6} =\frac{c_{4}}{6} =\frac{1}{2\cdot 4\cdot 6},\ldots,c2n=12nn!c_{2n} =\frac{1}{2^{n} n!}. The other given condition is y′(0)=0y'( 0) =0.

    But y′(0)=∑n=1∞ncn(0)n−1=c1+0+0+…=c1y'( 0) =\sum _{n=1}^{\infty } nc_{n}( 0)^{n-1} =c_{1} +0+0+\ldots=c_{1}. So, c1=0c_{1} =0.

    By the recursion relation, c3=c13=0c_{3} =\frac{c_{1}}{3} =0, c5=0c_{5} =0,\ldots,c2n+1=0c_{2n+1} =0 for n=0,1,2,…n=0,1,2,\ldots.

    Thus, the solution to the initial-value problem is,

    y(x)=∑n=0∞cnxn=∑n=0∞c2nx2n=∑n=0∞x2n2nn!=∑n=0∞(x2/2)nn!=cx2/2y( x) =\sum _{n=0}^{\infty } c_{n} x^{n} =\sum _{n=0}^{\infty } c_{2n} x^{2n} =\sum _{n=0}^{\infty }\frac{x^{2n}}{2^{n} n!} =\sum _{n=0}^{\infty }\frac{\left( x^{2} /2\right)^{n}}{n!} =c^{x^{2} /2}

    b. y′′+x2y′+xy=0, y(0)=0, y′(0)=1y''+x^2y'+xy=0, y(0)=0, y'(0)=1

    Solution

    Assuming that y(x)=∑n=0∞cnxny( x) =\sum _{n=0}^{\infty } c_{n} x^{n}, we have xy=x∑n=0∞cnxn=∑n=0∞cnxn+1xy=x\sum _{n=0}^{\infty } c_{n} x^{n} =\sum _{n=0}^{\infty } c_{n} x^{n+1},

    x2y′=x2∑n=1∞ncnxn−1=∑n=0∞ncnxn+1x^{2} y'=x^{2}\sum _{n=1}^{\infty } nc_{n} x^{n-1} =\sum _{n=0}^{\infty } nc_{n} x^{n+1}y′′(x)=∑n=2∞n(n−1)cnxn−2=∑n=−1∞(n+3)(n+2)cn+3xn+1=2c2+∑n=0∞(n+3)(n+2)cn+3xn+1\begin{aligned} y''( x) & =\sum _{n=2}^{\infty } n( n-1) c_{n} x^{n-2} =\sum _{n=-1}^{\infty }( n+3)( n+2) c_{n+3} x^{n+1}\\ & =2c_{2} +\sum _{n=0}^{\infty }( n+3)( n+2) c_{n+3} x^{n+1} \end{aligned}

    Thus, the equation y′′+x2y′+xy=0y''+x^{2} y'+xy=0 becomes

    2c2+∑n=0∞[(n+3)(n+2)cn+3+ncn+cn]xn+1=02c_{2} +\sum _{n=0}^{\infty }[( n+3)( n+2) c_{n+3} +nc_{n} +c_{n}] x^{n+1} =0

    So, c2=0c_{2} =0 and the recursion relation is,

    cn+3=−ncn−cn(n+3)(n+2)=−(n+1)cn(n+3)(n+2), n=0,1,2,…c_{n+3} =\frac{-nc_{n} -c_{n}}{( n+3)( n+2)} =-\frac{( n+1) c_{n}}{( n+3)( n+2)} ,\ n=0,1,2,\ldots

    Also, c1=y′(0)=1c_{1} =y'( 0) =1, so,

    c4=−2c14⋅3=−24⋅3,c7=−5c47⋅6=(−1)22⋅57⋅6⋅4⋅3=(−1)222527!,…c3n+1=(−1)n2252⋅⋅⋅⋅(3n−1)2(3n+1)!\begin{aligned} c_{4} &=-\frac{2c_{1}}{4\cdot 3} =-\frac{2}{4\cdot 3} ,\quad c_{7} =-\frac{5c_{4}}{7\cdot 6} =( -1)^{2}\frac{2\cdot 5}{7\cdot 6\cdot 4\cdot 3} =( -1)^{2}\frac{2^{2} 5^{2}}{7!} ,\ldots\\ \\ c_{3n+1} &=( -1)^{n}\frac{2^{2} 5^{2} \cdot \cdot \cdot \cdot ( 3n-1)^{2}}{( 3n+1) !} \end{aligned}

    Thus, the solution is,

    y(x)=∑n=0∞cmxn=x+∑n=1∞[(−1)n2252⋅⋅⋅⋅(3n−1)2x3n+1(3n+1)!]\mathbf{y( x) =\sum _{n=0}^{\infty } c_{m} x^{n} =x+\sum _{n=1}^{\infty }\left[( -1)^{n}\frac{2^{2} 5^{2} \cdot \cdot \cdot \cdot ( 3n-1)^{2} x^{3n+1}}{( 3n+1) !}\right]}

    c. (x+1)y′′−(2−x)y′+y=0, y(0)=2, y′(0)=−1(x+1)y''-(2-x)y'+y=0, y(0)=2, y'(0)=-1

    Solution

    Subtituting y=∑n=0∞cnxny=\sum _{n=0}^{\infty } c_{n} x^{n} into the differential equation, we have,

    (x+1)y′′−(2−x)y′+y=∑n=2∞n(n−1)cnxn−1⏟k=n−1+∑n=2∞n(n−1)cnxn−2⏟k=n−2−2∑n=2∞ncnxn−1⏟k=n−1+∑n=1∞ncnxn⏟k=n+∑n=0∞cnxn⏟k=n=∑k=1∞(k+1)kck+1xk+∑k=0∞(k+2)(k+1)ck+2xk−2∑k=0∞(k+1)ck+1xk+∑k=1∞kckxk+∑k=0∞ckxk=2c2−2c1+c0+∑k=1∞[(k+2)(k+1)ck+2−(k+1)ck+1+(k+1)ck]xk=0\begin{aligned} & ( x+1) y''-( 2-x) y'+y\\ = & \underbrace{ \sum _{n=2}^{\infty } n( n-1) c_{n} x^{n-1}}_{k=n-1} +\underbrace{ \sum _{n=2}^{\infty } n( n-1) c_{n} x^{n-2}}_{k=n-2} -2\underbrace{ \sum _{n=2}^{\infty } nc_{n} x^{n-1}}_{k=n-1} +\underbrace{ \sum _{n=1}^{\infty } nc_{n} x^{n}}_{k=n} +\underbrace{ \sum _{n=0}^{\infty } c_{n} x^{n}}_{k=n}\\ & = \sum _{k=1}^{\infty }( k+1) kc_{k+1} x^{k} +\sum _{k=0}^{\infty }( k+2)( k+1) c_{k+2} x^{k} -2\sum _{k=0}^{\infty }( k+1) c_{k+1} x^{k} +\sum _{k=1}^{\infty } kc_{k} x^{k} +\sum _{k=0}^{\infty } c_{k} x^{k}\\ & =2c_{2} -2c_{1} +c_{0} + \sum _{k=1}^{\infty }[( k+2)( k+1) c_{k+2} -( k+1) c_{k+1} +( k+1) c_{k}] x^{k} =0 \end{aligned}

    Thus,

    2c2−2c1+c0=0(k+2)(k+1)ck+2−(k+1)ck+1+(k+1)ck=0\begin{aligned} 2c_{2} -2c_{1} +c_{0} & =0\\ ( k+2)( k+1) c_{k+2} -( k+1) c_{k+1} +( k+1) c_{k} & =0 \end{aligned}

    and,

    c2=c1−12c0ck+2=1k+2ck+1−1k+2ck,k=1,2,3,…\begin{aligned} c_{2} & =c_{1} -\frac{1}{2} c_{0}\\ c_{k+2} & =\frac{1}{k+2} c_{k+1} -\frac{1}{k+2} c_{k} ,\quad k=1,2,3,\ldots \end{aligned}

    Choosing c0=1c_{0} =1 and c1=0c_{1} =0, we find

    c2=−12,c3=−16,c4=112c_{2} =-\frac{1}{2} ,\quad \quad c_{3} =-\frac{1}{6} ,\quad \quad c_{4} =\frac{1}{12}

    and so on. For c0=0c_{0} =0 and c1=1c_{1} =1, we obtain,

    c2=1,c3=0,c4=−14c_{2} =1 ,\quad \quad c_{3} =0,\quad \quad c_{4} =-\frac{1}{4}

    and do on. Thus,

    y=C1(1−12x2−16x3+112x4+…)+C2(x+x2−14x4+…)y=C_{1}\left( 1-\frac{1}{2} x^{2} -\frac{1}{6} x^{3} +\frac{1}{12} x^{4} +\ldots\right) +C_{2}\left( x+x^{2} -\frac{1}{4} x^{4} +\ldots\right)

    and

    y′=C1(−x−12x2+13x3+…)+C2(1+2x−x3+…)y'=C_{1}\left( -x-\frac{1}{2} x^{2} +\frac{1}{3} x^{3} +\ldots\right) +C_{2}\left( 1+2x-x^{3} +\ldots\right)

    The initital condition imply C1=2C_{1} =2 and C2=−1C_{2} =-1, so

    y=2(1−12x2−16x3+112x4+…)−(x+x2−14x4+…)=2−x−2x2−13x3+512x4+…\begin{aligned} y & =2\left( 1-\frac{1}{2} x^{2} -\frac{1}{6} x^{3} +\frac{1}{12} x^{4} +\ldots\right) -\left( x+x^{2} -\frac{1}{4} x^{4} +\ldots\right)\\ & =2-x-2x^{2} -\frac{1}{3} x^{3} +\frac{5}{12} x^{4} +\ldots \end{aligned}

  6. Determine the singular points of the given differential equation. Classify each singular point as regular or irregular.

    a. x3y′′−4x2y′+3y=0x^{3} y''-4x^{2} y'+3y=0

    Solution

    Dividing the equation with x3x^{3}

    y′′+4x2x3y′+3x3y=0y''+\frac{4x^{2}}{x^{3}} y'+\frac{3}{x^{3}} y=0P(x)=4xQ(x)=3x3P( x) =\frac{4}{x} \quad \quad Q( x) =\frac{3}{x^{3}}
    • The factor xx appears at most to the first power in the denominator of P(x)P( x), but more that the second power in the denominator of Q(x)Q( x)
    • x=0x=0 is irregular singular point

    b. (x2−9)2y′′+(x+3)y′+2y=0\left( x^{2} -9\right)^{2} y''+( x+3) y'+2y=0

    Solution

    Dividing the eqution with (x2−9)2=((x−3)(x+3))2\left( x^{2} -9\right)^{2} =(( x-3)( x+3))^{2},

    y′′+(x+3)((x−3)(x+3))2y′+2((x−3)(x+3))2y=0P(x)=1(x−3)2(x+3)Q(x)=2(x−3)2(x+3)2\begin{gather*} y''+\frac{( x+3)}{(( x-3)( x+3))^{2}} y'+\frac{2}{(( x-3)( x+3))^{2}} y=0\\ \\ P( x) =\frac{1}{( x-3)^{2}( x+3)} \quad \quad Q( x) =\frac{2}{( x-3)^{2}( x+3)^{2}} \end{gather*}
    • The factor x + 3x\ +\ 3 appears at most to the first power in the denominator of P(x)P( x) and at most to the second power in the denominator of Q(x)Q( x)
    • x = −3x\ =\ -3 is regular singular point
    • The factor x − 3x\ -\ 3 appears more than the first power in the denominator of P(x)P( x); but at most to the second power in the denominator of Q(x)Q( x)
    • x =3x\ =3 is irregular singular point

    c. (2x2−5x−3)y′′+(2x+1)y′+6(x−3)y=0\left( 2x^{2} -5x-3\right) y''+( 2x+1) y'+\frac{6}{( x-3)} y=0

    Solution

    Factor the equation (2x2−5x−3)=(x−3)(2x+1)\left( 2x^{2} -5x-3\right) =( x-3)( 2x+1),

    y′′+(2x+1)(x−3)(2x+1)y′+6(x−3)2(2x+1)y=0P(x)=1(x−3)Q(x)=6(x−3)2(2x+1)\begin{gather*} y''+\frac{( 2x+1)}{( x-3)( 2x+1)} y'+\frac{6}{( x-3)^{2}( 2x+1)} y=0\\ \\ P( x) =\frac{1}{( x-3)} \quad \quad Q( x) =\frac{6}{( x-3)^{2}( 2x+1)} \end{gather*}
    • The factor (x−3)( x-3) appears at most to the first power in P(x)P( x) and second power in the denominator of Q(x)Q( x)
    • x = 3x\ =\ 3 is regular singular point -The factor (2x+1)( 2x+1) appears at to the first power in the denominator of Q(x)Q( x)
    • x = −12x\ =\ -\frac{1}{2} is regular singular point

    d. (x3−2x2−3x)2y′′+x(x−3)2y′−(x+1)y=0\left( x^{3} -2x^{2} -3x\right)^{2} y''+x( x-3)^{2} y'-( x+1) y=0

    Solution

    Divide the equation with (x3−2x2−3x)2=(x(x2−2x−3))2=x2(x−3)2(x+1)2\left( x^{3} -2x^{2} -3x\right)^{2} =\left( x\left( x^{2} -2x-3\right)\right)^{2} =x^{2}( x-3)^{2}( x+1)^{2},

    y′′+x(x−3)2x2(x−3)2(x+1)2y′−(x+1)x2(x−3)2(x+1)2y=0P(x)=1x(x+1)2Q(x)=1x2(x−3)2(x+1)\begin{gather*} y''+\frac{x( x-3)^{2}}{x^{2}( x-3)^{2}( x+1)^{2}} y'-\frac{( x+1)}{x^{2}( x-3)^{2}( x+1)^{2}} y=0\\ \\ P( x) =\frac{1}{x( x+1)^{2}} \quad \quad Q( x) =\frac{1}{x^{2}( x-3)^{2}( x+1)} \end{gather*}
    • The factor xx appears at most to the first power in the denominator of P(x)P( x) and at most to the second power in the denominator of Q(x)Q( x)
    • x = 0x\ =\ 0 is regular singular point
    • The factor x − 3x\ -\ 3 appears to the second power in the denominator of Q(x)Q( x)
    • x = 3x\ =\ 3 is regular singular point
    • The factor x + 1x\ +\ 1 appears to the second power in the denominator of P(x)P(x) and to the first power in the denominator of Q(x)Q( x)
    • x = −1x\ =\ -1 is irregular singular point

  7. Find the indical roots for the given differential equations where x=0x=0 is a regular singular point.

    a. 2xy′′−y′+2y=02xy''-y'+2y=0

    Solution

    Divide the equation with 2x2x,

    y′′−12xy′+22xy=y′′−12xy′+1xy=y′′−12xy′+xx2y=0y''-\frac{1}{2x} y'+\frac{2}{2x} y=y''-\frac{\frac{1}{2}}{x} y'+\frac{1}{x} y=y''-\frac{\frac{1}{2}}{x} y'+\frac{x}{x^{2}} y=0

    Hence, b0=−12b_{0} =-\frac{1}{2} and r2=0r_{2} =0,

    7r(r−1)+b0r+c0=r2−r−12r+0=r2−32r=r(r−32)=07r( r-1) +b_{0} r+c_{0} =r^{2} -r-\frac{1}{2} r+0=r^{2} -\frac{3}{2} r=r\left( r-\frac{3}{2}\right) =0

    The indical roots are r1=32r_{1} =\frac{3}{2} and r2=0r_{2} =0.

    b. 3xy′′+(2−x)y′−y=03xy''+( 2-x) y'-y=0

    Solution

    Divide the equation with 3x3x,

    y′′+(2−x)3xy′−13xy=y′′+13(2−x)xy′−13xy=y′′+23−13xxy′−13xx2y=0y''+\frac{( 2-x)}{3x} y'-\frac{1}{3x} y=y''+\frac{\frac{1}{3}( 2-x)}{x} y'-\frac{\frac{1}{3}}{x} y=y''+\frac{\frac{2}{3} -\frac{1}{3} x}{x} y'-\frac{\frac{1}{3} x}{x^{2}} y=0

    Hence, b0=23b_{0} =\frac{2}{3} and c0=0c_{0} =0,

    r(r−1)+b0r+c0=r2−r+23r+0=r2−13r=r(r−13)=0r( r-1) +b_{0} r+c_{0} =r^{2} -r+\frac{2}{3} r+0=r^{2} -\frac{1}{3} r=r\left( r-\frac{1}{3}\right) =0

    The indical roots are r1=13r_{1} =\frac{1}{3} and r2=0r_{2} =0.

    c. 9x2y′′+9x2y′+2y=09x^{2} y''+9x^{2} y'+2y=0

    Solution

    Divide the equation with 9x29x^{2},

    y′′+9x29x2y′+29x2y=y′′+y′+29x2y=y′′+xxy′+29x2y=0y''+\frac{9x^{2}}{9x^{2}} y'+\frac{2}{9x^{2}} y=y''+y'+\frac{\frac{2}{9}}{x^{2}} y=y''+\frac{x}{x} y'+\frac{\frac{2}{9}}{x^{2}} y=0

    Hence, b0=0b_{0} =0 and c0=29c_{0} =\frac{2}{9},

    r(r−1)+b0r+c0=r2−r+29=(r−13)(r−23)=0r( r-1) +b_{0} r+c_{0} =r^{2} -r+\frac{2}{9} =\left( r-\frac{1}{3}\right)\left( r-\frac{2}{3}\right) =0

    The indical roots are r1=23r_{1} =\frac{2}{3} and r2=13r_{2} =\frac{1}{3}.

    d. x2y′′+xy′+(x2−49)y=0x^{2} y''+xy'+\left( x^{2} -\frac{4}{9}\right) y=0

    Solution

    Divide the equation with x2x^{2},

    y′′+xx2y′+(x2−49)x2y=y′′+1xy′+−49+x2x2y=y′′+xxy′+29x2y=0y''+\frac{x}{x^{2}} y'+\frac{\left( x^{2} -\frac{4}{9}\right)}{x^{2}} y=y''+\frac{1}{x} y'+\frac{-\frac{4}{9} +x^{2}}{x^{2}} y=y''+\frac{x}{x} y'+\frac{\frac{2}{9}}{x^{2}} y=0

    Hence, b0=1b_{0} =1 and c0=−49c_{0} =-\frac{4}{9},

    r(r−1)+b0r+c0=r2−r+r−49=r2−49=(r+23)(r−23)=0r( r-1) +b_{0} r+c_{0} =r^{2} -r+r-\frac{4}{9} =r^{2} -\frac{4}{9} =\left( r+\frac{2}{3}\right)\left( r-\frac{2}{3}\right) =0

    The indical roots are r1=23r_{1} =\frac{2}{3} and r2=−23r_{2} =-\frac{2}{3}.

    e. xy′′+(1−x)y′−y=0xy''+( 1-x) y'-y=0

    Solution

    Divide the equation with xx,

    y′′+(1−x)xy′−1xy=y′′+(1−x)xy′+xx2y=0y''+\frac{( 1-x)}{x} y'-\frac{1}{x} y=y''+\frac{( 1-x)}{x} y'+\frac{x}{x^{2}} y=0

    Hence, b0=1b_{0} =1 and c0=0c_{0} =0,

    r(r−1)+b0r+c0=r2−r+r=r2=0r( r-1) +b_{0} r+c_{0} =r^{2} -r+r=r^{2} =0

    The indical roots are r1=0r_{1} =0 and r2=0r_{2} =0.