Tutorial 13: Frobenius Method
Q1: Find the radius of convergence and interval of cinvergence for the given power series.
(a) \(\sum _{n=1}^{\infty }\frac{2^{n}}{n} x^{n}\)
(b) \(\sum _{n=1}^{\infty }\frac{( -1)^{n}}{4^{n}}( x+3)^{n}\)
(c) \(\sum _{n=0}^{\infty }\frac{( 100)^{n}}{n!}( x+7)^{n}\)
(d) \(\sum _{n=0}^{\infty } n!( 2x+1)^{n}\)
Solution
Question (a)
\[\lim _{n\rightarrow \infty }\left\vert \frac{a_{n} +1}{a_{n}}\right\vert =\lim _{n\rightarrow \infty }\left\vert \frac{2^{n+1} x^{n+1} /( n+1)}{2^{n} x^{n} /n}\right\vert =\lim _{n\rightarrow \infty }\frac{2n}{n+1}\vert x\vert =2\vert x\vert\]The series is absolutely convergent for \(2\vert x\vert < 1\) or \(\vert x\vert < \frac{1}{2}\). At \(x=-\frac{1}{2}\), the series \(\sum _{n=1}^{\infty }\frac{( -1)^{n}}{n}\) converges by the alternating series test. At \(x=\frac{1}{2}\), the series \(\sum _{n=1}^{\infty }\frac{1}{n}\) is the harmonic series which diverges. Thus, the given series converges on \(\left[ -\frac{1}{2} ,\frac{1}{2}\right]\).
Question (b)
\[\begin{aligned} L & =\lim _{n\rightarrow \infty }\left\vert \frac{( -1)^{n+1}( n+1)( x+3)^{n+1}}{4^{n+1}}\frac{4^{n}}{( -1)^{n}( n)( x+3)^{n}}\right\vert \\ & =\lim _{n\rightarrow \infty }\left\vert \frac{-( n+1)( x+3)}{4n}\right\vert \\ & =\vert x+3\vert \lim _{n\rightarrow \infty }\frac{n+1}{4n}\\ & =\frac{1}{4}\vert x+3\vert \end{aligned}\]\(\frac{1}{4}\vert x+3\vert < 1\Longrightarrow \vert x+3\vert < 4\) series converges \(\frac{1}{4}\vert x+3\vert >1\Longrightarrow \vert x+3\vert >4\) series diverges
The radius of convergence for this power series is \(R=4\).
\[\begin{aligned} -4 & < x+3< 4\\ -7 & < x< 1 \end{aligned}\]Question (c)
\[\lim _{n\rightarrow \infty }\left\vert \frac{a_{n+1}}{a_{n}}\right\vert =\lim _{n\rightarrow \infty }\left\vert \frac{100^{n+1}( x+7)^{n+1} /( n+1) !}{100^{n}( x+7)^{n} /n!}\right\vert =\lim _{n\rightarrow \infty }\frac{100}{n+1}\vert x+7\vert =0\]The series is absolutely convergent on \(( -\infty ,\infty )\).
Question (d)
\[\begin{aligned} L & =\lim _{n\rightarrow \infty }\left\vert \frac{( n+1) !( 2x+1)^{n+1}}{n!( 2x+1)^{n}}\right\vert \\ & =\lim _{n\rightarrow \infty }\left\vert \frac{( n+1) n!( 2x+1)}{n!}\right\vert \\ & =\vert 2x+1\vert \lim _{n\rightarrow \infty }( n+1) \end{aligned}\]At this point we need to be careful. Yje limit is infinite, but there is that term with the \(x\)’s in front of the limit. We will have \(L=\infty >1\) provided \(x\neq -\frac{1}{2}\) will only converge if \(x=-\frac{1}{2}\).
The radius of convergence is \(R=0\) and the interval of convergence is \(x=-\frac{1}{2}\).
Q2: Rewrite the given power series by shifting the index, so that its general term involves \(x^{k}\).
(a) \(\sum _{n=3}^{\infty }( 2n-1) c_{n} x^{n-3}\)
(b) \(\sum _{n=3}^{\infty }\frac{3^{n}}{( 2n) !} x^{n-3}\)
(c) \(\sum _{n=3}^{\infty }\frac{( -1)^{n}}{( 2n+1) !} x^{2n+1}\)
Solution
Question (a)
\[\sum _{n=3}^{\infty }( 2n-1) c_{n} x^{n-3} =\sum _{k=0}^{\infty }( 2( k+3) -1) c_{k+3} x^{k} =\mathbf{\sum _{k=0}^{\infty }( 2k+5) c_{k+3} x^{k}}\]Question (b)
\[\sum _{n=3}^{\infty }\frac{3^{n}}{( 2n) !} x^{n-2} =\sum _{k=1}^{\infty }\frac{3^{k+2}}{( 2( k+2)) !} x^{k} =\mathbf{\sum _{k=1}^{\infty }\frac{3^{k+2}}{( 2k+1) !} x^{k}}\]Question (c)
\[\sum _{n=3}^{\infty }\frac{( -1)^{n}}{( 2n+1) !} x^{2n+1} =\sum _{k=7}^{\infty }\frac{( -1)^{\frac{k-1}{2}}}{\left( 2\left(\frac{k-1}{2}\right) +1\right) !} x^{k} =\sum _{k=7}^{\infty }\frac{( -1)^{\frac{k-1}{2}}}{k!} x^{k} =\mathbf{\sum _{k=7}^{\infty }\frac{\sqrt{( -1)^{k-1}}}{k!} x^{k}}\]Q3: Rewrite the given expression as a single power series whose general term involves \(x^k\).
a. \(\sum _{n=2}^{\infty } n( n-1) C_{n} x^{n} +2\sum _{n=2}^{\infty } n( n-1) c_{n} x^{n-2} +3\sum _{n=1}^{\infty } nc_{n} x^{n}\)
b. \(3x^{2}\sum _{n=-2}^{\infty } n( n-1) x^{n-2} +x\sum _{n=1}^{\infty } nx^{n}\)
c. \(\sum _{n=1}^{\infty }\frac{3^{n}}{( 2n) !} x^{n-1} +3x^{3}\sum _{n=-1}^{\infty }\frac{( -1)^{n}}{( 2n+1) !} x^{2n+1}\)
Solution
Question (a)
\[\begin{align*} & \sum _{n=2}^{\infty } n( n-1) c_{n} x^{n} +2\sum _{n=2}^{\infty } n( n-1) c_{n} x^{n-2} +3\sum _{n=1}^{\infty } nc_{n} x_{n}\\ &= 2\cdot 2\cdot 1c_{2} x^{0} +2\cdot 3\cdot 2c_{3} x'+3\cdot 1\cdot c_{1} x'+\underbrace{ \sum _{n=2}^{\infty } n( n-1) c_{n}}_{k=n} x^{n}_{k=n} +2\underbrace{ \sum _{n=4}^{\infty } n( n-1) c_{n} x^{n-2}}_{k=n-2} +3\underbrace{ \sum _{n=2}^{\infty } nc_{n} x^{n}}_{k=n}\\ &= 4c_{2} +( 12c_{3} +( 12c_{3} +3c_{1}) x+\sum _{n=2}^{\infty } k( k-1) c_{k} x^{k} +2\sum _{n=2}^{\infty }( k+2)( k+1) x_{k+2} x^{k} +3\sum _{n=2}^{\infty } kc_{k} x^{k}\\ &= 4c_{2} +( 3c_{1} +12c_{3}) x+\sum _{n=2}^{\infty }([ k( k-1) +3k] c_{k} +2( k+2)( k+1) x_{k+2}) x^{k}\\ &= 4c_{2} +( 3c_{1} +12c_{3}) x+\sum _{n=2}^{\infty }[ k( k+2) c_{k} +2( k+1)( k+2) c_{k+2}] x^{k} \end{align*}\]Question (b)
\[3x^{2}\sum _{n=-2}^{\infty } n( n-1) x^{n-2} +x\sum _{n=1}^{\infty } nx^{n} =3\sum _{n=-2}^{\infty } n( n-1) x^{n} +\sum _{n=1}^{\infty } nx^{n+1} =3\sum _{k=-2}^{\infty } k( k-1) x^{k} +\sum _{k=2}^{\infty }( k-1) x^{k} \tag{1}\]Solving first part of Eq(1),
\[\begin{aligned} \sum _{k=-2}^{\infty } k( k-1) x^{k} & =-2( -2-1) x^{-2} +( -1)( -1-1) x^{-1} +0+1( 1-1) x^{1} +\sum _{k=2}^{\infty } k( k-1) x^{k}\\ & =\frac{6}{x^{2}} +\frac{2}{x} +\sum _{k=2}^{\infty } k( k-1) x^{k} \end{aligned}\]Substitute back to Eq (1),
\[\begin{aligned} \therefore 3\left[\frac{6}{x^{2}} +\frac{2}{x} +\sum _{k=2}^{\infty } k( k-1) x^{k}\right] +\sum _{k=2}^{\infty }( k-1) x^{k} & =\frac{18}{x^{2}} +\frac{6}{x} +\sum _{k=2}^{\infty }[ 3k( k-1) +( k-1)] x^{k}\\ & =\frac{18}{x^{2}} +\frac{6}{x} +\sum _{k=2}^{\infty }[( k-1)( 3k+1)] x^{k} \end{aligned}\]Question (c)
\[\begin{aligned} \sum _{n=1}^{\infty }\frac{3^{n}}{( 2n) !} x^{n-1} +2x^{3}\sum _{n=-1}^{\infty }\frac{( -1)^{n}}{( 2n+1) !} x^{2n+1} & =\sum _{n=1}^{\infty }\frac{3^{n}}{( 2n) !} x^{n-1} +2\sum _{n=-1}^{\infty }\frac{( -1)^{n}}{( 2n+1) !} x^{2n+4}\\ & =\sum _{k=0}^{\infty }\frac{3^{k+1}}{( 2( k+1)) !} x^{k} +2\sum _{k=2}^{\infty }\frac{( -1)^{\frac{k-4}{2}}}{\left( 2\left(\frac{k-4}{2}\right) +1\right) !} x^{k}\\ & =\sum _{k=0}^{\infty }\frac{3^{k+1}}{( 2k+2) !} x^{k} +2\sum _{k=2}^{\infty }\frac{( -1)^{\frac{k-4}{2}}}{( k-3) !} x^{k} \end{aligned} \tag{2}\]Solving first part of Eq (2),
\[\begin{aligned} \sum _{k=0}^{\infty }\frac{3^{k+1}}{( 2k+2) !} x^{k} & =\frac{3^{1}}{4!} x^{0} +\frac{3^{2}}{4!} x^{1} +\sum _{k=2}^{\infty }\frac{3^{k+1}}{( 2k+2) !}\\ & =\frac{3}{2} +\frac{3}{8} x+\sum _{k=2}^{\infty }\frac{3^{k+1}}{( 2k+2) !} x^{k} \end{aligned}\]Subtitute back to Eq (2),
\[\begin{aligned} \therefore \frac{3}{2} +\frac{3}{8} x+\sum _{k=2}^{\infty }\frac{3^{k+1}}{( 2k+2) !} x^{k} +2\sum _{k=2}^{\infty }\frac{( -1)^{\frac{k-4}{2}}}{( k-3) !} x^{k} & =\frac{3}{2} +\frac{3}{8} x+\sum _{k=2}^{\infty }\left[\frac{3^{k+2}}{( 2k+2) !} +\frac{2( -1)^{\frac{k-4}{2}}}{( k-3) !}\right] x^{k}\\ & =\frac{3}{2} +\frac{3}{8} x+\sum _{k=2}^{\infty }\left[\frac{3^{k+2}}{( 2k+2) !} +\frac{2\sqrt{( -1)^{k-4}}}{( k-3) !}\right] x^{k} \end{aligned}\]Q4: Find two power series solutions of given differential equation about the ordinary point \(x=0\).
a. \(y''+xy'+y=0\)
b. \(( x-1) y''+y'=0\)
c. \(y''+e^{x} y'-y=0\)
d. \(\left( x^{2} +1\right) y''+xy'-y=0\)
Solution
Question (a)
Let \(y( x) =\sum _{n=0}^{\infty } c_{n} x^{n} \Longrightarrow \ y'( x) =\sum _{n=1}^{\infty } nc_{n} x^{n-1}\) and \(y''( x) =\sum _{n=2}^{\infty }( n+2)( n+1) c_{n+2} x^{n}\).
The differential equation becomes
\[\begin{aligned} \sum _{n=0}^{\infty }( n+2)( n+1) c_{n+2} x^{n} +x\sum _{n=1}^{\infty } nc_{n} x^{n-1} +\sum _{n=0}^{\infty } c_{n} x^{n} & =0\\ \sum _{n=0}^{\infty }[( n+2)( n+1) c_{n+2} +nc_{n} +c_{n}] x^{n} & =0\\ \color{red} {\scriptsize { \text{[} \triangle \text{Since} \sum _{n=1}^{\infty }nc_{n}x^{n} = \sum _{n=0}^{\infty } nc_{n}x^{n} \text{]}}} & \end{aligned}\]Equating coefficients gives,
\[( n+2)( n+1) c_{n+2} +( n+1) c_{n} =0\]Thus, the recursion relation is,
\[c_{n+2} =\frac{-( n+1) c_{n}}{( n+2)( n+1)} =-\frac{c_{n}}{n+2} ,\quad n=0,1,2,...\]Then, the even coefficients are given by \(c_{2} =-\frac{c_{0}}{2} ,\quad c_{4} =-\frac{c_{2}}{4} =\frac{c_{0}}{2\cdot 4} ,\quad c_{6} =-\frac{c_{4}}{6} =-\frac{c_{0}}{2\cdot 4\cdot 6}\). Hence,
\[c_{2n} =( -1)^{n}\frac{c_{0}}{2\cdot 4\cdot \cdot \cdot 2n} =\frac{( -1)^{n} c_{0}}{2^{n} n!}\]The odd coefficients are \(c_{3} =-\frac{c_{1}}{3} ,\quad c_{5} =-\frac{c_{3}}{5} =\frac{c_{1}}{3\cdot 5} ,\quad c_{7} =-\frac{c_{5}}{7} =-\frac{c_{1}}{3\cdot 5\cdot 7}\). Hence,
\[c_{2n+1} =( -1)^{n}\frac{c_{1}}{3\cdot 5\cdot 7\cdot \cdot \cdot ( 2n+1)} =\frac{( -2)^{n} n!c_{1}}{( 2n+1) !}\]The solution is,
\[\mathbf{y( x) =c_{0}\sum _{n=0}^{\infty }\frac{( -1)^{n}}{2^{n} n!} x^{2n} +c_{1}\sum _{n=0}^{\infty }\frac{( -2)^{n} n!}{( 2n+1) !} x^{2n+1}}\]Question (b)
Subtituting \(y=\sum _{n=0}^{\infty } c_{n} x^{n}\) into the differential equation,
\[\begin{aligned} ( x-1) y''+y' & =\underbrace{ \sum _{n=2}^{\infty } n( n-1) c_{n} x^{n-1}}_{k=n-1} -\underbrace{ \sum _{n=2}^{\infty } n( n-1) c_{n} x^{n-2}}_{k=n-2} +\underbrace{ \sum _{n=1}^{\infty } nc_{n} x^{n-1}}_{k=n-1}\\ & ={ \sum _{k=1}^{\infty }( k+1) kc_{k+1} x^{k} -\sum _{k=0}^{\infty }( k+2)( k+1) c_{k+2} x^{k} +\sum _{k=0}^{\infty }( k+1) c_{k+1} x^{k}}\\ & =-2c_{2} +c_{1} +{ \sum _{k=1}^{\infty }[( k+1) kc_{k+1} -( k+2)( k+1) c_{k+2} +( k+1) c_{k+1}] x^{k} =0} \end{aligned}\]Thus,
\[\begin{aligned} -2c_{2} +c_{1} & =0\\ ( k+1)^{2} c_{k+1} -( k+2)( k+1) c_{k+2} & =0 \end{aligned}\]and
\[\begin{aligned} c_{2} & =\frac{1}{2} c_{1}\\ c_{k+2} & =\frac{k+1}{k+2} c_{k+1} ,\quad k=1,2,3,... \end{aligned}\]Choosing \(c_{0} =1\) and \(c_{1} =0\), we find \(c_{2} =c_{3} =c_{4} =...=0\). For \(c_{0} =0\) and \(c_{1} =1\), we obtain
\[c_{2} =\frac{1}{2} ,\quad c_{3} =\frac{1}{3} ,\quad c_{4} =\frac{1}{4}\]Thus, the two solutions are
\[\mathbf{y_{1} =1\quad \quad y_{2} =x+\frac{1}{2} x^{2} +\frac{1}{3} x^{3} +\frac{1}{4} x^{4} +...}\]Question (c)
Substituting \(y=\sum _{n=0}^{\infty } c_{n} x^{n}\) into the differential equation we have,
\[\begin{aligned} y''-( x+1) y'-y & =\underbrace{ \sum _{n=2}^{\infty } n( n-1) c_{n} x^{n-2}}_{k=n-2} -\underbrace{ \sum _{n=1}^{\infty } nc_{n} x^{n}}_{k=n} -\underbrace{ \sum _{n=1}^{\infty } nc_{n} x^{n-1}}_{k=n-1} -\underbrace{ \sum _{n=0}^{\infty } c_{n} x^{n}}_{k=n}\\ & =\sum _{k=0}^{\infty }( k+2)( k+1) c_{k+2} x^{k} -\sum _{k=1}^{\infty } kc_{k} x^{k} -\sum _{k=0}^{\infty }( k+1) c_{k+1} x^{k} -\sum _{k=0}^{\infty } c_{k} x^{k}\\ & =2c_{2} -c_{1} -c_{0} +\sum _{k=1}^{\infty }[( k+2)( k+1) c_{k+2} -( k+1) c_{k+1} -( k+1) c_{k}] x^{k} =0 \end{aligned}\]Thus,
\[\begin{aligned} 2c_{2} -c_{1} -c_{0} & =0\\ ( k+2)( k+1) c_{k+2} -( k-1)( c_{k+1} +c_{k}) & =0 \end{aligned}\]and,
\[\begin{aligned} c_{2} & =\frac{c_{1} +c_{0}}{2}\\ c_{k+2} & =\frac{c_{k+1} +c_{k}}{k+2} c_{k} ,\quad k=2,3,4,... \end{aligned}\]Choosing \(c_{0} =1\) and \(c_{1} =0\), we find,
\[c_{2} =\frac{1}{2} ,\quad \quad c_{3} =\frac{1}{6} ,\quad \quad c_{4} =\frac{1}{6}\]and so on. For \(c_{0} =0\) and \(c_{1} =1\), we obtain,
\[c_{2} =\frac{1}{2} ,\quad \quad c_{3} =\frac{1}{2} ,\quad \quad c_{4} =\frac{1}{4}\]and so on. Thus, the two solutions are,
\[y_{1} =1+\frac{1}{2} x^{2} +\frac{1}{6} x^{3} +\frac{1}{6} x^{4} +...\quad \quad \text{and} \quad \quad y_{2} =x+\frac{1}{2} x^{2} +\frac{1}{2} x^{3} +\frac{1}{4} x^{4} +...\]Question (d)
Let \(y( x) =\sum _{n=0}^{\infty } c_{n} x^{n} \Longrightarrow \ y''( x) =\sum _{n=0}^{\infty } n( n-1) c_{n} x^{n-2}\) and \(xy''=\sum _{n=0}^{\infty } nc_{n} x^{n}\). Thus,
\[\left( x^{2} +1\right) y''=\sum _{n=0}^{\infty } n( n-1c_{n} x^{n} +\sum _{n=0}^{\infty }( n+2)( n+1) c_{n+2} x^{n}\]The differential equation becomes,
\[\sum _{n=0}^{\infty }[( n+2)( n+1) c_{n+2} +[ n( n-1) +n-1] c_{n}] x^{n} =0\]The recursion relation is
\[c_{n+2} =-\frac{( n-1) c_{n}}{n+2} ,\quad n=0,1,2,...\]Given \(c_{0}\) and \(c_{1}\), \(c_{2} =\frac{c_{0}}{2}\), \(c_{4} =-\frac{c_{0}}{2^{2} \cdot 2!}\), \(c_{6} =-\frac{3c_{4}}{6} =( -1)^{2}\frac{3c_{0}}{2^{3} \cdot 3!}\),…
\[c_{2n} =( -1)^{n-1}\frac{1\cdot 3\cdot \cdot \cdot ( 2n-3) c_{0}}{2^{n} n!} =( -1)^{n-1}\frac{( 2n-3) !c_{0}}{2^{n} 2^{n-2} n!( n-2) !} =( -1)^{n-1}\frac{( 2n-3) !c_{0}}{2^{2n-2} n!( n-2) !} \ \text{for} \ n=2,3,...\]\(c_{3} =\frac{0\cdot c_{1}}{3} =0\Longrightarrow c_{2n+1} =0\ \text{for} \ n=1,2,...\) Thus the solution is,
\[\mathbf{y( x) =c_{0} +c_{1} x+c_{0}\frac{x^{2}}{2} +c_{0}\sum _{n=2}^{\infty }\frac{( -1)^{n-1}( 2n-3) !}{2^{2n-2} n!( n-2) !} x^{2n}}\]Q5: Use the power series method to solve the given initial-value problem.
a. \(y''-xy'-y=0, y(0)=1, y'(0)=0\)
b. \(y''+x^2y'+xy=0, y(0)=0, y'(0)=1\)
c. \((x+1)y''-(2-x)y'+y=0, y(0)=2, y'(0)=-1\)
Solution
Question (a)
Let \(y( x) =\sum _{n=0}^{\infty } c_{n} x^{n}\). Then \(-xy'( x) =-x\sum _{n=1}^{\infty } nc_{n} x^{n-1} =-\sum _{n=1}^{\infty } nc_{n} x^{n} =-\sum _{n=0}^{\infty } nc_{n} x^{n}\).
\[y''( x) =\sum _{n=0}^{\infty }( n+2)( n+1) c_{n+2} x^{n}\]The equation \(y''-xy'-y=0\) becomes
\[\sum _{n=0}^{\infty }[( n+2)( n+1) c_{n+2} -nc_{n} -c_{n}] x^{n} =0\]Thus, the recursion relation is
\[c_{n+2} =\frac{nc_{n} +c_{n}}{( n+2)( n+1)} =\frac{c_{n}( n+1)}{( n+2)( n+1)} =\frac{c_{n}}{n+2} \ \text{for} \ n=0,1,2,...\]One of the condition is, \(y( 0) =1\). But \(y( 0) =\sum _{n=0}^{\infty } c_{n}( 0)^{n} =c_{0} +0+0+...=c_{0}\), so \(c_{0} =1\).
Hence, \(c_{2} =\frac{c_{0}}{2} =\frac{1}{2}\), \(c_{4} =\frac{c_{2}}{4} =\frac{1}{2\cdot 4}\), \(c_{6} =\frac{c_{4}}{6} =\frac{1}{2\cdot 4\cdot 6}\),….,\(c_{2n} =\frac{1}{2^{n} n!}\). The other given condition is \(y'( 0) =0\).
But \(y'( 0) =\sum _{n=1}^{\infty } nc_{n}( 0)^{n-1} =c_{1} +0+0+...=c_{1}\). So, \(c_{1} =0\).
By the recursion relation, \(c_{3} =\frac{c_{1}}{3} =0\), \(c_{5} =0\),…,\(c_{2n+1} =0\) for \(n=0,1,2,...\).
Thus, the solution to the initial-value problem is,
\[y( x) =\sum _{n=0}^{\infty } c_{n} x^{n} =\sum _{n=0}^{\infty } c_{2n} x^{2n} =\sum _{n=0}^{\infty }\frac{x^{2n}}{2^{n} n!} =\sum _{n=0}^{\infty }\frac{\left( x^{2} /2\right)^{n}}{n!} =c^{x^{2} /2}\]Question (b)
Assuming that \(y( x) =\sum _{n=0}^{\infty } c_{n} x^{n}\), we have \(xy=x\sum _{n=0}^{\infty } c_{n} x^{n} =\sum _{n=0}^{\infty } c_{n} x^{n+1}\),
\[x^{2} y'=x^{2}\sum _{n=1}^{\infty } nc_{n} x^{n-1} =\sum _{n=0}^{\infty } nc_{n} x^{n+1}\] \[\begin{aligned} y''( x) & =\sum _{n=2}^{\infty } n( n-1) c_{n} x^{n-2} =\sum _{n=-1}^{\infty }( n+3)( n+2) c_{n+3} x^{n+1}\\ & =2c_{2} +\sum _{n=0}^{\infty }( n+3)( n+2) c_{n+3} x^{n+1} \end{aligned}\]Thus, the equation \(y''+x^{2} y'+xy=0\) becomes
\[2c_{2} +\sum _{n=0}^{\infty }[( n+3)( n+2) c_{n+3} +nc_{n} +c_{n}] x^{n+1} =0\]So, \(c_{2} =0\) and the recursion relation is,
\[c_{n+3} =\frac{-nc_{n} -c_{n}}{( n+3)( n+2)} =-\frac{( n+1) c_{n}}{( n+3)( n+2)} ,\ n=0,1,2,...\]Also, \(c_{1} =y'( 0) =1\), so,
\[\begin{gather*} c_{4} =-\frac{2c_{1}}{4\cdot 3} =-\frac{2}{4\cdot 3} ,\quad c_{7} =-\frac{5c_{4}}{7\cdot 6} =( -1)^{2}\frac{2\cdot 5}{7\cdot 6\cdot 4\cdot 3} =( -1)^{2}\frac{2^{2} 5^{2}}{7!} ,...\\ \\ c_{3n+1} =( -1)^{n}\frac{2^{2} 5^{2} \cdot \cdot \cdot \cdot ( 3n-1)^{2}}{( 3n+1) !} \end{gather*}\]Thus, the solution is,
\[\mathbf{y( x) =\sum _{n=0}^{\infty } c_{m} x^{n} =x+\sum _{n=1}^{\infty }\left[( -1)^{n}\frac{2^{2} 5^{2} \cdot \cdot \cdot \cdot ( 3n-1)^{2} x^{3n+1}}{( 3n+1) !}\right]}\]Question (c)
Subtituting \(y=\sum _{n=0}^{\infty } c_{n} x^{n}\) into the differential equation, we have,
\[\begin{aligned} & ( x+1) y''-( 2-x) y'+y\\ = & \underbrace{ \sum _{n=2}^{\infty } n( n-1) c_{n} x^{n-1}}_{k=n-1} +\underbrace{ \sum _{n=2}^{\infty } n( n-1) c_{n} x^{n-2}}_{k=n-2} -2\underbrace{ \sum _{n=2}^{\infty } nc_{n} x^{n-1}}_{k=n-1} +\underbrace{ \sum _{n=1}^{\infty } nc_{n} x^{n}}_{k=n} +\underbrace{ \sum _{n=0}^{\infty } c_{n} x^{n}}_{k=n}\\ & = \sum _{k=1}^{\infty }( k+1) kc_{k+1} x^{k} +\sum _{k=0}^{\infty }( k+2)( k+1) c_{k+2} x^{k} -2\sum _{k=0}^{\infty }( k+1) c_{k+1} x^{k} +\sum _{k=1}^{\infty } kc_{k} x^{k} +\sum _{k=0}^{\infty } c_{k} x^{k}\\ & =2c_{2} -2c_{1} +c_{0} + \sum _{k=1}^{\infty }[( k+2)( k+1) c_{k+2} -( k+1) c_{k+1} +( k+1) c_{k}] x^{k} =0 \end{aligned}\]Thus,
\[\begin{aligned} 2c_{2} -2c_{1} +c_{0} & =0\\ ( k+2)( k+1) c_{k+2} -( k+1) c_{k+1} +( k+1) c_{k} & =0 \end{aligned}\]and,
\[\begin{aligned} c_{2} & =c_{1} -\frac{1}{2} c_{0}\\ c_{k+2} & =\frac{1}{k+2} c_{k+1} -\frac{1}{k+2} c_{k} ,\quad k=1,2,3,... \end{aligned}\]Choosing \(c_{0} =1\) and \(c_{1} =0\), we find
\[c_{2} =-\frac{1}{2} ,\quad \quad c_{3} =-\frac{1}{6} ,\quad \quad c_{4} =\frac{1}{12}\]and so on. For \(c_{0} =0\) and \(c_{1} =1\), we obtain,
\[c_{2} =1 ,\quad \quad c_{3} =0,\quad \quad c_{4} =-\frac{1}{4}\]and do on. Thus,
\[y=C_{1}\left( 1-\frac{1}{2} x^{2} -\frac{1}{6} x^{3} +\frac{1}{12} x^{4} +...\right) +C_{2}\left( x+x^{2} -\frac{1}{4} x^{4} +...\right)\]and
\[y'=C_{1}\left( -x-\frac{1}{2} x^{2} +\frac{1}{3} x^{3} +...\right) +C_{2}\left( 1+2x-x^{3} +...\right)\]The initital condition imply \(C_{1} =2\) and \(C_{2} =-1\), so
\[\begin{aligned} y & =2\left( 1-\frac{1}{2} x^{2} -\frac{1}{6} x^{3} +\frac{1}{12} x^{4} +...\right) -\left( x+x^{2} -\frac{1}{4} x^{4} +...\right)\\ & =2-x-2x^{2} -\frac{1}{3} x^{3} +\frac{5}{12} x^{4} +... \end{aligned}\]Q6: Determine the singular points of the given differential equation. Classify each singular point as regular or irregular.
a. \(x^{3} y''-4x^{2} y'+3y=0\)
b. \(\left( x^{2} -9\right)^{2} y''+( x+3) y'+2y=0\)
c. \(\left( 2x^{2} -5x-3\right) y''+( 2x+1) y'+\frac{6}{( x-3)} y=0\)
d. \(\left( x^{3} -2x^{2} -3x\right)^{2} y''+x( x-3)^{2} y'-( x+1) y=0\)
Solution
Question (a)
Dividing the equation with \(x^{3}\)
\[\begin{gather*} y''+\frac{4x^{2}}{x^{3}} y'+\frac{3}{x^{3}} y=0\\ \\ P( x) =\frac{4}{x} \quad \quad Q( x) =\frac{3}{x^{3}} \end{gather*}\]- The factor \(x\) appears at most to the first power in the denominator of \(P( x)\), but more that the second power in the denominator of \(Q( x)\)
- \(x=0\) is irregular singular point
Question (b)
Dividing the eqution with \(\left( x^{2} -9\right)^{2} =(( x-3)( x+3))^{2}\),
\[\begin{gather*} y''+\frac{( x+3)}{(( x-3)( x+3))^{2}} y'+\frac{2}{(( x-3)( x+3))^{2}} y=0\\ \\ P( x) =\frac{1}{( x-3)^{2}( x+3)} \quad \quad Q( x) =\frac{2}{( x-3)^{2}( x+3)^{2}} \end{gather*}\]- The factor \(x\ +\ 3\) appears at most to the first power in the denominator of \(P( x)\) and at most to the second power in the denominator of \(Q( x)\)
- \(x\ =\ -3\) is regular singular point
- The factor \(x\ -\ 3\) appears more than the first power in the denominator of \(P( x)\); but at most to the second power in the denominator of \(Q( x)\)
- \(x\ =3\) is irregular singular point
Question (c)
Factor the equation \(\left( 2x^{2} -5x-3\right) =( x-3)( 2x+1)\),
\[\begin{gather*} y''+\frac{( 2x+1)}{( x-3)( 2x+1)} y'+\frac{6}{( x-3)^{2}( 2x+1)} y=0\\ \\ P( x) =\frac{1}{( x-3)} \quad \quad Q( x) =\frac{6}{( x-3)^{2}( 2x+1)} \end{gather*}\]- The factor \(( x-3)\) appears at most to the first power in \(P( x)\) and second power in the denominator of \(Q( x)\)
- \(x\ =\ 3\) is regular singular point -The factor \(( 2x+1)\) appears at to the first power in the denominator of \(Q( x)\)
- \(x\ =\ -\frac{1}{2}\) is regular singular point
Question (d)
Divide the equation with \(\left( x^{3} -2x^{2} -3x\right)^{2} =\left( x\left( x^{2} -2x-3\right)\right)^{2} =x^{2}( x-3)^{2}( x+1)^{2}\),
\[\begin{gather*} y''+\frac{x( x-3)^{2}}{x^{2}( x-3)^{2}( x+1)^{2}} y'-\frac{( x+1)}{x^{2}( x-3)^{2}( x+1)^{2}} y=0\\ \\ P( x) =\frac{1}{x( x+1)^{2}} \quad \quad Q( x) =\frac{1}{x^{2}( x-3)^{2}( x+1)} \end{gather*}\]- The factor \(x\) appears at most to the first power in the denominator of \(P( x)\) and at most to the second power in the denominator of \(Q( x)\)
- \(x\ =\ 0\) is regular singular point
- The factor \(x\ -\ 3\) appears to the second power in the denominator of \(Q( x)\)
- \(x\ =\ 3\) is regular singular point
- The factor \(x\ +\ 1\) appears to the second power in the denominator of \(P(x)\) and to the first power in the denominator of \(Q( x)\)
- \(x\ =\ -1\) is irregular singular point
Q7: ind the indical roots for the given differential equations where \(x=0\) is a regular singular point.
a. \(2xy''-y'+2y=0\)
b. \(3xy''+( 2-x) y'-y=0\)
c. \(9x^{2} y''+9x^{2} y'+2y=0\)
d. \(x^{2} y''+xy'+\left( x^{2} -\frac{4}{9}\right) y=0\)
e. \(xy''+( 1-x) y'-y=0\)
Solution
Question (a)
Divide the equation with \(2x\),
\[y''-\frac{1}{2x} y'+\frac{2}{2x} y=y''-\frac{\frac{1}{2}}{x} y'+\frac{1}{x} y=y''-\frac{\frac{1}{2}}{x} y'+\frac{x}{x^{2}} y=0\]Hence, \(b_{0} =-\frac{1}{2}\) and \(r_{2} =0\),
\[7r( r-1) +b_{0} r+c_{0} =r^{2} -r-\frac{1}{2} r+0=r^{2} -\frac{3}{2} r=r\left( r-\frac{3}{2}\right) =0\]The indical roots are \(r_{1} =\frac{3}{2}\) and \(r_{2} =0\).
Question (b)
Divide the equation with \(3x\),
\[y''+\frac{( 2-x)}{3x} y'-\frac{1}{3x} y=y''+\frac{\frac{1}{3}( 2-x)}{x} y'-\frac{\frac{1}{3}}{x} y=y''+\frac{\frac{2}{3} -\frac{1}{3} x}{x} y'-\frac{\frac{1}{3} x}{x^{2}} y=0\]Hence, \(b_{0} =\frac{2}{3}\) and \(c_{0} =0\),
\[r( r-1) +b_{0} r+c_{0} =r^{2} -r+\frac{2}{3} r+0=r^{2} -\frac{1}{3} r=r\left( r-\frac{1}{3}\right) =0\]The indical roots are \(r_{1} =\frac{1}{3}\) and \(r_{2} =0\).
Question (c)
Divide the equation with \(9x^{2}\),
\[y''+\frac{9x^{2}}{9x^{2}} y'+\frac{2}{9x^{2}} y=y''+y'+\frac{\frac{2}{9}}{x^{2}} y=y''+\frac{x}{x} y'+\frac{\frac{2}{9}}{x^{2}} y=0\]Hence, \(b_{0} =0\) and \(c_{0} =\frac{2}{9}\),
\[r( r-1) +b_{0} r+c_{0} =r^{2} -r+\frac{2}{9} =\left( r-\frac{1}{3}\right)\left( r-\frac{2}{3}\right) =0\]The indical roots are \(r_{1} =\frac{2}{3}\) and \(r_{2} =\frac{1}{3}\).
Question (d)
Divide the equation with \(x^{2}\),
\[y''+\frac{x}{x^{2}} y'+\frac{\left( x^{2} -\frac{4}{9}\right)}{x^{2}} y=y''+\frac{1}{x} y'+\frac{-\frac{4}{9} +x^{2}}{x^{2}} y=y''+\frac{x}{x} y'+\frac{\frac{2}{9}}{x^{2}} y=0\]Hence, \(b_{0} =1\) and \(c_{0} =-\frac{4}{9}\),
\[r( r-1) +b_{0} r+c_{0} =r^{2} -r+r-\frac{4}{9} =r^{2} -\frac{4}{9} =\left( r+\frac{2}{3}\right)\left( r-\frac{2}{3}\right) =0\]The indical roots are \(r_{1} =\frac{2}{3}\) and \(r_{2} =-\frac{2}{3}\).
Question (e)
Divide the equation with \(x\),
\[y''+\frac{( 1-x)}{x} y'-\frac{1}{x} y=y''+\frac{( 1-x)}{x} y'+\frac{x}{x^{2}} y=0\]Hence, \(b_{0} =1\) and \(c_{0} =0\),
\[r( r-1) +b_{0} r+c_{0} =r^{2} -r+r=r^{2} =0\]The indical roots are \(r_{1} =0\) and \(r_{2} =0\).