When the Method is Used
Use the power-series method when \(x=0\) is an ordinary point of the differential equation.
Power Series Method
Recurrence Relations and Even-Odd Series
The power series method follows the same coefficient recurrence presentation used in the class notes. Form the recurrence relation, put successive values of k and separate the two series.
UnitSeries Solutions and Special Functions
TopicPower Series Solutions
Question Type4 Mark Problems
Theory Required for These Problems
Required Power-Series Solution
Begin with
\[y=a_0+a_1x+a_2x^2+\cdots\]
or
\[y=\sum_{m=0}^{\infty}a_mx^m.\]
Same Power of x
After substitution, change the indices so that every term contains \(x^k\). Write each index change before equating coefficients.
Recurrence Relation
Equate the coefficient of \(x^k\) on both sides and isolate the later coefficient. Then put \(k=0,1,2,\ldots\) one value at a time.
Final Series
Substitute the calculated coefficients into
\[y=a_0+a_1x+a_2x^2+\cdots.\]
Collect the terms containing \(a_0\) and \(a_1\) separately.
Method used: Write the required series, differentiate, substitute, show the index changes, equate coefficients of \(x^k\), form the recurrence relation, put successive values of \(k\), and substitute back.
Continue through the IOE Engineering Mathematics II Digital Workbook. This power series method page follows the Engineering Mathematics II course sequence in the Tribhuvan University curriculum.
63
4 MarksPower-Series Solution with an Exponential Pattern
PAST-PAPER REFERENCE: TU IOE • 2083 Baisakh • Back (New Course) • ENSH 151 • Q. 15 • 4 Marks
Solve
\[y''-4xy'+(4x^2-2)y=0\]
by the power-series method.
Solution
Coefficient recurrence methodLet the required power-series solution be
\[y=a_0+a_1x+a_2x^2+a_3x^3+\cdots.\]In sigma notation,
\[y=\sum_{m=0}^{\infty}a_mx^m.\]Differentiating,
\[y'=\sum_{m=1}^{\infty}a_m m x^{m-1}.\]Again differentiating,
\[y''=\sum_{m=2}^{\infty}a_m m(m-1)x^{m-2}.\]Write the given equation as
\[y''-4xy'+4x^2y-2y=0.\]Substituting the series,
\[\sum_{m=2}^{\infty}a_m m(m-1)x^{m-2}-4\sum_{m=1}^{\infty}a_m m x^m+4\sum_{m=0}^{\infty}a_mx^{m+2}-2\sum_{m=0}^{\infty}a_mx^m=0.\]
\[\sum_{m=2}^{\infty}a_m m(m-1)x^{m-2}\]\[{}-4\sum_{m=1}^{\infty}a_m m x^m\]\[{}+4\sum_{m=0}^{\infty}a_mx^{m+2}\]\[{}-2\sum_{m=0}^{\infty}a_mx^m=0.\]
To obtain the same power of x, use
\[m-2=k,\quad m=k,\quad m+2=k.\]
\[m-2=k\quad\Rightarrow\quad m=k+2.\]\[m=k.\]\[m+2=k\quad\Rightarrow\quad m=k-2.\]
Equating the coefficient of \(x^k\) on both sides,
\[a_{k+2}(k+2)(k+1)-4ka_k+4a_{k-2}-2a_k=0.\]
\[a_{k+2}(k+2)(k+1)-4ka_k\]\[{}+4a_{k-2}-2a_k=0.\]
Therefore, the recurrence relation is
\[a_{k+2}=\frac{(4k+2)a_k-4a_{k-2}}{(k+1)(k+2)},\quad a_{-2}=a_{-1}=0.\]
\[a_{k+2}=\frac{(4k+2)a_k-4a_{k-2}}{(k+1)(k+2)}.\]\[a_{-2}=a_{-1}=0.\]
Put \(k=0\):
\[a_2=\frac{2a_0}{1\cdot2}=a_0.\]Put \(k=1\):
\[a_3=\frac{6a_1}{2\cdot3}=a_1.\]Put \(k=2\):
\[a_4=\frac{10a_2-4a_0}{3\cdot4}=\frac{a_0}{2}.\]
\[a_4=\frac{10a_2-4a_0}{3\cdot4}.\]\[a_4=\frac{10a_0-4a_0}{12}=\frac{a_0}{2}.\]
Put \(k=3\):
\[a_5=\frac{14a_3-4a_1}{4\cdot5}=\frac{a_1}{2}.\]
\[a_5=\frac{14a_3-4a_1}{4\cdot5}.\]\[a_5=\frac{14a_1-4a_1}{20}=\frac{a_1}{2}.\]
Put \(k=4\):
\[a_6=\frac{18a_4-4a_2}{5\cdot6}=\frac{a_0}{6}.\]
\[a_6=\frac{18a_4-4a_2}{5\cdot6}.\]\[a_6=\frac{9a_0-4a_0}{30}=\frac{a_0}{6}.\]
Put \(k=5\):
\[a_7=\frac{22a_5-4a_3}{6\cdot7}=\frac{a_1}{6}.\]
\[a_7=\frac{22a_5-4a_3}{6\cdot7}.\]\[a_7=\frac{11a_1-4a_1}{42}=\frac{a_1}{6}.\]
Substituting in the original power series,
\[y=a_0\left(1+x^2+\frac{x^4}{2!}+\frac{x^6}{3!}+\cdots\right)+a_1\left(x+x^3+\frac{x^5}{2!}+\frac{x^7}{3!}+\cdots\right).\]
\[y=a_0\left(1+x^2+\frac{x^4}{2!}+\frac{x^6}{3!}+\cdots\right)\]\[{}+a_1\left(x+x^3+\frac{x^5}{2!}+\frac{x^7}{3!}+\cdots\right).\]
Recognising the exponential series,
\[y=(a_0+a_1x)e^{x^2}.\]\[y=(C_1+C_2x)e^{x^2}\]
64
4 MarksPower-Series Solution with a Two-Step Recurrence
PAST-PAPER REFERENCE: TU IOE • 2082 Bhadra • Regular (New Course) • ENSH 151 • Q. 15 • 4 Marks
Solve the differential equation
\[y''+xy'+y=0\]
by the power-series method.
Solution
Coefficient recurrence methodLet the required power-series solution be
\[y=a_0+a_1x+a_2x^2+a_3x^3+\cdots.\]In sigma notation,
\[y=\sum_{m=0}^{\infty}a_mx^m.\]Differentiating,
\[y'=\sum_{m=1}^{\infty}a_m m x^{m-1}.\]Again differentiating,
\[y''=\sum_{m=2}^{\infty}a_m m(m-1)x^{m-2}.\]Substituting in the given equation,
\[\sum_{m=2}^{\infty}a_m m(m-1)x^{m-2}+\sum_{m=1}^{\infty}a_m m x^m+\sum_{m=0}^{\infty}a_mx^m=0.\]
\[\sum_{m=2}^{\infty}a_m m(m-1)x^{m-2}\]\[{}+\sum_{m=1}^{\infty}a_m m x^m\]\[{}+\sum_{m=0}^{\infty}a_mx^m=0.\]
To obtain the same power of x, use
\[m-2=k\quad\mathrm{and}\quad m=k.\]Equating the coefficient of \(x^k\),
\[a_{k+2}(k+2)(k+1)+ka_k+a_k=0.\]Therefore,
\[(k+1)\left[(k+2)a_{k+2}+a_k\right]=0.\]Hence, the recurrence relation is
\[a_{k+2}=-\frac{a_k}{k+2}.\]Put \(k=0\):
\[a_2=-\frac{a_0}{2}.\]Put \(k=1\):
\[a_3=-\frac{a_1}{3}.\]Put \(k=2\):
\[a_4=-\frac{a_2}{4}=\frac{a_0}{2\cdot4}.\]Put \(k=3\):
\[a_5=-\frac{a_3}{5}=\frac{a_1}{3\cdot5}.\]Put \(k=4\):
\[a_6=-\frac{a_4}{6}=-\frac{a_0}{2\cdot4\cdot6}.\]Put \(k=5\):
\[a_7=-\frac{a_5}{7}=-\frac{a_1}{3\cdot5\cdot7}.\]Substituting in the original power series,
\[y=a_0\left(1-\frac{x^2}{2}+\frac{x^4}{2\cdot4}-\frac{x^6}{2\cdot4\cdot6}+\cdots\right)+a_1\left(x-\frac{x^3}{3}+\frac{x^5}{3\cdot5}-\frac{x^7}{3\cdot5\cdot7}+\cdots\right).\]
\[y=a_0\left(1-\frac{x^2}{2}+\frac{x^4}{2\cdot4}-\frac{x^6}{2\cdot4\cdot6}+\cdots\right)\]\[{}+a_1\left(x-\frac{x^3}{3}+\frac{x^5}{3\cdot5}-\frac{x^7}{3\cdot5\cdot7}+\cdots\right).\]
\[y=C_1\left(1-\frac{x^2}{2}+\frac{x^4}{2\cdot4}-\cdots\right)+C_2\left(x-\frac{x^3}{3}+\frac{x^5}{3\cdot5}-\cdots\right).\]
\[y=C_1\left(1-\frac{x^2}{2}+\frac{x^4}{2\cdot4}-\cdots\right)\]\[{}+C_2\left(x-\frac{x^3}{3}+\frac{x^5}{3\cdot5}-\cdots\right).\]
65
4 MarksPower-Series Solution with an Exponential Pattern
PAST-PAPER REFERENCE: TU IOE • 2081 Ashwin • Regular (New Course, 2080 Batch) • SH 151 • Q. 15 • 4 Marks
Solve
\[y''-4xy'+(4x^2-2)y=0\]
by the power-series method.
Solution
Coefficient recurrence methodLet the required power-series solution be
\[y=a_0+a_1x+a_2x^2+a_3x^3+\cdots.\]In sigma notation,
\[y=\sum_{m=0}^{\infty}a_mx^m.\]Differentiating,
\[y'=\sum_{m=1}^{\infty}a_m m x^{m-1}.\]Again differentiating,
\[y''=\sum_{m=2}^{\infty}a_m m(m-1)x^{m-2}.\]Write the given equation as
\[y''-4xy'+4x^2y-2y=0.\]Substituting the series,
\[\sum_{m=2}^{\infty}a_m m(m-1)x^{m-2}-4\sum_{m=1}^{\infty}a_m m x^m+4\sum_{m=0}^{\infty}a_mx^{m+2}-2\sum_{m=0}^{\infty}a_mx^m=0.\]
\[\sum_{m=2}^{\infty}a_m m(m-1)x^{m-2}\]\[{}-4\sum_{m=1}^{\infty}a_m m x^m\]\[{}+4\sum_{m=0}^{\infty}a_mx^{m+2}\]\[{}-2\sum_{m=0}^{\infty}a_mx^m=0.\]
To obtain the same power of x, use
\[m-2=k,\quad m=k,\quad m+2=k.\]
\[m-2=k\quad\Rightarrow\quad m=k+2.\]\[m=k.\]\[m+2=k\quad\Rightarrow\quad m=k-2.\]
Equating the coefficient of \(x^k\) on both sides,
\[a_{k+2}(k+2)(k+1)-4ka_k+4a_{k-2}-2a_k=0.\]
\[a_{k+2}(k+2)(k+1)-4ka_k\]\[{}+4a_{k-2}-2a_k=0.\]
Therefore, the recurrence relation is
\[a_{k+2}=\frac{(4k+2)a_k-4a_{k-2}}{(k+1)(k+2)},\quad a_{-2}=a_{-1}=0.\]
\[a_{k+2}=\frac{(4k+2)a_k-4a_{k-2}}{(k+1)(k+2)}.\]\[a_{-2}=a_{-1}=0.\]
Put \(k=0\):
\[a_2=\frac{2a_0}{1\cdot2}=a_0.\]Put \(k=1\):
\[a_3=\frac{6a_1}{2\cdot3}=a_1.\]Put \(k=2\):
\[a_4=\frac{10a_2-4a_0}{3\cdot4}=\frac{a_0}{2}.\]
\[a_4=\frac{10a_2-4a_0}{3\cdot4}.\]\[a_4=\frac{10a_0-4a_0}{12}=\frac{a_0}{2}.\]
Put \(k=3\):
\[a_5=\frac{14a_3-4a_1}{4\cdot5}=\frac{a_1}{2}.\]
\[a_5=\frac{14a_3-4a_1}{4\cdot5}.\]\[a_5=\frac{14a_1-4a_1}{20}=\frac{a_1}{2}.\]
Put \(k=4\):
\[a_6=\frac{18a_4-4a_2}{5\cdot6}=\frac{a_0}{6}.\]
\[a_6=\frac{18a_4-4a_2}{5\cdot6}.\]\[a_6=\frac{9a_0-4a_0}{30}=\frac{a_0}{6}.\]
Put \(k=5\):
\[a_7=\frac{22a_5-4a_3}{6\cdot7}=\frac{a_1}{6}.\]
\[a_7=\frac{22a_5-4a_3}{6\cdot7}.\]\[a_7=\frac{11a_1-4a_1}{42}=\frac{a_1}{6}.\]
Substituting in the original power series,
\[y=a_0\left(1+x^2+\frac{x^4}{2!}+\frac{x^6}{3!}+\cdots\right)+a_1\left(x+x^3+\frac{x^5}{2!}+\frac{x^7}{3!}+\cdots\right).\]
\[y=a_0\left(1+x^2+\frac{x^4}{2!}+\frac{x^6}{3!}+\cdots\right)\]\[{}+a_1\left(x+x^3+\frac{x^5}{2!}+\frac{x^7}{3!}+\cdots\right).\]
Recognising the exponential series,
\[y=(a_0+a_1x)e^{x^2}.\]\[y=(C_1+C_2x)e^{x^2}\]
A Few Important Questions
Power-Series Solution of a Standard Equation
Solve
\[y''+y=0\]
by the power-series method.
Show Solution
The recurrence relation is
\[a_{k+2}=-\frac{a_k}{(k+1)(k+2)}.\]Putting \(k=0,1,2,3,\ldots\),
\[a_2=-\frac{a_0}{2!},\quad a_3=-\frac{a_1}{3!},\quad a_4=\frac{a_0}{4!},\quad a_5=\frac{a_1}{5!}.\]
\[a_2=-\frac{a_0}{2!},\quad a_3=-\frac{a_1}{3!}.\]\[a_4=\frac{a_0}{4!},\quad a_5=\frac{a_1}{5!}.\]
\[y=C_1\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots\right)+C_2\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right).\]
\[y=C_1\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots\right)\]\[{}+C_2\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right).\]
Power-Series Solution with a Constant Coefficient
Solve
\[y''+2y=0\]
by the power-series method.
Show Solution
The recurrence relation is
\[a_{k+2}=-\frac{2a_k}{(k+1)(k+2)}.\]Putting \(k=0,1,2,3,\ldots\),
\[a_2=-a_0,\quad a_3=-\frac{a_1}{3},\quad a_4=\frac{a_0}{6},\quad a_5=\frac{a_1}{30}.\]\[y=C_1\left(1-x^2+\frac{x^4}{6}-\cdots\right)+C_2\left(x-\frac{x^3}{3}+\frac{x^5}{30}-\cdots\right).\]
\[y=C_1\left(1-x^2+\frac{x^4}{6}-\cdots\right)\]\[{}+C_2\left(x-\frac{x^3}{3}+\frac{x^5}{30}-\cdots\right).\]