In the following exercises, when asked to solve an equation using power series methods, you should find the first few terms of the series, and if possible find a general formula for the \(k^{\text{th}}\) coefficient.
Exercise \(\PageIndex{7.2.1}\)
Use power series methods to solve \(y''+y = 0\) at the point \(x_0 = 1\).
Exercise \(\PageIndex{7.2.2}\)
Use power series methods to solve \(y''+4xy = 0\) at the point \(x_0 = 0\).
Exercise \(\PageIndex{7.2.3}\)
Use power series methods to solve \(y''xy = 0\) at the point \(x_0 = 1\).
Exercise \(\PageIndex{7.2.4}\)
Use power series methods to solve \(y''+x^2y = 0\) at the point \(x_0 = 0\).
Exercise \(\PageIndex{7.2.5}\)
The methods work for other orders than second order. Try the methods of this section to solve the first order system \(y'xy = 0\) at the point \(x_0 = 0\).
Exercise \(\PageIndex{7.2.6}\)
Chebyshev’s equation of order \(p\):
 Solve \((1x^2)y''xy' + p^2y = 0\) using power series methods at \(x_0=0\).
 For what \(p\) is there a polynomial solution?
Exercise \(\PageIndex{7.2.7}\)
Find a polynomial solution to \((x^2+1) y''2xy'+2y = 0\) using power series methods.
Exercise \(\PageIndex{7.2.8}\)
 Use power series methods to solve \((1x)y''+y = 0\) at the point \(x_0 = 0\).
 Use the solution to part a) to find a solution for \(xy''+y=0\) around the point \(x_0=1\).
Exercise \(\PageIndex{7.2.9}\)
Use power series methods to solve \(y'' + 2 x^3 y = 0\) at the point \(x_0 =0\).
 Answer

\(a_{2}=0\), \(a_{3}=0\), \(a_{4}=0\), recurrence relation (for \(k\geq 5\)): \(a_{k}=\frac{2a_{k5}}{k(k1)}\), so \(y(x)=a_{0}+a_{1}x\frac{a_{0}}{10}x^{5}\frac{a_{1}}{15}x^{6}+\frac{a_{0}}{450}x^{10}+\frac{a_{1}}{825}x^{11}\frac{a_{0}}{47250}x^{15}\frac{a_{1}}{99000}x^{16}+\cdots\)
Exercise \(\PageIndex{7.2.10}\): (challenging)
We can also use power series methods in nonhomogeneous equations.
 Use power series methods to solve \(y''  x y = \frac{1}{1x}\) at the point \(x_0 = 0\). Hint: Recall the geometric series.
 Now solve for the initial condition \(y(0)=0\), \(y'(0) = 0\).
 Answer

 \(a_{2}=\frac{1}{2}\), and for \(k\geq 1\) we have \(a_{k}=\frac{a_{k3}+1}{k(k1)}\), so \(y(x)=a_{0}+a_{1}x+\frac{1}{2}x^{2}+\frac{a_{0}+1}{6}x^{3}+\frac{a_{1}+1}{12}x^{4}+\frac{3}{40}x^{5}+\frac{a_{0}+2}{30}x^{6}+\frac{a_{1}+2}{42}x^{7}+\frac{5}{112}x^{8}+\frac{a_{0}+3}{72}x^{9}+\frac{a_{1}+3}{90}x^{10}+\cdots\)
 \(y(x)=\frac{1}{2}x^{2}+\frac{1}{6}x^{3}+\frac{1}{12}x^{4}+\frac{3}{40}x^{5}+\frac{1}{15}x^{6}+\frac{1}{21}x^{7}+\frac{5}{112}x^{8}+\frac{1}{24}x^{9}+\frac{1}{30}x^{10}+\cdots\)
Exercise \(\PageIndex{7.2.11}\)
Attempt to solve \(x^2 y''  y = 0\) at \(x_0 = 0\) using the power series method of this section (\(x_0\) is a singular point). Can you find at least one solution? Can you find more than one solution?
 Answer

Applying the method of this section directly we obtain \(a_{k}=0\) for all \(k\) and so \(y(x)=0\) is the only solution we find.