Question

\(x = 0\) is an ordinary point of a certain linear differential equation. After the assumed solution \(y = \sum\limits_{n = 0}^\infty {{c_n}} {x^n}\) is substituted into the DE, the following algebraic system is obtained by equating the coefficients of \({x^0},{x^1},{x^2}\), and \({x^3}\) to zero:

\(\begin{aligned}{{}{r}}{2{c_2} + 2{c_1} + {c_0} = 0}\\{6{c_3} + 4{c_2} + {c_1} = 0}\\{12{c_4} + 6{c_3} + {c_2} - \frac{1}{3}{c_1} = 0}\\{20{c_5} + 8{c_4} + {c_3} - \frac{2}{3}{c_2} = 0}\end{aligned}\)

Bearing in mind that \({c_0}\) and \({c_1}\) are arbitrary, write down the first five terms of two power series solutions of the differential equation.

Short answer

The two particular solutions are

\({y_1}(x) = \;\;\;1 - \frac{1}{2}{x^2} + \frac{1}{3}{x^3} - \frac{1}{8}{x^4} + \frac{1}{{60}}{x^5} + ...\)

\({y_2}(x) = x - {x^2} + \frac{1}{2}{x^3} - \frac{5}{{36}}{x^4} - \frac{1}{{360}}{x^5} + ...\)

Step-by-step solution

To Find the differential equation

The assumed solution of the unknown differential equation have a power series from, which is

\(y = \sum\limits_{n = 0}^\infty {{c_n}} {x^n}\)

By substituting\(y(x)\)and its higher derivatives and equating the coefficients of\({x^0},{x^1},{x^2}\)and\({x^3}\), yields

\(2{c_2} + 2{c_1} + {c_0} = 0\quad \Rightarrow \quad {c_2} = \quad - {c_1} - \frac{1}{2}{c_0}\)

\(6{c_3} + 4{c_2} + {c_1} = 0\quad \Rightarrow \quad {c_3} = \quad - \frac{2}{3}{c_2} - \frac{1}{6}{c_1}\)

\( = \quad - \frac{2}{3}\left( { - {c_1} - \frac{1}{2}{c_0}} \right) - \frac{1}{6}{c_1}\)

\( = \quad \frac{1}{2}{c_1} + \frac{1}{3}{c_0}\)

\(12{c_4} + 6{c_3} + {c_2} - \frac{1}{3}{c_1} = 0\quad \Rightarrow \quad {c_4} = \quad - \frac{1}{2}{c_3} - \frac{1}{{12}}{c_2} + \frac{1}{{36}}{c_1}\)

\( = \quad - \frac{1}{2}\left( {\frac{1}{2}{c_1} + \frac{1}{3}{c_0}} \right) - \frac{1}{{12}}\left( { - {c_1} - \frac{1}{2}{c_0}} \right) + \frac{1}{{36}}{c_1}\)

\( = \quad - \frac{5}{{36}}{c_1} - \frac{1}{8}{c_0}\)

\(20{c_5} + 8{c_4} + {c_3} - \frac{2}{3}{c_2} = 0\quad \to \quad {c_5} = \quad - \frac{2}{5}{c_4} - \frac{1}{{20}}{c_3} + \frac{1}{{30}}{c_2}\)

\( = - \frac{2}{5}\left( { - \frac{5}{{36}}{c_1} - \frac{1}{8}{c_0}} \right) - \frac{1}{{20}}\left( {\frac{1}{2}{c_1} + \frac{1}{3}{c_0}} \right) + \frac{1}{{30}}\left( { - {c_1} - \frac{1}{2}{c_0}} \right)\)

\( = \quad - \frac{1}{{360}}{c_1} + \frac{1}{{60}}{c_0}\)

To expand the power series

Now, we expand the power series, yields the general solution to have the form

\(y(x) = \)\({c_0} + {c_1}x + {c_2}{x^2} + {c_3}{x^3} + {c_4}{x^4} + {c_5}{x^5} + ...\)

\( = {c_0} + {c_1}x + \left( { - {c_1} - \frac{1}{2}{c_0}} \right){x^2} + \left( {\frac{1}{2}{c_1} + \frac{1}{3}{c_0}} \right){x^3} + \left( { - \frac{5}{{36}}{c_1} - \frac{1}{8}{c_0}} \right){x^4} + \left( { - \frac{1}{{360}}{c_1} + \frac{1}{{60}}{c_0}} \right){x^5} + ....\)\( = {c_0}\underbrace {\left( {1 - \frac{1}{2}{x^2} + \frac{1}{3}{x^3} - \frac{1}{8}{x^4} + \frac{1}{{60}}{x^5} + ...} \right)}_{{y_1}(x)} + \underbrace {{c_1}}_{{y_2}(x)}\underbrace {\left( {x - {x^2} + \frac{1}{2}{x^3} - \frac{5}{{36}}{x^4} - \frac{1}{{360}}{x^5} + ...} \right)}_{}\)

where \({c_0}\) and \({c_1}\)are arbitrary constants. \({y_1}(x){y_2}(x)\)are the two particular solutions.

Final Answer

The two particular solutions are

\({y_1}(x) = \;\;\;1 - \frac{1}{2}{x^2} + \frac{1}{3}{x^3} - \frac{1}{8}{x^4} + \frac{1}{{60}}{x^5} + ...\)

\({y_2}(x) = x - {x^2} + \frac{1}{2}{x^3} - \frac{5}{{36}}{x^4} - \frac{1}{{360}}{x^5} + ...\)