Question

In Problems 9-14 use an appropriate infinite series method aboutto find two solutions of the given differential equation.

\[ 9. 2x{y''} + {y'} + y = 0\]

Short answer

\({y_1}(x) = {C_1}\left( {1 - x + \frac{1}{6}{x^2} - \frac{1}{{90}}{x^3} + ...} \right)\;\;\;{y_2}(x) = {C_2}{x^{\frac{1}{2}}}\left( {1 - \frac{1}{3}x + \frac{1}{{30}}{x^2} - \frac{1}{{630}}{x^3} + ...} \right)\)

Step-by-step solution

To Find a linear second-order differential equation

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

Differentiating\((*)\), we get

\({y'}\)\( = \sum\limits_{n = 0}^\infty {(n + r)} {c_n}{x^{n + r - 1}}\)

\( = \sum\limits_{n = 0}^\infty {(n + r)} (n + r - 1){c_n}{x^{n + r - 2}}\)

Replacing the equation (1) and equation (2) in the initial differential equation, we get

\(0 = 2x\sum\limits_{n = 0}^\infty {(n + r)} (n + r - 1){c_n}{x^{n + r - 2}} + \sum\limits_{n = 0}^\infty {(n + r)} {c_n}{x^{n + r - 1}} + \sum\limits_{n = 0}^\infty {{c_n}} {x^{n + r}}\)

\( = 2\sum\limits_{n = 0}^\infty {(n + r)} (n + r - 1){c_n}{x^{n + r - 1}} + \sum\limits_{n = 0}^\infty {(n + r)} {c_n}{x^{n + r - 1}} + \sum\limits_{n = 0}^\infty {{c_n}} {x^{n + r}}\)

\( = {x^r}\left( {r(2r - 1){c_0}{x^{ - 1}} + 2\sum\limits_{n = 1}^\infty {(n + r)} (n + r - 1){c_n}{x^{n - 1}} + \sum\limits_{n = 1}^\infty {(n + r)} {c_n}{x^{n - 1}} + \sum\limits_{n = 0}^\infty {{c_n}} {x^n}} \right)\)

For the first and the second summation, we let\(k = n - 1\). For the third summation, we let\(k = n\). Hence

\( = {x^r}\left( {r(2r - 1){c_0}{x^{ - 1}} + 2\sum\limits_{k = 0}^\infty {(k + 1 + r)} (k + r){c_{k + 1}}{x^k} + \sum\limits_{k = 0}^\infty {(k + 1 + r)} {c_{k + 1}}{x^k} + \sum\limits_{k = 0}^\infty {{c_k}} {x^k}} \right)\)

\( = {x^r}\left( {r(2r - 1){c_0}{x^{ - 1}} + 2\sum\limits_{k = 0}^\infty {\left( {(k + 1 + r)(2k + 2r + 1){c_{k + 1}} + {c_k}} \right)} {x^k}} \right)\)

From this, we can see that

\(\begin{aligned}{{}{r}}{r(2r - 1) = 0}\\{{r_1} = 0'U {r_2} = \frac{1}{2}}\end{aligned}\)

and

\({c_{k + 1}} = \frac{{{c_k}}}{{(k + 1 + r)(2k + 2r + 1)}}\)

- For\({r_1} = 0\), we have

\({c_{k + 1}} = - \frac{{{c_k}}}{{(k + 1)(2k + 1)}}\;\;\;k = 0,1,2 \ldots \)

\(k = 0 \Rightarrow {c_1} = - \frac{{{c_0}}}{1} = - {c_0}\)

\(k = 1 \Rightarrow {c_2} = - \frac{{{c_1}}}{6} = \frac{{{c_0}}}{6}\)

\(k = 2 \Rightarrow {c_3} = - \frac{{{c_2}}}{{15}} = - \frac{{{c_0}}}{{90}}\)

For\({r_2} = \frac{1}{2}\), we have

\({c_{k + 1}} = - \frac{{{c_k}}}{{(2k + 3)(k + 1)}}\;\;\;k = 0,1,2 \ldots \)

\(k = 0 \Rightarrow {c_1} = - \frac{{{c_0}}}{3}\)

\(k = 1 \Rightarrow {c_2} = \frac{{{c_0}}}{{30}}\)

\(k = 2 \Rightarrow {c_3} = - \frac{{{c_0}}}{{630}}\)

Final Answer

For the indicial root\({r_1} = 0\), we get the solution

\({y_1}(x)\)\( = {x^0}\left( {{c_0} - {c_0}x + \frac{1}{6}{c_0}{x^2} - \frac{1}{{90}}{c_0}{x^3} + ...} \right)\)

\( = {C_1}\left( {1 - x + \frac{1}{6}{x^2} - \frac{1}{{90}}{x^3} + ...} \right)\)

For the indicial root\({r_2} = \frac{1}{2}\), we get the solution

\({y_2}(x) \)\( = {x^{\frac{1}{2}}}\left( {{c_0} - \frac{1}{3}{c_0}x + \frac{1}{{30}}{c_0}{x^2} - \frac{1}{{630}}{c_0}{x^3} + ...} \right)\)

\( = {C_2}{x^{\frac{1}{2}}}\left( {1 - \frac{1}{3}x + \frac{1}{{30}}{x^2} - \frac{1}{{630}}{x^3} + ...} \right)\)

\({y_1}(x) = {C_1}\left( {1 - x + \frac{1}{6}{x^2} - \frac{1}{{90}}{x^3} + ...} \right)\;\;\;{y_2}(x) = {C_2}{x^{\frac{1}{2}}}\left( {1 - \frac{1}{3}x + \frac{1}{{30}}{x^2} - \frac{1}{{630}}{x^3} + ...} \right)\)