Question

In Problems 9-14 use an appropriate infinite series method about \(x=0\) to find two solutions of the given differential equation.

\({{y}^{''}}-x{{y}^{'}}-y=0\)

Short answer

Therefore the solution is

\(\begin{align} & {{y}_{1}}(x)=x+\frac{1}{3}{{x}^{3}}+\frac{1}{15}{{x}^{5}}+\frac{1}{105}{{x}^{7}}+.. \\ & {{y}_{2}}(x)=1+\frac{1}{2}{{x}^{2}}+\frac{1}{8}{{x}^{4}}+\frac{1}{48}{{x}^{6}}+.. \\ \end{align}\)

Step-by-step solution

Given Information

The given value is

\({{y}^{''}}-x{{y}^{'}}-y=0\)

The given value is\({{y}^{''}}-x{{y}^{'}}-y=0\)

Differentiate \(y(x)=\sum\limits_{n=0}^{\infty }{{{c}_{n}}}{{x}^{n}}\) to get \({{y}^{\prime }}(x)\) and \({{y}^{\prime \prime }}(x).\)

We have

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

\({{y}^{\prime }}(x)=\sum\limits_{n=1}^{\infty }{{{c}_{n}}}n{{x}^{n-1}}\)

\({{y}^{\prime \prime }}(x)=\sum\limits_{n=2}^{\infty }{{{c}_{n}}}n(n-1){{x}^{n-2}}\)

Given the problem with the original value

\({{y}^{\prime \prime }}-x{{y}^{\prime }}-y=0\)

\(\sum\limits_{n=2}^{\infty }{{{c}_{n}}}n(n-1){{x}^{n-2}}-x\sum\limits_{n=1}^{\infty }{{{c}_{n}}}n{{x}^{n-1}}-\sum\limits_{n=0}^{\infty }{{{c}_{n}}}{{x}^{n}}=0\)

\(\sum\limits_{n=2}^{\infty }{{{c}_{n}}}n(n-1){{x}^{n-2}}-\sum\limits_{n=1}^{\infty }{{{c}_{n}}}n{{x}^{n}}-\sum\limits_{n=0}^{\infty }{{{c}_{n}}}{{x}^{n}}=0\)

Now for the first series we substitute \(k=\left( n-2 \right),\) for second series \(k=n\)and for third \(k=n.\)

\(\sum\limits_{k=0}^{\infty }{{{c}_{k+2}}}(k+2)(k+1){{x}^{k}}-\sum\limits_{k=1}^{\infty }{{{c}_{k}}}k{{x}^{k}}-\sum\limits_{k=0}^{\infty }{{{c}_{k}}}{{x}^{k}}=0\)

\(2{{c}_{2}}-{{c}_{0}}+\sum\limits_{k=1}^{\infty }{\left( (k+2)(k+1){{c}_{k+2}}-{{c}_{k}}k-{{c}_{k}} \right)}{{x}^{k}}=0\)

\(2{{c}_{2}}-{{c}_{0}}+\sum\limits_{k=1}^{\infty }{\left( (k+2)(k+1){{c}_{k+2}}-(k+1){{c}_{k}} \right)}{{x}^{k}}=0\)

Using identity property

\(2{{c}_{2}}-{{c}_{0}}=0\to {{c}_{2}}=\frac{{{c}_{0}}}{2}\)

\(\begin{align} & (k+2)(k+1){{c}_{k+2}}-(k+1){{c}_{k}}=0 \\ & {{c}_{k+2}}=\frac{1}{k+2}{{c}_{k}} \\ \end{align}\)

For \({{\mathbf{c}}_{\mathbf{0}}}=\mathbf{0}\)and \({{\mathbf{c}}_{\mathbf{1}}}=\mathbf{1}\) we have:

\({{c}_{2}}=0\)

\({{c}_{3}}=\frac{1}{3}\)

\({{c}_{4}}=0\)

\({{c}_{5}}=\frac{1}{5}{{c}_{3}}=\frac{1}{15}\)

\({{c}_{6}}=0\)

\({{c}_{7}}=\frac{1}{7}{{c}_{5}}=\frac{1}{105}\)

\({{y}_{1}}(x)=\mathbf{x}+\frac{1}{3}{{\mathbf{x}}^{3}}+\frac{1}{15}{{\mathbf{x}}^{5}}+\frac{1}{105}{{\mathbf{x}}^{7}}+\ldots \)

For \({{c}_{0}}=\mathbf{1}\)and \({{c}_{1}}=\mathbf{0}\)we have:

\({{c}_{2}}=\frac{1}{2}\)

\({{c}_{3}}=0\)

\({{c}_{4}}=\frac{1}{4}{{c}_{2}}=\frac{1}{8}\)

\({{c}_{5}}=0\)

\({{c}_{6}}=\frac{1}{6}{{c}_{4}}=\frac{1}{48}\)

\({{y}_{2}}(x)=1+\frac{1}{2}{{x}^{2}}+\frac{1}{8}{{x}^{4}}+\frac{1}{48}{{x}^{6}}+\ldots \)