2.4.3.1 Solution using Frobenius series method

Entering first order ode series solver frobenius solver

\[\begin {aligned} y^{\prime } x +y&=0\\ y \left (0\right ) &= 1\\ \end {aligned}\]

Series expansion around \(x=0\). Let the homogeneous solution be represented as Frobenius power series of the form

\begin{align*} y &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \end{align*}

Then

\begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \end{align*}

Substituting the above back into the ode gives

\begin{align*} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x +\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0\tag {1} \end{align*}

Which simplifies to

\begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} \left (n +r \right ) a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation}

The next step is to make all powers of \(x\) be \(n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r}\) and adjusting the power and the corresponding index gives Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\).

\begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} \left (n +r \right ) a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation}

The indicial equation is obtained from \(n=0\). From Eq (2) this gives

\begin{align*} x^{n +r} \left (n +r \right ) a_{n}+a_{n} x^{n +r} = 0 \end{align*}

When \(n=0\) the above becomes

\begin{align*} x^{r} r a_{0}+a_{0} x^{r} = 0 \end{align*}

Since \(a_{0}\neq 0\) then the indicial equation becomes

\begin{align*} \left (r +1\right ) x^{r} = 0 \end{align*}

Since the above is true for all \(x\) then the indicial equation simplifies to

\begin{align*} r +1 = 0 \end{align*}

Solving for \(r\) gives the root of the indicial equation as

\[ r=-1 \]

Replacing \(r=-1\) found above results in

\begin{align*} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} \left (n -1\right ) a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n -1}\right ) = 0 \end{align*}

From the above we see that there is no recurrence relation since there is only one summation term. Therefore all \(a_{n}\) terms are zero except for \(a_{0}\). Hence

\begin{align*} y_h &= a_{0} \left (\frac {1}{x}+O\left (x^{6}\right )\right ) \end{align*}

The solution is

\begin{gather*} y = c_1 \left (\frac {1}{x}+O\left (x^{6}\right )\right ) \end{gather*}

Applying initial conditions shows that no solution is possible, (Steps will be added soon)