Verify that the differential equation in Problem$\mathbf{11}.\mathbf{13}$is not Fuchsian. Solve it by separation of variables to find the obvious solution$y=$const. and a second solution in the form of an integral. Show that the second solution is not expandable in a Frobenius series.

Fuchs's theorem:

If the differential equation of the form

$y^{\text{'}\text{'}}+f\left(x\right)y^{\text{'}}+g\left(x\right)y=0$

Satisfies the conditions of the Fuchs's theorem then it will be two Frobenius series or one solution$S_1\left(x\right)$which is a Frobenius series, and a secondsolution which islocalid="1654850496708" $S_1\left(x\right)$In $x+S_2\left(x\right)$Wherelocalid="1656757763274" $S_2\left(x\right)$is another Frobenius series.

The differential equation, $y^{\text{'}\text{'}}+y^{\text{'}}/x^2=0$.

Solve the equation by separation of variables as follows:

$\begin{array}{c}{y}^{\text{'}\text{'}}+y^{\text{'}}/x^2=0\\ y^{\text{'}\text{'}}=-y^{\text{'}}/x^2\\ \frac{d}{dx}\left(y^{\text{'}}\right)=-y^{\text{'}}/x^2\\ \frac{d}{y^{\text{'}}}\left(y^{\text{'}}\right)=-dx/x^2\end{array}$

Simplify further as follows:

$\begin{array}{c}\ln y^{\text{'}}=\frac{1}{x}+k\\ y^{\text{'}}=c_1{e}^{1/x}\\ \frac{dy}{dx}=c_1{e}^{1/x}\\ dy=c_1{e}^{1/x}dx\end{array}$

So,

$\begin{array}{c}dy=c_1{e}^{1/x}dx\\ y=c_1\int e^{1/x}dx+c_2\end{array}$

Therefore, the two solutions are $y_1=const.$and $y_2=\int e^{1/x}dx$

Suppose the solution $y_2=\int e^{1/x}dx$is expandable in Frobenius series.

Let,$y_2=\sum _{n=0}{a}_n{x}^{n+r}$.

The recurrence relation for this will be $a_{n+1}=-\frac{n(n-1)}{n+1}{a}_n$.

Now, check if the series converges or not by ratio test.

$\begin{array}{c}\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}\left|\frac{a_{n+1}{x}^{n+1}}{a_n{x}^n}\right|\mathbf{=}\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}\left|\frac{-\frac{m(n-1)}{n+1}{a}_n{x}^{n+1}}{a_n{x}^n}\right|\\ \mathbf{=}\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}\left|-\frac{n(n-1)}{n+1}x\right|\mathbf{=}\mathbf{\infty }\end{array}$

This means that the series diverges.

So, the solution $y_2$is not expandable in Frobenius series.

Therefore, the solutions do not satisfy the Fuchsian's conditions.

Hence, the differential equation is not Fuchsian.