Question

x= 0 is a regular singular point of the given differential equation. Show that the indicial roots of the singularity differ by an integer. Use the method of Frobenius to obtain at least one series solution about. Use (23) where necessary and a CAS, if instructed, to find a second solution. Form the general solution on$\left(0,\infty \right)$

Question: $x{y}^{\text{'}\text{'}}+(1-x)y^{\text{'}}-y=0$

Short answer

$y=C_1{e}^x+C_2{e}^x\ln x+\sum _{n=1}^{\infty }\frac{{(-1)}^n{x}^n}{n·n!}$

Step-by-step solution

To find the differential equation

We state, without a proof, that if we have a differential equation, with the following standard form;

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

which has a regular singular point at x =x₀ then we can use Forbenious Method to solve this differential equation.

We have

$x{y}^{\text{'}\text{'}}+(1-x)y^{\text{'}}-y=0$

which has a singular point at x=0 We assume the solution, according to Forbenious, to be as follows;

$y{=}_{n=0}^{\infty }{c}_n{x}^{n+r}\dots \left(1\right)$

We find the first and second derivative.

$$\begin{gathered} y^{\text{'}}=c_n(n+r)x^{n+r-1}\dots \left(2\right) \\ {y}^{\text{'}\text{'}}=c_{n=0}(n+r)(n+r-1)x^{n+r-2}\dots \left(3\right) \end{gathered}$$

We substitute (1), (2) and (3) into the given differential equation, yields

$$\begin{gathered} f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right){=}_{n=0}{c}_n(n+r)(n+r-1)x^{n+r-2}+(1-x{)}_{n=0}^{\infty }{c}_n(n+r)x^{n+r-1}-c_n^{\infty }{x}^{n+r} \\ =\underset{a_0\to x^{r-1}}{\underbrace{n_{n=0}{c}_n(n+r)(n+r-1)x^{n+r-1}}}+\underset{a_0\to x^{r-1}}{\underbrace{n_n(n+r)x^{n+r-1}}}-\underset{a_0\to x^r}{\underbrace{c_n{c}_n(n+r)x^{n+r}-c_{n=0}^{n=0}{c}_n^{n+r}}}\underset{a_0\to x^r}{\underbrace{n=0}} \end{gathered}$$

Our aim, now, is to make the summation index and the power of $x$, for the four series, in the same phase. We, first, have to assure that the first terms, for the four series, are raised to the same power, which is not the case. Therefore, we do the following

$$\begin{gathered} f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right)=c_{0}r(r-1)x^{r-1}+\infty c_n(n+r)(n+r-1)x^{n+r-1}+c_{0}r{x}^{r-1}+\infty \\ f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right)=c_{0}r(r-1)x^{r-1}+\underset{a_1\to x^r}{\underbrace{\underset{n=1}{\underbrace{}}{c}_n(n+r)(n+r-1)x^{n+r-1}}}+c_{a_0}r{x}^{r-1}+\underset{x^r}{\underbrace{}}{c}_n(n+r)x^{n+r-1} \\ \underset{a_0\to x^r}{\underbrace{n_{n=0}^{\infty }{c}_n(n+r)x^{n+r}}}-\underset{a_0\to x^r}{\underbrace{c_n{x}^{n+r}}} \end{gathered}$$

To find the third and fourth series

Now, we do the shift for the third and fourth series from n to n-1 yields

$$\begin{gathered} f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right)=c_{0}r(r-1)x^{r-1}+c_{n=1}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r-1}+c_{0}r{x}^{r-1}+c_{n=1}^{\infty }(n+r)x^{n+r-1} \\ {-}_{n-1=0}^{\infty }{c}_{n-1}\left(\right(n-1)+r)x^{(n-1)+r}{-}_{n-1=0}^{\infty }{c}_{n-1}{x}^{(n-1)+r} \\ =c_{0}r(r-1)x^{r-1}+c_{n=1}^{\infty }{c}_n(n+r)(n+r-1)x^{n+r-1}+c_{0}r{x}^{r-1}{+}_{n=1}^{\infty }{c}_n(n+r)x^{n+r-1} \\ -c_{n=1}^{\infty }{c}_{n-1}(n+r-1)x^{n+r-1}-c_{n=1}^{\infty }{c}_{n-1}^{n+r-1} \end{gathered}$$

Now, we sum the coefficients of the terms, which are raised to the same power together, yields

$$\begin{gathered} f\left(y^{\text{'}\text{'}},y^{\text{'}},y\right)=r(r-1)+r{c}_0{x}^{r-1}{+}_{n=1}^{\infty }{c}_n(n+r)(n+r-1)+c_n(n+r)-c_{n-1}(n+r-1)-c_{n-1}{x}^{n+r-1} \\ =r^2{c}_0{x}^{r-1}{+}_{n=1}^{\infty }{c}_n(n+r)(n+r-1+11)-c_{n-1}(n+r-1+1)x^{n+r-1} \\ =r^2{c}_0{x}^{r-1}+c_{n=1}{c}_n(n+r{)}^2-c_{n-1}(n+r)x^{n+r-1} \end{gathered}$$

which implies that

r²= 0

r₁ = r₂ = 0

$c_n(n+r{)}^2-c_{n-1}(n+r)=0\Rightarrow c_n=\frac{c_{n-1}(n+r)}{{(n+r)}^2}=\frac{c_{n-1}}{(n+r)}[where\text{n}=1,2,3,\dots ]$

Note that, you can't set to be zero as all the coefficients of the series are involved with . And the two roots of singularity, and differ by an integer.

For r=0 we have

$y=\sum _{n=0}^{\infty }{c}_n{x}^n$

and the recurrence relation is

$c_n=\frac{c_{n-1}}{n}$

Now, we find the coefficients of the series, using the recurrence relation.

$$\begin{gathered} n=1,c_1=c_0 \\ n=2,c_2=\frac{c_1}{2} \\ =\frac{c_0}{2!} \\ =\frac{c_2}{3} \\ =\frac{\left(c_0/2\right)}{3} \\ =\frac{c_0}{3!} \end{gathered}$$

$$\begin{gathered} n=4,c_4=\frac{c_3}{4} \\ =\frac{c_0}{4!} \\ {c}_n=\frac{c_0}{n!} \end{gathered}$$

Substitute the second solution

Therefore, the solution for r = 0 becomes as follows

$y_1=c_0\sum _{n=0}^{\infty }\frac{x^n}{n!}=c_0{e}^x\dots \left(4\right)$

Therefore, we will not be able to obtain a second solution, using Forbinous Method. Recall (23) from the book, we can obtain a second solution as follows

$y_2\left(x\right)=y_1\left(x\right)\int \frac{e^{-\int P\left(x\right)dx}}{y_1^2\left(x\right)}dx\dots \left(5\right)$

For the given differential equation, we have

$\frac{x}{x}{y}^{\text{'}\text{'}}+\frac{(1-x)}{x}{y}^{\text{'}}-\frac{1}{x}y=0$

$$\begin{gathered} y^{\text{'}\text{'}}+\left(\frac{1}{x}-1\right)y^{\text{'}}-\frac{1}{x}y=0 \\ P\left(x\right)=\frac{1}{x}-1\dots \left(6\right)\text{and}Q\left(x\right)=-2n \end{gathered}$$

We substitute (6) and (4) into (5), after omitting $c_{0}$.

$$\begin{gathered} y_2\left(x\right)=e^x\int \frac{e^{-\int \left(\frac{1}{x}-1\right)dx}}{e^{2x}}dx \\ =e^x\int \frac{e^{\int \left(-\frac{1}{x}+1\right)dx}}{e^{2x}}dx \\ =e^x\int \frac{e^{-\ln x+x}}{e^{2x}}dx \\ =e^x\int \frac{e^{\ln x^{-1}}·e^x}{e^{2x}}dx\left[Recally\ln x=\ln x^y\right] \\ =e^x\int e^{\ln x^{-1}{e}^{-x}}dx \end{gathered}$$

Let

$$\begin{gathered} u=e^{\ln x^{-1}}\Rightarrow \ln u=\ln e^{\ln x^{-1}} \\ =\ln x^{-1}\ln e^{-1} \\ =\ln x^{-1}[\text{Recall}\ln e=1] \\ u=x^{-1}n \end{gathered}$$

Final proof

Therefore,

$$\begin{gathered} y_2\left(x\right)=e^x\int x^{-1}\left[1-x+\frac{1}{2!}{x}^2-\frac{1}{3!}{x}^3+\dots \right]dx\left[\text{Using the Maclaurin series for}{e}^{-x}\right] \\ =e^x\int \left[\frac{1}{x}-1+\frac{1}{2!}x-\frac{1}{3!}{x}^2+\dots \right]dx \\ =e^x\left[\ln x-x+\frac{1}{2·2!}{x}^2-\frac{1}{3·3!}{x}^3+\dots \right] \\ =e^x\left[\ln x+\sum _{n=1}^{\infty }\frac{{(-1)}^n{x}^n}{n·n!}\right] \end{gathered}$$

We can obtain a general solution by using Superposition Principle as follows

$y=C_1{e}^x+C_2{e}^x\left[\ln x+\sum _{n=1}^{\infty }\frac{{(-1)}^n{x}^n}{n·n!}\right]$