For Problems 1 to 4, find one (simple) solution of each differential equation by series, and then find the second solution by the "reduction of order" method, Chapter 8, Section .

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

Reduction of order is a technique in mathematics for solving second-order linear ordinary differential equations. It is employed when one solutionY₁(x) is known and the second linearly independent solution y₂(x) s desired. The method also applies to n-th order equations. In this case (n-1)th the ansatz will yield an order equation for v

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

Consider the differential equation is $y=(x-1)y^{\text{'}\text{'}}-x{y}^{\text{'}}+y=0$

Its need find one (simple) solution of the differential equation by series, and then find the Second solution by the "reduction of order" method.

$\begin{array}{l}lety=\sum _{n=0}^{\infty }{c}_n(x-1{)}^{n}bethesolutionofthegivenequation\\ asy=\sum _{n=0}^{\infty }{c}_n(x-1{)}^n,obtain:\\ y^{\text{'}}=\sum _{n=0}^{\infty }n{c}_n(x-1{)}^{n-1}and{y}^{\text{'}}=\sum _{n=0}^{\infty }n{c}_n(x-1{)}^{n-1}\end{array}$

Then, calculate:

$\begin{array}{c}=(x-1)y^{\text{'}\text{'}}-x{y}^{\text{'}}+y\\ =(x-1)\sum _{n=0}^{\infty }n(n-1)c_n(x-1{)}^{n-2}-x\sum _{n=0}^{\infty }n{c}_n(x-1{)}^{n-1}+\sum _{n=0}^{\infty }{c}_n(x-1{)}^n\\ =(x-1)\sum _{n=0}^{\infty }n(n-1)c_n(x-1{)}^{n-2}-x\sum _{n=0}^{\infty }n{c}_n(x-1{)}^{n-1}+\sum _{n=0}^{\infty }n{c}_n(x-1{)}^{n-1}+\sum _{n=0}^{\infty }{c}_n(x-1{)}^n\\ =\sum _{n=0}^{\infty }n(n-1)c_n(x-1{)}^{n-1}-(x-1)\sum _{n=0}^{\infty }n{c}_n(x-1{)}^{n-1}-\sum _{n=0}^{\infty }n{c}_n(x-1{)}^{n-1}+\sum _{n=0}^{\infty }{c}_n(x-1{)}^n\end{array}$

Simplify further as follows:

$\begin{array}{c}(x-1)y^{\text{'}\text{'}}-x{y}^{\text{'}}+y\\ =\sum _{n=0}^{\infty }n(n-1)c_n(x-1{)}^{n-1}-\sum _{n=0}^{\infty }n{c}_n(x-1{)}^n-\sum _{n=0}^{\infty }n{c}_n(x-1{)}^{n-1}n+\sum _{n=0}^{\infty }{c}_n(x-1{)}^{n}n\\ =\underset{k=n-1}{\underbrace{\sum _{n=0}^{\infty }\left(n^2-2n\right)c_n{(x-1)}^{n-1}}}-\underset{k=n}{\underbrace{\sum _{n=0}^{\infty }(n-1)c_n{(x-1)}^n}}n\\ =\underset{}{\underbrace{\sum _{k=0}^{\infty }\left({(k+1)}^2-2(k+1)\right)}}{c}_{k+1}(x-1{)}^k-\sum _{k=0}^{\infty }(k-1)c_k(x-1{)}^k\\ then,=\sum _{k=0}^{\infty }\left[\left(k^2-1\right)c_{k+1}+(k-1)c_k\right](x-1{)}^k\end{array}$

This implies that:

$\begin{array}{c}\left(k^2-1\right)c_{k+1}+(k-1)c_k=0\\ k=0,1,2,\dots \\ \Rightarrow c_{k+1}=\frac{1}{k+1}{c}_k\\ k=0,1,2,\dots \end{array}$

From (2) obtain:

$\begin{array}{c}{c}_1=\frac{1}{0+1}{c}_0=c_0\\ c_2=\frac{1}{1+1}{c}_1=\frac{1}{2}{c}_0\\ c_3=\frac{1}{2+1}{c}_2=\frac{1}{3}·\frac{1}{2}{c}_0\\ =\frac{1}{3!}{c}_0+c_n=\frac{1}{n!}{c}_0\end{array}$

Then the series solution becomes:

$\begin{array}{c}y=\sum _{n=0}^m{c}_n(x-1{)}^n\\ =c_0+c_1(x-1)+c_2(x-1{)}^2+c_3(x-1{)}^3+\dots +c_n(x-1{)}^n+\dots \\ =c_0+\frac{1}{1!}{c}_0(x-1)+\frac{1}{2!}{c}_0(x-1{)}^2+\frac{1}{3!}{c}_0(x-1{)}^3+\dots +\frac{1}{n!}{c}_0(x-1{)}^n+\dots \\ =c_0\left[1+\frac{1}{1!}(x-1)+\frac{1}{2!}{(x-1)}^2+\frac{1}{3!}{(x-1)}^3+\dots +\frac{1}{n!}{(x-1)}^n+\dots \right]\end{array}$

Therefore, $=co{e}^{x-1}$.

Thus, $y=e^{x-1}$ is one solution of the differential equation (1).

Reduction of order:

To find the second solution of, $y^{\text{'}\text{'}}+f\left(x\right)y^{\text{'}}+g\left(x\right)y=0$

Given one solution u(x), substitute$\mathrm{y}=\mathrm{u}\left(\mathrm{x}\right)\mathrm{v}\left(\mathrm{x}\right)$

Put into Equation (3) and solve for v.

On rewriting the differential equation (1), obtain:

$\begin{array}{c}{y}^{\text{'}\text{'}}+\left(\frac{-x}{x-1}\right)y^{\text{'}}+\frac{1}{x-1}y=0\\ asy=e^{x-1}v\left(x\right)obtain:\\ y^{\text{'}}=e^{x-1}{v}^{\text{'}}+v{e}^{x-1}\\ y^{\text{'}\text{'}}=e^{x-1}\left(v^{\text{'}\text{'}}+2v^{\text{'}}+v\right)\end{array}$

From equation (4), obtain:

$\begin{array}{c}{e}^{x-1}\left(v^{\text{'}\text{'}}+2v^{\text{'}}+v\right)+\left(\frac{-x}{x-1}\right)e^{x-1}\left(v^{\text{'}}+v\right)+\frac{1}{x-1}{e}^{x-1}v=0\\ e^{x-1}(x-1)\left(v^{\text{'}\text{'}}+2v^{\text{'}}+v\right)-x{e}^{x-1}\left(v^{\text{'}}+v\right)+e^{x-1}v=0\\ e^{x-1}\left[(x-1)\left(v^{\text{'}\text{'}}+2v^{\text{'}}+v\right)-x\left(v^{\text{'}}+v\right)+v\right]=0\\ (x-1)\left(v^{\text{'}\text{'}}+2v^{\text{'}}+v\right)v-x\left(v^{\text{'}}+v\right)+v=0\end{array}$

Simplify further as follows:

$\begin{array}{c}x{v}^{\text{'}\text{'}}+(x-2)v^{\text{'}}=0a\\ \frac{d{v}^{\text{'}}}{v^{\text{'}}}=-\frac{x-2}{x}\\ \ln v^{\text{'}}=-x+2\ln x+\ln k\\ v^{\text{'}}=e^{-x}{x}^{2}C\end{array}$

Then, calculate:

$\begin{array}{c}=C\left[x^2\left(-e^{-x}\right)-\int 2x\left(-e^{-x}\right)dx\right]\\ =C\left[-x^2{e}^{-x}-\int 2x\left(-e^{-x}\right)dx\right]\\ =C\left[-x^2{e}^{-x}+2\left[x\left(-e^{-x}\right)-\int -e^{-x}dx\right]\right]\\ =C\left(-x^2{e}^{-x}-2x{e}^{-x}-2e^{-x}\right)\\ then,=C{e}^{-x}\left(-x^2-2x-2\right)\\ Hence,thesecondsolutionis{y}_2\left(x\right)=C{e}^{-x}\left(-x^2-2x-2\right)\end{array}$