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 7 (e).

${\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{(}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{)}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{.}$

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) is desired. The method also applies to n-th order equations. In this case (n-1)the ansatz will yield an order equation for v

Fuchs’s theorem:

If the differential equation is in the form

Satisfies the conditions of Fuchs's Theorem then it has${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{y}\mathbf{=}\mathbf{0}$

1. Two Frobenius series or
2. One solution S₁ x which is a Frobenius series, and a second solution which is
3. ${\mathit{s}}_{\mathbf{1}}\mathit{x}\mathit{I}\mathit{n}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{+}{\mathit{s}}_{\mathbf{2}}\mathit{x}\mathit{I}\mathit{n}\mathbf{\left(}\mathit{x}\mathbf{\right)}$where S₂(x) is another Frobenius series.

Given differential equation is. $x^2{y}^{\text{'}\text{'}}+(x+1)y^{\text{'}}-y=0$ …… (1)

Observe that,

x=0,-1 are the singular points of the given differential equations.

After dividing the equation byx² , obtain:

$y^{\text{'}\text{'}}+\left(\frac{x+1}{x^2}\right)y^{\text{'}}+\left(-\frac{1}{x^2}\right)y=0$ …… (2)

Conditions of Fuchs's theorem:

Consider the differential of the form:

$y^{\text{'}\text{'}}+f\left(x\right)y^{\text{'}}+g\left(x\right)y=0$ …… (3)

If xf(x) and x² g(x) are expandable in convergent power series $\sum _{n=0}^{\infty }{a}_n{x}^n,$ we say that

the differential equation (3) is a regular (or has a nonessential singularity) on the origin of these conditions is called Fuchs’s theorem.

By comparing the differential equation (2) with (3), to get $f\left(x\right)=\frac{x+1}{x^2}$ and $g\left(x\right)=-\frac{1}{x^2}$

Now, consider xf(x)

$\begin{array}{c}xf\left(x\right)=x·\frac{x+1}{x^2}\\ =\frac{x+1}{x}\end{array}$.

Which is analytic everywhere exceptx=0 and hence xf(x) does not have a convergent

Power series representation at x=0

Hence, the given differential equation does not satisfy the Conditions of Fuchs's theorem.

Suppose that, $y=\sum _{n=0}^m{c}_n{x}^{n+r}$be the solution of the given differential equation.

As,$y=\sum _{n=1}^{\infty }{c}_n{x}^{n+r}$ we have that:

$y^{\text{'}}=\sum _{n=0}^{\infty }(n+r)c_n{x}^{n+r-1}and{y}^{\text{'}\text{'}}=\sum _{n=0}^{\infty }(n+r)(n+r-1)c_n{x}^{n+r-2}$

Then calculate as follows:

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

Simplify further as follows:

$\begin{array}{rcl}{x}^2{y}^{\text{'}\text{'}}+(x+1)y^{\text{'}}-y& =& \sum _{n=0}^{\infty }(n+r)C_n{x}^{n+r-1}+\underset{n=0}{\overset{\infty }{\sum [}}(n+r)(n+r-1+1)-1]C_n{x}^{n+r}+\\ & =& \sum _{n=0}^{\infty }(n+r)C_n{x}^{n+r-1}+\underset{n=0}{\overset{\infty }{\sum [}}{(n+r)}^2-1]C_n{x}^{n+r}+\\ & =& x^r[\underset{}{\underset{}{r{c}_0+\sum _{n=1}^{\infty }(n+r)c_n{x}^{n+r-1}}}+\underset{}{\underset{}{\sum _{n=0}^{\infty }\left[{(n+r)}^2-1\right]c_n{x}^{n+r}}}]\\ & =& x^r[\underset{k=n-1}{\underbrace{r{c}_0+\sum _{n=1}^{\infty }(n+r)c_n{x}^{n-1}}}+\underset{k=n}{\underbrace{\sum _{n=0}^{\infty }\left[{(n+r)}^2-1\right]c_n{x}^n}}]\\ & & \end{array}$

Simplify further as follows:

$\begin{array}{c}{x}^2{y}^{\text{'}\text{'}}+(x+1)y^{\text{'}}-y=x^r\left[r{c}_0+\sum _{k=0}^{\infty }(k+r+1)c_{k+1}{x}^k+\sum _{k=0}^{\infty }\left[{(k+r)}^2-1\right]c_k{x}^k\right]\\ =x^r\left[r_{c0}+\sum _{k=0}^{\infty }\left[(k+r+1)c_{k+1}+\left[{(k+r)}^2-1\right]c_k\right]x^k\right]\\ Thisimpliesthat,r{c}_0=0\\ and(k+r+1)c_{k+1}+\left[{(k+r)}^2-1\right]c_k=0\\ k=0,1,2,3,\dots \dots .\\ Becausenothingisgainedbytaking\mathrm{C}0=0,wemustthenhave\mathrm{r}=0.\\ Theseindicialrootsare,\mathrm{r}=0\\ C_{k+1}=-\frac{\left[{(k+r)}^2-1\right]}{(k+r+1)}{C}_k\\ andk=0,1,2,3,\dots \dots .\\ For\mathrm{r}=0from\left(4\right),obtain:\\ C_{k+1}=-\frac{k^2-1}{(k+1)}{c}_k\\ (1-k)c_k,k=0,1,2,\\ Fromequation\left(5\right)obtain:\\ c_1=(1-0)c_0\\ =c_0\\ c_2=(1-1)c_1\\ =0\\ so,c_n=0\\ Thus,theindicialrootsr=0weobtainthesolution.\\ y=\sum _{n=0}^{\infty }{c}_n{x}^{n+0}\\ =x^0\sum _{n=0}^{\infty }{c}_n{x}^n\\ =x^0\left[c_0+c_{1}x+c_2{x}^2+\dots +c_n{x}^n+\dots \right]\\ =\left[c_0+c_{0}x+0·x^2+\dots +0·x^n+\dots \right]\\ so,=c_0(1+x)\end{array}$uncaught exception: Invalid chunk

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#0 /var/www/html/integration/lib/php/Boot.class.php(769): com_wiris_plugin_impl_HttpImpl_1(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Invalid chunk') #1 /var/www/html/integration/lib/haxe/Http.class.php(532): _hx_lambda->execute('Invalid chunk') #2 /var/www/html/integration/lib/php/Boot.class.php(769): haxe_Http_5(true, Object(com_wiris_plugin_impl_HttpImpl), Object(com_wiris_plugin_impl_HttpImpl), Array, Object(haxe_io_BytesOutput), true, 'Invalid chunk') #3 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(30): _hx_lambda->execute('Invalid chunk') #4 /var/www/html/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Invalid chunk') #5 /var/www/html/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), Object(sys_net_Socket), NULL) #6 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(43): haxe_Http->request(true) #7 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(268): com_wiris_plugin_impl_HttpImpl->request(true) #8 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(307): com_wiris_plugin_impl_RenderImpl->showImage('683c740e2602f4a...', NULL, Object(PhpParamsProvider)) #9 /var/www/html/integration/createimage.php(17): com_wiris_plugin_impl_RenderImpl->createImage('uncaught exception: Invalid chunk

in file: /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 68
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This implies that, rc₀ =0.$\begin{array}{c}{x}^2{y}^{\text{'}\text{'}}+(x+1)y^{\text{'}}-y=x^r\left[r{c}_0+\sum _{k=0}^{\infty }(k+r+1)c_{k+1}{x}^k+\sum _{k=0}^{\infty }\left[{(k+r)}^2-1\right]c_k{x}^k\right]\\ =x^r\left[r_{c0}+\sum _{k=0}^{\infty }\left[(k+r+1)c_{k+1}+\left[{(k+r)}^2-1\right]c_k\right]x^k\right]\\ Thisimpliesthat,r{c}_0=0\\ and(k+r+1)c_{k+1}+\left[{(k+r)}^2-1\right]c_k=0\\ k=0,1,2,3,\dots \dots .\\ Becausenothingisgainedbytaking\mathrm{C}0=0,wemustthenhave\mathrm{r}=0.\\ Theseindicialrootsare,\mathrm{r}=0\\ C_{k+1}=-\frac{\left[{(k+r)}^2-1\right]}{(k+r+1)}{C}_k\\ andk=0,1,2,3,\dots \dots .\\ For\mathrm{r}=0from\left(4\right),obtain:\\ C_{k+1}=-\frac{k^2-1}{(k+1)}{c}_k\\ (1-k)c_k,k=0,1,2,\\ Fromequation\left(5\right)obtain:\\ c_1=(1-0)c_0\\ =c_0\\ c_2=(1-1)c_1\\ =0\\ so,c_n=0\\ Thus,theindicialrootsr=0weobtainthesolution.\\ y=\sum _{n=0}^{\infty }{c}_n{x}^{n+0}\\ =x^0\sum _{n=0}^{\infty }{c}_n{x}^n\\ =x^0\left[c_0+c_{1}x+c_2{x}^2+\dots +c_n{x}^n+\dots \right]\\ =\left[c_0+c_{0}x+0·x^2+\dots +0·x^n+\dots \right]\\ so,=c_0(1+x)\end{array}$uncaught exception: Invalid chunk

in file: /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 68
#0 /var/www/html/integration/lib/php/Boot.class.php(769): com_wiris_plugin_impl_HttpImpl_1(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Invalid chunk') #1 /var/www/html/integration/lib/haxe/Http.class.php(532): _hx_lambda->execute('Invalid chunk') #2 /var/www/html/integration/lib/php/Boot.class.php(769): haxe_Http_5(true, Object(com_wiris_plugin_impl_HttpImpl), Object(com_wiris_plugin_impl_HttpImpl), Array, Object(haxe_io_BytesOutput), true, 'Invalid chunk') #3 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(30): _hx_lambda->execute('Invalid chunk') #4 /var/www/html/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Invalid chunk') #5 /var/www/html/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), Object(sys_net_Socket), NULL) #6 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(43): haxe_Http->request(true) #7 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(268): com_wiris_plugin_impl_HttpImpl->request(true) #8 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(307): com_wiris_plugin_impl_RenderImpl->showImage('683c740e2602f4a...', NULL, Object(PhpParamsProvider)) #9 /var/www/html/integration/createimage.php(17): com_wiris_plugin_impl_RenderImpl->createImage('uncaught exception: Invalid chunk

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$\begin{array}{l}and(k+r+1)c_{k+1}+\left[{(k+r)}^2-1\right]c_k=0\\ k=0,1,2,3,\dots \dots .\end{array}$

Because nothing is gained by taking c₀ =0.we must then have r=0.

$\begin{array}{c}{C}_{k+1}=-\frac{\left[{(k+r)}^2-1\right]}{(k+r+1)}{C}_k\\ k=0,1,2,3,\dots \dots .\\ Forr=0from\left(4\right),obtain:\\ C_{k+1}=-\frac{k^2-1}{(k+1)}{c}_k\\ (1-k)c_k,k=0,1,2,\\ Fromequation\left(5\right)obtain:\\ c_1=(1-0)c_0\\ =c_0\\ c_2=(1-1)c_1\\ =0\\ So,c_n=0.\end{array}$

Thus, the indicial roots r=0 we obtain the solution.

$\begin{array}{c}y=\sum _{n=0}^{\infty }{c}_n{x}^{n+0}\\ =x^0\sum _{n=0}^{\infty }{c}_n{x}^n\\ =x^0\left[c_0+c_{1}x+c_2{x}^2+\dots +c_n{x}^n+\dots \right]\\ =\left[c_0+c_{0}x+0·x^2+\dots +0·x^n+\dots \right]\\ =c_0(1+x)\end{array}$

Thus, one solution of the given differential equation is,$\mathrm{y}1\left(\mathrm{x}\right)=\mathrm{c}0(1+\mathrm{x})$ .

Reducing the order

To find the second solution of, $y^{\text{'}\text{'}}+f\left(x\right)y^{\text{'}}+g\left(x\right)y=0$ . …… (6)

Give one solution u(x), substitute.

$\mathrm{y}=\mathrm{u}\left(\mathrm{x}\right)\mathrm{v}\left(\mathrm{x}\right)$

Into (5) the solve for, .

$y^{\text{'}\text{'}}+\left(\frac{x+1}{x^2}\right)y^{\text{'}}+\left(-\frac{1}{x^2}\right)y=0$

On rewriting the differential equation (1), obtain:

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

$\begin{array}{c}=\frac{2}{1+x}+\frac{x+1}{x^2}\\ =\frac{2}{1+x}+\frac{1}{x}+\frac{1}{x^2}\\ ln{v}^{\text{'}}=2\int \frac{1}{1+x}dx+\int \frac{1}{x}dx+\int \frac{1}{x^2}dx\\ =2\ln (x+1)+\ln x-\frac{1}{x}+\ln K\\ Simplifyfurtherasfollows:\\ =\ln Kx(x+1{)}^2-\frac{1}{x}{v}^{\text{'}}\\ =\int Kx(x+1{)}^2{e}^{-\frac{-1}{x}}dx\\ =K\int x\left(x^2+2x+1\right)e^{\frac{-1}{x}}dx\\ =K\int \left(x^3+2x^2+1\right)\left(1-\frac{1}{x}+\frac{1}{2x^2}-\frac{1}{3x^2}+\dots \right)dx\\ Therefore,\\ =K\left(-\frac{1}{3x^2}-\frac{1}{6x}-\frac{1}{3}-\frac{1}{2}x+x^2+x^3+\dots \right)\\ y_2\left(x\right)=K\left(-\frac{1}{3x^2}-\frac{1}{6x}-\frac{1}{3}-\frac{1}{2}x+x^2+x^3+\dots \right)\end{array}$