Tutorial 7: Frobenius Method

Tutorial Question

Part 1: Using method of reduction of order, find $y_2$ such that $y_1, y_2$ form a basis.

1. $2t^2y''+ty'-3y=0$; $y_1(t)=t^{-1}, t\neq0$

   Solution

   $$\begin{aligned} y_2(t)&=v(t)y_1(t)\\ y_2(t)&=t^{-1}v \\ y_2'(t)&=-t^{-2}v+t^{-1}v'\\ y_2''(t)&=2t^{-3}v-2t^{-2}v'+t^{-1}v'' \end{aligned}$$

   Substituting back yields,

   $$\begin{aligned} 2t^2(2t^{-3}v-2t^{-2}v'+t^{-1}v'')+t(-t^{-2}v+t^{-1}v')-3(t^{-1}v)&=0 \\ 2tv''+(-4+1)v'+(4t^{-1}-t^{-1}-3t^{-1})v&=0 \\ \color{red} {\scriptsize {[\triangle \text{ Ignoring the non-differentiated terms}]}}\\ 2tv''-3v'&=0 \end{aligned}$$

   Changing variables, $w=v'\rightarrow w'=v''$, Thus,

   $$\begin{aligned} 2tw'-3w&=0\\ 2t\frac{dw}{dt}&=3w\\ \int\frac{1}{3w}dw&=\int\frac{1}{2t}dt\\ \frac{1}{3}\ln|w|&=\frac{1}{2}\ln|t|+C\\ \ln|w|&=\ln t^{\frac{3}{2}}+C\\ w&=Ct^{\frac{3}{2}} \end{aligned}$$

   And, $v'=w=Ct^{\frac{3}{2}} \rightarrow v=\int Ct^{\frac{3}{2}}dt=\frac{2}{5}Ct^{\frac{2}{5}}+K$.

   Letting $C=\frac{5}{2}$ and $K=0$, $v(t)=t^{\frac{5}{2}}\rightarrow y_2(t)=t^{-1}t^{\frac{5}{2}}=t^{\frac{3}{2}}$

   $\therefore \boldsymbol{y_2(t)=t^{\frac{3}{2}}}$.

   The $y(t)$ is: $y(t)=C_1t^{-1}+C_2t^{\frac{3}{2}}$

2. $t^2y''-(t^2+2t)y'+(t+2)y=0$; $y_1(t)=t$

   Solution

   $$\begin{aligned} y_2(t)&=v(t)y_1(t)\\ y_2(t)&=tv \\ y_2'(t)&=v+tv'\\ y_2''(t)&=v'+v'+tv''=2v'+tv'' \end{aligned}$$

   Substituting back yields,

   $$\begin{aligned} t^2(2v'+tv'')-(t^2+2t)(v+tv')+(t+2)(tv)&=0 \\ 2t^2v'+t^3v''-t^2v-t^3v'-2tv-2t^2v'+t^2v+2tv&=0 \\ v'(2t^2-t^3-2t^2)+v''(t^3)-t^2v+t^2v-2tv+2tv&=0\\ t^3v''-t^3v'&=0\\ v''-v'&=0 \end{aligned}$$

   A solution to the equation is $v'=e^t$. Integrate this yield $v(t)=e^t$.

   $\therefore \boldsymbol{y_2(t)=te^t}$.

   The $y(t)$ is: $y(t)=C_1t+C_2te^{t}$

3. $y''+6y'+9y=0$; $y_1(t)=e^{-3t}$

   Solution

   $$\begin{aligned} y_2(t)&=v(t)y_1(t)\\ y_2(t)&=ve^{-3t}\\ y_2'(t)&=v'e^{-3t}-3e^{-3t}v\\ y_2''(t)&=v''e^{-3t}-6v'e^{-3t}+9ve^{-3t} \end{aligned}$$

   Substituting back yields,

   $$\begin{aligned} (v''e^{-3t}-6v'e^{-3t}+9ve^{-3t})+6(v'e^{-3t}-3e^{-3t}v)+9(ve^{-3t}) &= 0 \\ \color{red} {\scriptsize{[\triangle \text{ Ignoring the non-differentiated terms}]}}\\ v''e^{-3t}&=0\\ \end{aligned}$$

   This gives two possibilities, $v''=0$ or $e^{-3t}=0$. However, $e^{-3t}=0$ only true if $t=\infty$.

   $$\begin{aligned} v''&=0\\ \frac{d}{dt}(\frac{dv}{dt})&=0\\ \int\int d(dv)&=\int\int dt\\ \int v&=\int Adt\\ v&=At+B \end{aligned}$$

   Pick $A$ and $B$ so that second linearly independent solution is obtained.

   • If $A=0$, $y_2=Be^{-3t}$. $y_1$ and $y_2$ are linearly dependent.

   • If $B=0$, $y_2=Ate^{-3t}$. $y_1$ and $y_2$ are linearly independent. This is chosen and let $A=1$.

   $\boldsymbol{\therefore y_2(t)=te^{-3t}}$

   The $y(t)$ is: $y(t)=(C_1t+C_2)e^{-3t}$

4. $(x-1)y''-xy'+y=0$; $y_1(x)=e^x,x>1$

   Solution

   $$\begin{aligned} y_2(x)&=v(x)y_1(x)\\ y_2(x)&=ve^{x}\\ y_2'(x)&=v'e^{x}+ve^{x}\\ y_2''(x)&=v''e^x+2v'e^x+ve^x \end{aligned}$$

   Substituting back yields,

   $$\begin{aligned} (x-1)(v''e^x+2v'e^x+ve^x)-x(v'e^{x}+ve^{x})+ve^x&=0\\ v''xe^x+2xv'e^x+vxe^x-v''e^x-2v'e^x-ve^x-v'xe^x+vxe^x+ve^x&=0\\ v''(xe^x-e^x)+v'(2xe^x-2e^x-xe^x)+vxe^x-ve^x+ve^x&=0 \\ \scriptsize{[\triangle \text{ Ignoring the non-differentiated terms}]} \\ e^x[v''(x-1)+v'(x-2)]&=0\\ v''(x-1)+ v'(x-2)&=0 \end{aligned}$$

   Let $u=v'$ and that $u'=v''$,

   $$\begin{aligned} u'(x-1)+u(x-2)&=0\\ u'(x-1)&=-u(x-2)\\ u'\frac{x-1}{x-2}&=-u\\ \frac{du}{dx}\frac{x-1}{x-2}&=-u\\ \frac{du}{u}&=-\frac{x-2}{x-1}dx\\ \int\frac{du}{u}&=-\int\frac{x-2}{x-1}dx\\ \ln|u|&=-\int\frac{x-1-1}{x-1}dx\\ \ln|u|&=\int(-1+\frac{1}{x-1})dx\\ \ln|u|&=-x+\ln|x-1|+C\\ u&=e^{-x+\ln|x-1|+C}\\ u&=C_1(x-1)e^{-x} \end{aligned}$$$$\begin{aligned} v(x)&=C_1\int(x-1)e^{-x}dx\\ &=C_1[-(x-1)e^{-x}+\int e^{-x}dx]\\ &=C_1[-xe^{-x}+e^{-x}-e^{-x}]+C_2\\ &=C_1xe^{-x}+C_2 \end{aligned}$$

   Let$C_1=1$ and $C_2=0$,

   $v(x)=xe^{-x}$

   Thus, $y_2(t)=v(x)y_1(x)=xe^{-x}e^{x}$.

   $\boldsymbol{\therefore y_2(x)=x}$

   The $y(x)$ is: $y(x)=C_1e^x+C_2x$

5. $xy''-y'+4x^3y=0$; $y_1(x)=\sin x^2, x>0$.

   Solution

   $$\begin{aligned} y_2(t)&=v(t)y_1(t)\\ y_2(t)&=v\sin x^2\\ y_2'(t)&=v'\sin x^2 +2xv\cos x^2\\ y_2''(t)&=v''\sin x^2+2xv'\cos x^2 +(2v+2xv')\cos x^2-4x^2v\sin x^2 \\ &=v''\sin x^2+4xv'\cos x^2 +2v\cos x^2 -4x^2v\sin x^2 \end{aligned}$$

   Substituting back yields,

   $$\begin{aligned} x(v''\sin x^2+4xv'\cos x^2+2v\cos x^2-4x^2v\sin x^2)&-\\v'\sin x^2-2xv\cos x^2+4x^3(v\sin x^2)&=0\\ xv''\sin x^2+(4x^2\cos x^2-\sin x^2)v'&=0 \end{aligned}$$

   Let $w=v'$ and so $w'=v''$.

   $$\begin{aligned} xw'sin x^2+(4x^2\cos x^2-\sin x^2)w&=0\\ w'+\frac{(4x^2\cos x^2-\sin x^2)w}{x\sin x^2}&=0\\ \frac{dw}{dx}&=\frac{(4x^2\cos x^2-\sin x^2)w}{x\sin x^2}\\ \frac{dw}{w}&=\int \frac{1}{x}dx -\int 4x\frac{\cos x^2}{\sin x^2}dx\\ \ln|w|&=\ln|x|-\ln|(\sin x^2)^2|+C\\ &=\ln\left|\frac{x}{(\sin x^2)^2}\right|\\ w&=C\frac{x}{(\sin x^2)^2} \end{aligned}$$

   Therefore,

   $$\begin{aligned} v'(x)&=w=C\frac{x}{(\sin x^2)^2}\\ v(x)&=\int C\frac{x}{(\sin x^2)^2}dx\\ &=-\frac{1}{2}C\cot x^2+A \end{aligned}$$

   Let $C=-2$ and $K=0$,

   $v(x)=\cot x^2$

   $\boldsymbol{\therefore y_2(t)=\cot x^2\sin x^2=\cos x^2}$

   The $y(t)$ is: $y(t)=C_1\sin x^2+C_2\cos x^2$

Part 2: Discuss whether two Frobenius series solutions exist or do not exist for the following equations.

1. $2x^2y''+x(x+1)y'-(\cos x)y=0$

   Solution

   $$\begin{aligned} 2x^2y''+x(x+1)y'-(\cos x)y&=0\\ x^2y''+\frac{(x+1)xy'}{2}-\frac{\cos x}{2}y&=0\\ \end{aligned}$$

   $p(x)=\frac{x+1}{2} \Rightarrow p(0)=\frac{1}{2}$

   $q(x)=\frac{-\cos x}{2} \Rightarrow q(0)=-\frac{1}{2}$

   Substituting $r$ to find $r_1$ and $r_2$.

   $r^2+(p(0)-1)r+q(0)=2r^2-r-1=0$

   $\Rightarrow r_1=1, r_2=-\frac{1}{2}$.

   $\therefore r_1-r_2=1-(-\frac{1}{2})=\frac{3}{2} \text{ }\triangleright \text{ not a zero or positive integer}$.

   Two Frobenius series solutions exist.

2. $x^4y''-(x^2\sin x)y'+2(1-\cos x)y=0$

   Solution

   $$\begin{aligned} x^4y''-(x^2\sin x)y'+2(1-\cos x)y&=0\\ x^2y''-\frac{\sin x}{x}xy'+2\frac{1-\cos x}{x^2}y&=0\\ \end{aligned}$$

   $p(x)=\frac{-\sin x}{x} \Rightarrow p(0)=-1$

   $q(x)=\frac{2(1-\cos x)}{x^2} \Rightarrow q(0)=1$

   Substituting $r$ to find $r_1$ and $r_2$.

   $r^2+(p(0)-1)r+q(0)=r^2-2r-1=0$

   $\Rightarrow r_1=1, r_2=1$.

   One Frobenius series solution exists since $r_1=r_2$.

Part 3: Apply Frobenius Method to find the basis of solutions of the following differential equations.

1. $2xy''+y'+y=0$

   Solution

   $x=0$ is a regular singular solution.

   Assuming the solution is $y=\sum \limits_{k=0}^\infty a_kx^{k+r}$.

   $$\begin{aligned} y&=\sum \limits_{k=0}^\infty a_kx^{k+r}\\ y'&=\sum \limits_{k=0}^\infty (k+r)a_kx^{k+r-1}\\ y''&=\sum \limits_{k=0}^\infty (k+r)(k+r-1)a_kx^{k+r-2} \end{aligned}$$

   Substituting back yields,

   $$\begin{aligned} 2x\sum \limits_{k=0}^\infty (k+r)(k+r-1)a_kx^{k+r-2}+\sum \limits_{k=0}^\infty (k+r)a_kx^{k+r-1}+\sum \limits_{k=0}^\infty a_kx^{k+r}&=0\\ 2\sum \limits_{k=0}^\infty (k+r)(k+r-1)a_kx^{k+r-1}+\sum \limits_{k=0}^\infty (k+r)a_kx^{k+r-1}+\sum \limits_{k=0}^\infty a_kx^{k+r}&=0 \text{ }\\ \color{red} {\scriptsize {[\triangle { \text{ Removing the } x \text{ from the first term}}}]}\\ 2\sum \limits_{k=0}^\infty (k+r)(k+r-1)a_kx^{k-1}+\sum \limits_{k=0}^\infty (k+r)a_kx^{k-1}+\sum \limits_{k=0}^\infty a_kx^{k}&=0 \\ \color{red}{\scriptsize{[\triangle { \text{ Diving out } x^r}}]} \\ 2\sum \limits_{n=-1}^\infty (n+r+1)(n+r)a_{n+1}x^n+\sum \limits_{n=-1}^\infty (n+r+1)a_{n+1}x^n+\sum \limits_{n=0}^\infty a_nx^n&=0 \\ \color{red} {\scriptsize{[\triangle { \text{ Substituting } k \text{ with }n \text{ (change limits)}}]}} \\ 2r(r-1)a_0x^{-1}+2\sum \limits_{n=0}^\infty (n+r+1)(n+r)a_{n+1}x^n&+\\ ra_0x^{-1}+\sum \limits_{n=0}^\infty (n+r+1)a_{n+1}x^n+\sum \limits_{n=0}^\infty a_nx^n&=0 \\ \color{red} {\scriptsize{[\triangle { \text{ Taking out all terms to make sure } n=0}]}} \\ \end{aligned}$$

   For $a_0$, $[2(r^2-r)+r]a_0=0$.

   Since $a_0\neq0\Rightarrow r(2r-1)=0\Rightarrow r_2=0, r_1=\frac{1}{2}$

   The recurrence relation is then,

   $$\begin{aligned} [2(n+r+1)(n+r)+(n+r+1)]a_{n+1}+a_n&=0\\ (n+r+1)(2n+2r+1)a_{n+1}&=-a_n\\ a_{n+1}&=\frac{-a_n}{(n+r+1)(2n+2r+1)}, n=0,1,2,... \end{aligned}$$

   For $r=\frac{1}{2}$,

   $$\begin{aligned} a_{n+1}&=-\frac{a_n}{(n+\frac{3}{2})(2n+2)} \\ \\ n&=0, a_1=-\frac{a_0}{(\frac{3}{2})(2)}=-\frac{a_0}{3}\\ n&=1, a_2=\frac{a_1}{(\frac{5}{2})(4)}=\frac{a_0}{10\cdot3}\\ n&=2, a_3=-\frac{a_2}{(\frac{7}{2})(6)}=-\frac{a_0}{21\cdot10\cdot3}\\ \\ y_1(x)&=\sum\limits_{k=0}^\infty a_kx^{k+r}=a_0x^{\frac{1}{2}}+a_1x^{\frac{3}{2}}+... \\ y_1(x)&=a_0x^{\frac{1}{2}}-\frac{a_0}{3}x^{\frac{3}{2}}+\frac{a_0}{10\cdot3}x^{\frac{5}{2}}-\frac{a_0}{21\cdot10\cdot3}x^{\frac{7}{2}}+...\\ \boldsymbol{y_1(x)}&=\boldsymbol{a_0x^{\frac{1}{2}}\left[1-\frac{x}{3}+\frac{x^2}{10\cdot3}-\frac{x^3}{21\cdot10\cdot3}+...\right]} \end{aligned}$$

   For $r=0$,

   $$\begin{aligned} a_{n+1}&=-\frac{a_n}{(n+1)(2n+1)} \\ \\ n&=0, a_1=-\frac{a_0}{(1)(1)}=-a_0\\ n&=1, a_2=\frac{a_1}{(2)(3)}=\frac{a_0}{6}\\ n&=2, a_3=-\frac{a_2}{(3)(5)}=-\frac{a_0}{15\cdot6}\\ \\ y_2(x)&=\sum\limits_{k=0}^\infty a_kx^k=a_0x^0+a_1x^1+... \\ y_2(x)&=a_0-a_0x+\frac{a_0}{6}x^2-\frac{a_0}{15\cdot6}x^3+...\\ \boldsymbol{y_2(x)}&=\boldsymbol{a_0\left[1-x+\frac{x^2}{6}-\frac{x^3}{15\cdot6}+...\right]} \end{aligned}$$

2. $xy''+2y'+xy=0$

   Solution

   $x=0$ is a regular singular solution.

   Assuming the solution is $y=\sum \limits_{k=0}^\infty a_kx^{k+r}$.

   $$\begin{aligned} y&=\sum \limits_{k=0}^\infty a_kx^{k+r}\\ y'&=\sum \limits_{k=0}^\infty (k+r)a_kx^{k+r-1}\\ y''&=\sum \limits_{k=0}^\infty (k+r)(k+r-1)a_kx^{k+r-2} \end{aligned}$$

   Substituting back yields,

   $$\begin{aligned} x\sum \limits_{k=0}^\infty (k+r)(k+r-1)a_kx^{k+r-2}+2\sum \limits_{k=0}^\infty (k+r)a_kx^{k+r-1}+x\sum \limits_{k=0}^\infty a_kx^{k+r}&=0\\ \sum \limits_{k=0}^\infty (k+r)(k+r-1)a_kx^{k+r-1}+2\sum \limits_{k=0}^\infty (k+r)a_kx^{k+r-1}+\sum \limits_{k=0}^\infty a_kx^{k+r+1}&=0 \\ \color{red}{\scriptsize{[\triangle { \text{ Removing the } x \text{ from the first term]}}}}\\ \sum \limits_{k=0}^\infty (k+r)(k+r-1)a_kx^{k-1}+2\sum \limits_{k=0}^\infty (k+r)a_kx^{k-1}+\sum \limits_{k=0}^\infty a_kx^{k+1}&=0 \\ \color{red}{\scriptsize{[\triangle {\text{ Diving out } x^r]}}} \\ \sum \limits_{n=-1}^\infty (n+r+1)(n+r)a_{n+1}x^n+2\sum \limits_{n=-1}^\infty (n+r+1)a_{n+1}x^n+\sum \limits_{n=1}^\infty a_{n-1}x^n&=0 \\ \color{red}{\scriptsize{[\triangle { \text{ Substituting } k \text{ with }n \text{ (change limits)}}]}} \\ r(r-1)a_0x^{-1}+r(r+1)a_1x^0+\sum \limits_{n=1}^\infty (n+r+1)(n+r)a_{n+1}x^n&+\\ 2ra_0x^{-1}+2(r+1)a_1x^0+2\sum \limits_{n=1}^\infty (n+r+1)a_{n+1}x^n+\sum \limits_{n=1}^\infty a_{n-1}x^n&=0\\ (r^2-r+2r)a_0x^{-1}+(r^2+r+r+1)a_1x^0&+\\ \sum\limits_{n=1}^\infty \left[((n+r+1)(n+r)+2(n+r+1))a_{n+1}+a_{n-1}\right]x^n&=0 \\ \color{red}{\scriptsize{[\triangle { \text{ Taking out all terms to make sure } n=0}]}} \\ \end{aligned}$$

   For $a_0$, $(r^2-r+2r)a_0=0$.

   Since $a_0\neq0\Rightarrow r^2+r=0\Rightarrow r_1=0, r_2=-1$

   The recurrence relation is then,

   $$\begin{aligned} [(n+r+1)(n+r)+2(n+r+1)]a_{n+1}+a_{n-1}&=0\\ (n+r+1)(n+r+2)a_{n+1}&=-a_{n-1}\\ a_{n+1}&=\frac{-a_{n-1}}{(n+r+1)(n+r+2)}, n=0,1,2,... \end{aligned}$$

   For $r=0$,

   $$\begin{aligned} a_{n+1}&=-\frac{a_{n-1}}{(n+1)(n+2)} \\ \\ n&=0, [(r^2+r)+2(r+1)]a_1=0\Rightarrow a_1=0\\ n&=1, a_2=\frac{-a_0}{3\cdot2}=\frac{-a_0}{3!}\\ n&=2, a_3=-\frac{-a_1}{12}=0\\ n&=3, a_4=\frac{-a_2}{20\cdot3!}=\frac{-a_0}{5!}\\ n&=4, a_5=-\frac{-a_3}{30}=0\\ \\ y_1(x)&=\sum\limits_{k=0}^\infty a_kx^{k+r}=a_0x^0+a_1x^1+... \\ y_1(x)&=a_0+0-\frac{a_0}{3!}x^2+0+\frac{a_0}{5!}+...\\ \boldsymbol{y_1(x)}&=\boldsymbol{a_0(1-\frac{1}{3!}x^2+\frac{1}{5!}x^4+...)} \end{aligned}$$

   For $r=-1$,

   $$\begin{aligned} a_{n+1}&=\frac{-a_{n-1}}{n(n+1)}, n=1,2,3,...\\ \\ n&=0, [(r^2+r)+2(r+1)]a_1=0 \Rightarrow 0\cdot a_1=0\text{ }\triangleright a_1\neq0\text{(unknown)}\\ n&=1, a_2=\frac{-a_0}{1\cdot2}=\frac{-a_0}{2!}\\ n&=2, a_3=-\frac{-a_1}{2\cdot3}=0\\ n&=3, a_4=\frac{-a_2}{3\cdot4}=\frac{a_0}{4!}\\ n&=4, a_5=-\frac{-a_3}{4\cdot5}=\frac{a_1}{5!}\\ \\ y_2(x)&=\sum\limits_{k=0}^\infty a_kx^{k+r}=a_0x^{-1}+a_1x^0+a_2x^1+... \\ y_2(x)&=\frac{1}{x}a_0+a_1-\frac{a_0}{2!}x-\frac{a_1}{3!}x^2+...\\ y_2(x)&=a_0(\frac{1}{x}-\frac{1}{2!}x+\frac{1}{4!}x^3+...)+a_1(1-\frac{x^2}{3!}+\frac{x^4}{5!}+...)\\ \boldsymbol{y_2(x)}&=\boldsymbol{a_0\frac{1}{x}\cos x+a_1\frac{1}{x}\sin x} \end{aligned}$$

3. $xy''+(1-2x)y'+(x-1)y=0$

   Solution

   $x=0$ is a regular singular solution.

   Assuming the solution is $y=\sum \limits_{k=0}^\infty a_kx^{k+r}$.

   $$\begin{aligned} y&=\sum \limits_{k=0}^\infty a_kx^{k+r}\\ y'&=\sum \limits_{k=0}^\infty (k+r)a_kx^{k+r-1}\\ y''&=\sum \limits_{k=0}^\infty (k+r)(k+r-1)a_kx^{k+r-2} \end{aligned}$$

   Substituting back yields,

   $$\begin{aligned} x\sum \limits_{k=0}^\infty (k+r)(k+r-1)a_kx^{k+r-2}+(1-2x)\sum \limits_{k=0}^\infty (k+r)a_kx^{k+r-1}&+\\ (x-1)\sum \limits_{k=0}^\infty a_kx^{k+r}&=0\\ \sum \limits_{k=0}^\infty (k+r)(k+r-1)a_kx^{k+r-1}+\sum \limits_{k=0}^\infty (k+r) a_k x^{k+r-1}&-\\ 2\sum \limits_{k=0}^\infty (k+r)a_kx^{k+r-1}+(x-1)\sum \limits_{k=0}^\infty a_kx^{k+r}&=0 \\ \color{red}{\scriptsize{[\triangle { \text{ Removing the } x \text{ from the first term]}}}}\\ \sum \limits_{k=0}^\infty (k+r)(k+r-1)a_kx^{k-1}+\sum\limits_{k=0}^\infty (k+r)a_kx^{k-1}&-\\ 2\sum \limits_{k=0}^\infty (k+r)a_kx^{k-1}+\sum \limits_{k=0}^\infty a_kx^{k+1}-\sum\limits_{k=0}^\infty a_kx^k&=0 \\ \color{red}{\scriptsize{[\triangle {\text{ Diving out } x^r]}}} \\ \sum \limits_{n=-1}^\infty (n+r+1)(n+r)a_{n+1}x^n+2\sum \limits_{n=-1}^\infty (n+r+1)a_{n+1}x^n+\sum \limits_{n=1}^\infty a_{n-1}x^n&=0 \\ \color{red}{\scriptsize{[\triangle { \text{ Substituting } k \text{ with }n \text{ (change limits)}}]}} \\ r(r-1)a_0x^{-1}+r(r+1)a_1x^0+\sum \limits_{n=1}^\infty (n+r+1)(n+r)a_{n+1}x^n&+\\ ra_0x^{-1}+(r+1)a_1x^0+\sum \limits_{n=1}^\infty (n+r+1)a_{n+1}x^n-2ra_0x^0&-\\ 2\sum \limits_{n=1}^\infty (n+r)a_{n}x^n+\sum\limits_{n=1}^\infty a_{n-1}x^n-a_0x^0-\sum\limits_{n=1}^\infty a_nx^n&=0\\ \end{aligned}$$

   For $a_0$, $[(r^2-r)+r]a_0=0$.

   Collecting the non-summation terms,

   $$\begin{aligned} ([(r^2+r)+(r+1)]a_1+(-2r-1)a_0)&=0\\ (r^2+2r+1)a_1-(2r+1)a_0&=0 \triangleright \text{will be used to find }a_0, a_1 \text{(Eq 1)} \end{aligned}$$

   Since $a_0\neq0\Rightarrow r^2=0\Rightarrow r_1=0, r_2=0$

   The recurrence relation is then,

   $$\begin{aligned} ([(n+r)(n+r+1)+(n+r+1)]a_{n+1}+a_{n-1}-[2(n+r)+1]a_n)&=0\\ ([(n+r+1)(n+r+1)]a_{n+1}+a_{n-1}-[2(n+r+1)a_n])&=0\\ a_{n+1}=\frac{[2(n+r)+1]a_n-a_{n-1}}{(n+r+1)^2}, n=1,2,3... & \end{aligned}$$

   For $r=0$,

   $$\begin{aligned} a_{n+1}&=\frac{(2n+1)a_n-a_{n-1}}{(n+1)^2}, n=1,2,3,... \\ \\ &\text{From Eq(1)}, (0+2(0)+1)a_1-(2(0)+1)a_0=0 \Rightarrow a_1=a_0\\\\ n&=1, a_2=\frac{3a_1-a_0}{2^2}=\frac{3a_0-a_0}{2^2}=\frac{a_0}{2}\\ n&=2, a_3=-\frac{5a_2-a_1}{3^2}=\frac{5(\frac{a_0}{2})-a_1}{3^2}=\frac{a_0}{3!}\\ n&=3, a_4=\frac{7a_3-a_2}{4^2}=\frac{7(\frac{a_0}{3!})-\frac{a_0}{2}}{4^2}=\frac{a_0}{4!}\\ \\ y_1(x)&=\sum\limits_{k=0}^\infty a_kx^{k+r}=a_0x^0+a_1x^1+... \\ y_1(x)&=a_0+a_0x+\frac{a_0}{2!}x^2+\frac{a_0}{3!}x^3<u>...\\ y_1(x)&=a_0(1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^4+...)\\ \boldsymbol{y_1(x)}&=\boldsymbol{a_0e^x} \end{aligned}$$

   For the second $r=0$, using reduction of order.

   Knowing $y_1=e^x$,

   $$\begin{aligned} y_2&=ve^x\\ y_2'&=ve^x+v'e^x\\ y_2''&=ve^x+2v'e^x+v''e^x \end{aligned}$$

   Substituting into main equation,

   $$\begin{aligned} x(v+2v'+v'')+(1-2x)(ve^x+v'e^x)+(x-1)ve^x&=0\\ x(v+2v'+v'')+(1-2x)(v+v')+(x-1)v&=0 \\ \color{red}{\scriptsize{\triangle[\text{ Dividing out }e^x]}}\\ xv+2xv'+xv''+v-2xv'+v'-2xv+xv-v&=0\\ v'+xv''&=0\\ xv''&=-v'\\ \frac{v''}{v'}&=-\frac{1}{x}\\ \int\frac{v''}{v'}dv&=-\int\frac{1}{x}dx\\ \ln v'&=-\ln x=\ln x^{-1}\\ v'&=x^{-1}\\ \int v' dv&=\int \frac{1}{x}dx\\ v&=\ln x \end{aligned}$$

   $\therefore \boldsymbol{y_2(x)=(\ln x)e^x}$

4. $2ty''+(1+t)y'+y=0$

   Solution

   $x=0$ is a regular singular solution.

   Assuming the solution is $y=\sum \limits_{k=0}^\infty a_kx^{k+r}$.

   $$\begin{aligned} y&=\sum \limits_{k=0}^\infty a_kx^{k+r}\\ y'&=\sum \limits_{k=0}^\infty (k+r)a_kx^{k+r-1}\\ y''&=\sum \limits_{k=0}^\infty (k+r)(k+r-1)a_kx^{k+r-2} \end{aligned}$$

   Substituting back yields,

   $$\begin{aligned} 2t\sum \limits_{k=0}^\infty (k+r)(k+r-1)a_kt^{k+r-2}+(1-t)\sum \limits_{k=0}^\infty (k+r)a_kt^{k+r-1}&+\\ \sum \limits_{k=0}^\infty a_kt^{k+r}&=0\\ 2\sum \limits_{k=0}^\infty (k+r)(k+r-1)a_kt^{k+r-1}+\sum \limits_{k=0}^\infty (k+r) a_k t^{k+r-1}&+\\ \sum \limits_{k=0}^\infty (k+r)a_kt^{k+r}+\sum \limits_{k=0}^\infty a_kt^{k+r}&=0 \\ \color{red}{\scriptsize{[\triangle { \text{ Removing the } t \text{ from the first term]}}}}\\ 2\sum \limits_{k=0}^\infty (k+r)(k+r-1)a_kt^{k-1}+\sum\limits_{k=0}^\infty (k+r)a_kt^{k-1}&+\\ \sum \limits_{k=0}^\infty (k+r)a_kt^{k}+\sum \limits_{k=0}^\infty a_kt^{k}&=0 \\ \color{red}{\scriptsize{[\triangle {\text{ Diving out } t^r]}}} \\ 2\sum \limits_{n=-1}^\infty (n+r+1)(n+r)a_{n+1}t^n+\sum \limits_{n=-1}^\infty (n+r+1)a_{n+1}t^n&+\\ \sum \limits_{n=0}^\infty (n+r)a_{n}t^n+\sum\limits_{n=0}^\infty a_nt^n&=0 \\ \color{red}{\scriptsize{[\triangle { \text{ Substituting } k \text{ with }n \text{ (change limits)}}]}} \\ 2r(r-1)a_0t^{-1}+2\sum \limits_{n=0}^\infty (n+r+1)(n+r)a_{n+1}t^n&+\\ ra_0t^{-1}+\sum \limits_{n=0}^\infty (n+r+1)a_{n+1}t^n+\sum \limits_{n=0}^\infty (n+r)a_{n}t^n+\sum\limits_{n=0}^\infty a_{n}t^n&=0\\ \end{aligned}$$

   For $a_0$, $[2(r^2-r)+r]a_0=0$.

   Since $a_0\neq0\Rightarrow 2r^2-r=0\Rightarrow r_1=\frac{1}{2}, r_2=0$

   The recurrence relation is then,

   $$\begin{aligned} [2(n+r)(n+r+1)+(n+r+1)]a_{n+1}+[(n+r)+1]a_n&=0\\ (n+r+1)(2n+2r+1)a_{n+1}&=-[(n+r)+1]a_n\\ a_{n+1}&=\frac{-[(n+r)+1]a_n} {(n+r+1)(2n+2r+1)}, n=0,1,2,... \end{aligned}$$

   For $r=-\frac{1}{2}$,

   $$\begin{aligned} a_{n+1}&=\frac{a_n}{2n+2}, n=0,1,2,3,... \\ \\ n&=0, a_1=-\frac{a_0}{2}\\ n&=1, a_2=-\frac{a_1}{2\cdot2}=\frac{a_0}{2^2\cdot 2!}\\ n&=2, a_3=-\frac{a_2}{2\cdot3}=-\frac{a_0}{2^3\cdot3!}\\ n&=3, a_4=-\frac{a_3}{2\cdot4}=\frac{a_0}{2^4\cdot4!}\\ \\ y_1(t)&=\sum\limits_{k=0}^\infty a_kt^{k+r}=a_0t^{\frac{1}{2}}+a_1t^{\frac{3}{2}}+... \\ y_1(t)&=a_0t^{\frac{1}{2}}-\frac{a_0}{2!}t^{\frac{3}{2}}+\frac{a_0}{2^2\cdot2!}t^{\frac{5}{2}}-\frac{a_0}{2^3\cdot3!}t^{\frac{7}{2}}+...\\ \boldsymbol{y_1(t)}&=\boldsymbol{a_0t^{\frac{1}{2}} \left(1-\frac{1}{2!}t+\frac{1}{2^2\cdot2!}t^2-\frac{1}{2^3\cdot3!}t^3+... \right)} \end{aligned}$$

   For $r=0$,

   $$\begin{aligned} a_{n+1}&=-\frac{a_n}{2n+1}, n=0,1,2,3,... \\ \\ n&=0, a_1=-a_0\\ n&=1, a_2=-\frac{a_1}{3}=\frac{a_0}{3}\\ n&=2, a_3=-\frac{a_2}{5}=-\frac{a_0}{5\cdot3}\\ n&=3, a_4=-\frac{a_3}{7}=\frac{a_0}{7\cdot5\cdot3}\\ \\ y_2(t)&=\sum\limits_{k=0}^\infty a_kt^{k+r}=a_0+a_1t+a_2t^2+a_3t^3+... \\ y_2(t)&=a_0-a_0t+\frac{a_0}{3}t^2-\frac{a_0}{5\cdot3}t^3+...\\ \boldsymbol{y_2(t)}&=\boldsymbol{a_0\left(1-t+\frac{1}{3}t^2-\frac{1}{5\cdot3}t^3+\frac{1}{7\cdot5\cdot3}t^4-... \right)} \end{aligned}$$

5. $x(1-x)y''-3xy'-y=0$

   Solution

   The motivation behind Frobenius method is to seek a power series solution to ordinary differential equations.

   Let $y(x) = \displaystyle \sum_{n=0}^{\infty} a_n x^n$.

   Then we get that

   $y'(x) = \sum_{n=0}^{\infty} na_n x^{n-1}$

   $3xy'(x) = \sum_{n=0}^{\infty} 3na_n x^{n}$

   $y''(x) = \sum_{n=0}^{\infty} n(n-1)a_n x^{n-2}$

   $xy''(x) = \sum_{n=0}^{\infty} n(n-1)a_n x^{n-1} = \sum_{n=0}^{\infty} n(n+1)a_{n+1} x^{n}$

   $x^2y''(x) = \sum_{n=0}^{\infty} n(n-1)a_n x^{n}$

   The ODE is $xy'' - x^2 y'' -3xy' - y = 0$

   Plugging in the appropriate series expansions, we get that

   $\sum_{n=0}^{\infty} \left(n(n+1)a_{n+1} - n(n-1)a_n - 3na_n - a_n\right)x^n = 0$

   Hence, we get that

   $n(n+1)a_{n+1} = (n(n-1) +3n+1)a_n = (n+1)^2 a_n \implies a_{n+1} = \dfrac{n+1}{n}a_n$

   First note that $a_0 = 0$. Choose $a_1$ arbitrarily. Then we get that $a_2 = 2a_1$, $a_3 = 3a_1$, $a_4 = 4a_1$ and in general, $a_{n} = na_1$.

   Hence, the solution is given by $y_1(x) = a_1 \left(x+2x^2 + 3x^3 + \cdots\right)$

   This power series is valid only within $\vert x \vert <1$. In this region, we can simplify the power series to get

   $$\begin{aligned} y_1(x) & = a_1 x \left(1 + 2x + 3x^2 + \cdots \right)\\ & = a_1 x \dfrac{d}{dx} \left(x + x^2 + x^3 + \cdots \right)\\ & = a_1 x \dfrac{d}{dx} \left(\dfrac{x}{1-x}\right)\\ & = a_1 \dfrac{x}{(1-x)^2} \end{aligned}$$

   Taking $a_1=1$, $\boldsymbol{y_1(x)=\frac{x}{(1-x)^2}}$.

   The order reduction method seeks a second basis solution in the form $y=y_1u$, where $y_1(x)=\frac{x}{(1-x)^2}$ is the already found basis solution.

   $$x(1-x)[y_1u''+2y_1'u']-3x[y_1u']=0 \implies \frac{u''}{u'}=\frac{3y_1-2(1-x)y_1'}{(1-x)y_1}$$

   Insert

   $y_1(x)=\dfrac1{(1-x)^2}-\dfrac1{1-x}$,

   $y_1'=\dfrac2{(1-x)^3}-\dfrac1{(1-x)^2}$,

   $y_1''=\dfrac{6}{(1-x)^4}-\dfrac{2}{(1-x)^3}$

   into that formula to find

   $$\begin{aligned} \frac{u''}{u'}&=\frac{-\frac1{(1-x)^2}-\frac1{1-x}}{\frac{x}{1-x}}=-\frac{2-x}{x(1-x)}=-\frac{2}x+\frac1{1-x} \\ \implies u'&=\frac1{x^2(1-x)}=\frac{1+x}{x^2}-\frac1{1-x} \\ \implies u&=-\frac1x+\ln|x(1-x)| \end{aligned}$$

   So that the second basis solution is

   $$\boldsymbol{y_2=\frac{x\ln|x(1-x)|-1}{(1-x)^2}}$$