# Thread: Second order DE (Frobenius Method)

1. ## Second order DE (Frobenius Method)

Hello everyone!

Here's the problem. Equation

$(x^2+3x+2)y''(x)+3y'(x)-2y(x)=0$

needs to be solved using Frobenius method at $x=-1$.

First, I've translated the origin, $t=x+1\Rightarrow x=t-1$, so that the original equation becomes

$(t^2+t)y''(t)+3y'(t)-2y(t)=0$,

derivatives staying the same,

$\displaystyle{\frac{\mathrm{d}\,y}{\mathrm{d}\,x}= \frac{\mathrm{d}\,y}{\mathrm{d}\,t}\frac{\mathrm{d }\,t}{\mathrm{d}\,t}=\frac{\mathrm{d}\,y}{\mathrm{ d}\,t}\cdots}$

Now, we seek solution in the form

$\displaystyle{y(t)=\sum_{n=0}^\infty a_n t^{n+r}.}$

Upon differentiating and substituting, we get

$\displaystyle{\sum_{n=0}^\infty [(n+r)(n+r-1)-2]a_nt^{n+r}+\sum_{n=0}^\infty [(n+r)(n+r-1)+3(n+r)]a_nt^{n+r-1}=0.}$

I hope I've done everything right by now. Now, I'm not sure if my indicial equation is right. First, I shifted the second sum so it becomes

$\displaystyle{\sum_{n=0}^\infty [(n+r)(n+r-1)-2]a_nt^{n+r}+}$
$\displaystyle{\sum_{n=1}^\infty [(n+r)(n+r-1)+3(n+r+1)]a_{n+1}t^{n+r}=0.}$

$\displaystyle{[r(r-1)-2]a_0t^r+}$
$\displaystyle{\sum_{n=1}^\infty \{[(n+r)(n+r-1)-2]a_n+[(n+r)(n+r-1)+3(n+r+1)]a_{n+1}\}t^{n+r}=0}$

I've taken $r(r-1)-2=0$ to be indicial eqation and then plugged the solutions ( $r=2$ and $r=-1$). Recurrence formula for $r=2$ gives trivial solution $y=0$ and formula for $r=-1$ gives polynomial solution.

But, if I use

$\displaystyle{y(t)=\sum_{n=0}^\infty a_n t^{n+r}}$

as a solution, I get $1/t$ in the polynomial and it diverges for $t=0$. However, if I start the sum with index 1, the solution seems right.

So, I'm just wondering if I did everyting right, especially with indicial equation.

2. Originally Posted by Dr. Jekyll
Hello everyone!

Here's the problem. Equation

$(x^2+3x+2)y''(x)+3y'(x)-2y(x)=0$

needs to be solved using Frobenius method at $x=-1$.

First, I've translated the origin, $t=x+1\Rightarrow x=t-1$, so that the original equation becomes

$(t^2+t)y''(t)+3y'(t)-2y(t)=0$,

derivatives staying the same,

$\displaystyle{\frac{\mathrm{d}\,y}{\mathrm{d}\,x}= \frac{\mathrm{d}\,y}{\mathrm{d}\,t}\frac{\mathrm{d }\,t}{\mathrm{d}\,t}=\frac{\mathrm{d}\,y}{\mathrm{ d}\,t}\cdots}$

Now, we seek solution in the form

$\displaystyle{y(t)=\sum_{n=0}^\infty a_n t^{n+r}.}$

Upon differentiating and substituting, we get

$\displaystyle{\sum_{n=0}^\infty [(n+r)(n+r-1)-2]a_nt^{n+r}+\sum_{n=0}^\infty [(n+r)(n+r-1)+3(n+r)]a_nt^{n+r-1}=0.}$

I hope I've done everything right by now. Now, I'm not sure if my indicial equation is right. First, I shifted the second sum so it becomes

$\displaystyle{\sum_{n=0}^\infty [(n+r)(n+r-1)-2]a_nt^{n+r}+}$
$\displaystyle{\sum_{n=1}^\infty [(n+r)(n+r-1)+3(n+r+1)]a_{n+1}t^{n+r}=0.}$

$\displaystyle{[r(r-1)-2]a_0t^r+}$
$\displaystyle{\sum_{n=1}^\infty \{[(n+r)(n+r-1)-2]a_n+[(n+r)(n+r-1)+3(n+r+1)]a_{n+1}\}t^{n+r}=0}$

I've taken $r(r-1)-2=0$ to be indicial eqation and then plugged the solutions ( $r=2$ and $r=-1$). Recurrence formula for $r=2$ gives trivial solution $y=0$ and formula for $r=-1$ gives polynomial solution.

But, if I use

$\displaystyle{y(t)=\sum_{n=0}^\infty a_n t^{n+r}}$

as a solution, I get $1/t$ in the polynomial and it diverges for $t=0$. However, if I start the sum with index 1, the solution seems right.

So, I'm just wondering if I did everyting right, especially with indicial equation.

$\displaystyle{\sum_{n=-1}^\infty [(n+r)(n+r+1)+3(n+r+1)]a_{n+1}t^{n+r}}$