Question

(X^2) y"-2(x^2+x) y'+(x^2+2x+2) y=0

Answer 1

Solution;

Assume;

$y=\displaystyle\sum_{n=0}^{\infin}a_nx^{r+n}$

$y'=\displaystyle\sum_{n=0}^{\infin}a_n(r+n)x^{r+n-1}$

$y''=\displaystyle\sum_{n=0}^{\infin}a_n(r+n)(r+n-1)x^{r+n-2}$

Substitute y,y' and y'' into the given equation to obtain;

$\displaystyle\sum_{n=0}^{\infin}a_n(r+n)(r+n-1)x^{r+n}-\displaystyle\sum_{n=0}^{\infin}2a_n(r+n)x^{r+n+1}-\displaystyle\sum_{n=0}^{\infin}2a_n(r+n)x^{r+n}+\displaystyle\sum_{n=0}^{\infin}a_nx^{r+n+2}+\displaystyle\sum_{n=0}^{\infin}2a_nx^{r+n+1} +\displaystyle\sum_{n=0}^{\infin}2a_nx^{r+n}$

We shift the terms as follows;

$\displaystyle\sum_{n=0}^{\infin}a_n(r+n)(r+n-1)x^{r+n} -\displaystyle\sum_{n=1}^{\infin}2a_{n-1}(r+n)x^{r+n} -\displaystyle\sum_{n=0}^{\infin}2a_n(r+n)x^{r+n}+\displaystyle\sum_{n=2}^{\infin}a_{n-2}x^{r+n}+\displaystyle\sum_{n=1}^{\infin}2a_{n-1}x^{r+n}+\displaystyle\sum_{n=0}^{\infin}2a_nx^{r+n}$

At n=0;

$a_0[r(r-1)-2r+2]=0$

But $a_0\neq0$

Hence;

$r^2-3r+2=0$

$(r-2)(r-1)=0$

Therefore;

r=2 or 1

At n=1;

$a_1(r+1)r-2a_0(r+1)-2a_1(r+1)+2a_0+2a_1=0$

$a_1r^2+a_1r-2a_0r-2a_0-2a_1r-2a_1+2a_0+2a_1=0$

Simplify as;

$a_1(r^2+r-2r)-2a_0r=0$

$a_1=\frac{2a_0r}{r(r-1)}$

To solve ,we take r=2;

$a_1=2a_0$

At n=2 ,all terms produces,hence we have a general formula;

$a_n(r+n)(r+n-1)-2a_{n-1}(r+n)-2a_n(r+n)+a_{n-2}+2a_{n-1}+2a_n=0$

Combine as follows;

$a_n[(2+n)(n+1)-2(2+n)+2]=2a_{n-1}[2+n+2]+a_{n-2}$

Hence;

$a_n=\frac{2a_{n-1}(4+n)+a_{n-2}}{(2+n)(n+1)-2(n+2)+2}$

This is the recurrence relationship using r=2;

Hence we find other terms;

n=2;

$a_2=\frac{2a_1(6)+a_0}{4.3-8+2}$ =$\frac{12a_1+a_0}{6}$

Since $a_1=2a_0$ ,by substitution;

$a_2=\frac{25a_0}{6}$

When n=3;

$a_3=\frac{2a_2(8)+a_1}{5.4-10+2}=\frac{14a_2+a_1}{12}$

By substitution ;

$a_3=\frac{181a_0}{36}$

When n=4;

$a_4=\frac{2a_3(8)+a_2}{6.5-16+2}=\frac{16a_3+a_2}{20}$

By substitution;

$a_4=\frac{1523}{360}a_0$

The solution of the equation becomes;

$y=x^2\displaystyle\sum_{n=0}^{\infin}a_nx^{n}$

Hence ;

$y=x^2[a_0+2a_0x+\frac{25a_0}{6}x^2+\frac{181}{36}a_0x^3+\frac{1523}{360}a_0x^4+...$