Answer to Question #265421 in Differential Equations for Garima

Question

Solve by the power series method d²y/dx² -y =0

Answer 1

$\frac{d^2y}{dx^2}-y=0...(1)$

Using power series;

$y=\displaystyle\sum_{n=0}^{\infin}a_nx^n$$=a_0x^0+a_1x+a_2x^2+...$

Differentiate;

$y'=\displaystyle\sum_{n=1}^{\infin}na_nx^{n-1}$

Differentiate further;

$y''=\displaystyle\sum_{n=2}^{\infin}n(n-1)a_nx^{n-2}$

Rewrite as;

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

Substitute into equation (1);

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

Equate the coefficients to zero since $x\neq0$ ;

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

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

If;

n=0;

$a_0=a_2=0$

When n=1;

$a_3=\frac{a_1}{3.2}$

When n=2;

$a_4=\frac{a_2}{4.3}=0$

When n=3;

$a_5=\frac{a_3}{5.3}$

Taking $a_1=1$ ,we have;

$a_3=\frac{1}{6}$

$a_5=\frac{1}{6.5.3}=\frac{1}{90}$

The power series solution is;

$y=x+\frac16x^3+\frac{1}{90}x^5+...$