Question #267486

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


1
Expert's answer
2021-11-19T05:11:58-0500

Solution;

d2ydx2y=0...(1)\frac{d^2y}{dx^2}-y=0...(1)

Using power series;

y=n=0anxny=\displaystyle\sum_{n=0}^{\infin}a_nx^n=a0x0+a1x+a2x2+...=a_0x^0+a_1x+a_2x^2+...

Differentiate;

y=n=1nanxn1y'=\displaystyle\sum_{n=1}^{\infin}na_nx^{n-1}

Differentiate further;

y=n=2n(n1)anxn2y''=\displaystyle\sum_{n=2}^{\infin}n(n-1)a_nx^{n-2}

Rewrite as;

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

Substitute into equation (1);

n=0(n+2)(n+1)an+2xnn=0anxn=0\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 x0x\neq0 ;

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

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

If;

n=0;

a0=a2=0a_0=a_2=0

When n=1;

a3=a13.2a_3=\frac{a_1}{3.2}

When n=2;

a4=a24.3=0a_4=\frac{a_2}{4.3}=0

When n=3;

a5=a35.3a_5=\frac{a_3}{5.3}

Taking a1=1a_1=1 ,we have;

a3=16a_3=\frac{1}{6}

a5=16.5.3=190a_5=\frac{1}{6.5.3}=\frac{1}{90}

The power series solution is;

y=x+16x3+190x5+...y=x+\frac16x^3+\frac{1}{90}x^5+...


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