Question #221429

Find the power series solution of the differential equation

𝑦``+𝑥𝑦`+(𝑥2+2)𝑦=0


1
Expert's answer
2021-07-29T16:44:42-0400

Solution

Let

y(x)=n=0anxny\left(x\right)=\sum_{n=0}^{\infty}{a_nx^n}

dydx=n=1nanxn1=n=0(n+1)an+1xn\frac{dy}{dx}=\sum_{n=1}^{\infty}{na_nx^{n-1}}=\sum_{n=0}^{\infty}{\left(n+1\right)a_{n+1}x^n}

d2ydx2=n=0n(n+1)an+1xn1=n=0(n+1)(n+2)an+2xn\frac{d^2y}{dx^2}=\sum_{n=0}^{\infty}{n\left(n+1\right)a_{n+1}x^{n-1}}=\sum_{n=0}^{\infty}{\left(n+1\right)\left(n+2\right)a_{n+2}x^n}


Substitution into equation:

n=0(n+1)(n+2)an+2xn+n=0(n+1)an+1xn+1+n=0anxn+2+2n=0anxn=0\sum_{n=0}^{\infty}{\left(n+1\right)\left(n+2\right)a_{n+2}x^n}+\sum_{n=0}^{\infty}{\left(n+1\right)a_{n+1}x^{n+1}}+\sum_{n=0}^{\infty}{a_nx^{n+2}}+2\sum_{n=0}^{\infty}{a_nx^n}=0

n=0(n+1)(n+2)an+2xn+n=1nanxn+n=2an2xn+2n=0anxn=0\sum_{n=0}^{\infty}{\left(n+1\right)\left(n+2\right)a_{n+2}x^n}+\sum_{n=1}^{\infty}{na_nx^n}+\sum_{n=2}^{\infty}{a_{n-2}x^n}+2\sum_{n=0}^{\infty}{a_nx^n}=0

2a2+6a3x+2a0+3a1x+n=2(n+1)(n+2)an+2xn+n=2nanxn+n=2an2xn+2n=2anxn=02a_2+6a_3x+2a_0+3a_1x+\sum_{n=2}^{\infty}{\left(n+1\right)\left(n+2\right)a_{n+2}x^n}+\sum_{n=2}^{\infty}{na_nx^n}+\sum_{n=2}^{\infty}{a_{n-2}x^n}+2\sum_{n=2}^{\infty}{a_nx^n}=0

Coefficient near xn are equal to zero.

n=0: 2a2 + 2a0=0   =>  a2 = -a0 

n=1: 6a3+3a1=0   =>  a3=-a1 /2

n>1: (n+1)(n+2)an+2+nan+an-2+2an=0 =>  (n+1)(n+2)an+2+(n+2)an+an-2=0

an+2 =-an/(n+1)-an-2/[(n+1) (n+2)]

Therefore for arbitrary a0 and a1 and last recurrent formula solution is

y(x)=a0+a1x a0x2a12x3+n=2an+2xn+2y\left(x\right)=a_0+a_1x\ -a_0x^2-\frac{a_1}{2}x^3+\sum_{n=2}^{\infty}{a_{n+2}x^{n+2}}


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