Answer to Question #245955 in Differential Equations for Thanu

Question

Solve the following IVP by power series method

xy′′ + y′ + 2y = 0, y(1) = 2,y′(1) = 4.

Answer 1

Solution

New variable t = x-1  => x = t+1 and IVP is (t+1)y′′ + y′ + 2y = 0, y(0) = 2,y′(0) = 4.

Let

$y\left(t\right)=\sum_{n=0}^{\infty}{a_nt^n}$

$\frac{dy}{dt}=\sum_{n=1}^{\infty}{na_nt^{n-1}}=\sum_{n=0}^{\infty}{\left(n+1\right)a_{n+1}t^n}$

$\frac{d^2y}{dt^2}=\sum_{n=0}^{\infty}{n\left(n+1\right)a_{n+1}t^{n-1}}=\sum_{n=0}^{\infty}{\left(n+1\right)\left(n+2\right)a_{n+2}t^n}$

Substitution into equation:

$\left(t+1\right)\sum_{n=0}^{\infty}{\left(n+1\right)\left(n+2\right)a_{n+2}t^n}+\sum_{n=0}^{\infty}{\left(n+1\right)a_{n+1}t^n}+2\sum_{n=0}^{\infty}{a_nt^n}=0$

$\sum_{n=0}^{\infty}{\left(n+1\right)\left(n+2\right)a_{n+2}t^n}+\sum_{n=1}^{\infty}{\left(n+1\right)na_{n+1}t^n}+\sum_{n=0}^{\infty}{\left(n+1\right)a_{n+1}t^n}+2\sum_{n=0}^{\infty}{a_nt^n}=0$

$2a_2+a_1+2a_0+\sum_{n=1}^{\infty}{\left(n+1\right)\left(n+2\right)a_{n+2}t^n}+\sum_{n=1}^{\infty}{\left(n+1\right)^2a_{n+1}t^n}+2\sum_{n=1}^{\infty}{a_nt^n}=0$

Coefficients near tⁿ are equal to zero.

n=0: 2a₂ +a₁ + 2a₀=0   =>  a₂ = -a₁ /2 - a₀ 

n>0: (n+1)(n+2)a_n+2+(n+1)²a_n+1+2a_n=0 =>  a_n+2 =-(n+1)a_n+1/(n+2)-2a_n/[(n+1) (n+2)]

From initial values for t: y(0) = 2,y′(0) = 4 => a₀ = 2, a₁ = 4

Therefore for recurrent formula  a_n+2 =-(n+1)a_n+1/(n+2)-2a_n/[(n+1) (n+2)] with  a₀ = 2, a₁ = 4 solution is

$y\left(x\right)=2+4\left(x-1\right)\ +\sum_{n=0}^{\infty}{a_{n+2}\left(x-1\right)^{n+2}}$