Question

solve y'+y= 1 by power series solution method

Answer 1

Let power series solution of given differential equation be $y= \sum_{n=0}^{\infin} c_nx^n$ .

$\implies y'= \sum_{n=1}^{\infin} nc_nx^{n-1}$ .

By putting in given differential equation, we got

$\sum_{n=1}^{\infin}n c_nx^{n-1}+ \sum_{n=0}^{\infin} c_nx^n = 1$

$\implies \sum_{n=0}^{\infin}(n+1) c_{n+1}x^{n}+ \sum_{n=0}^{\infin} c_nx^n = 1 \\ \implies \sum_{n=0}^{\infin}((n+1) c_{n+1}+c_n)x^{n}=1$

Now, by comparing the coefficient on both sides we get,

$c_1+c_0 = 1 , \ and \ (n+1) c_{n+1} + c_n = 0 \ \forall n\geq 2 \implies c_{n+1} = -\frac{c_n}{n+1}$

$\implies c_1=1-c_0, c_2 =- \frac{c_1}{2} = \frac{c_0-1}{2}, c_3 = \frac{c_2}{3} = \frac{c_0-1}{3!}, c_4= \frac{c_0-1}{4!}$ and so on.

So, power series solution of given differential equation is

$y= c_0+(1-c_0)x + \frac{c_0-1}{2!} x^2 + \frac{c_0-1}{3!} x^3+ ...$