Question

Find a power series solution of xy'=y​

Answer 1

Let

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

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

Substitution into equation:

$\sum_{n=0}^{\infty}{\left(n+1\right)a_{n+1}x^{n+1}}=\sum_{n=0}^{\infty}{a_nx^n}$

$\sum_{n=1}^{\infty}{na_nx^n}-\sum_{n=0}^{\infty}{a_nx^n}=0$

$-a_0+\sum_{n=1}^{\infty}{na_nx^n}-\sum_{n=1}^{\infty}{a_nx^n}=0$

Coefficients near xⁿ are equal to zero.

n=0: a₀=0  

n=1: a₁ - a₁=0   =>  a₁ is arbitrary constant C

n>1: na_n - a_n=0  =>  (n-1)a_n=0  =>  a_n = 0 

Therefore for arbitrary a₁ = C and a_n = 0 for any n≠1 we will get solution y(x) = Cx

Answer

y(x) = Cx