Question

How do you use Power Series to solve the differential equation `y'-y=0` ?

Answer 1

The solution is

`y=c_0sum_{n=0}^infty{x^n}/{n!}=c_0e^x`,

where `c_0` is any constant.

Let us look at some details.

Let
`y=sum_{n=0}^infty c_n x^n`
`y'=sum_{n=1}^infty nc_n x^{n-1}=sum_{n=0}^infty(n+1)c_{n+1}x^n`

So, we can rewrite `y'-y=0` as

`sum_{n=0}^infty (n+1)c_{n+1} x^n-sum_{n=0}^infty c_n x^n=0`

by combining the summations,

`Rightarrow sum_{n=0}^infty[(n+1)c_{n+1}-c_n]x^n=0`

so, we have

`(n+1)c_{n+1}-c_n=0 Rightarrow c_{n+1}=1/{n+1}c_n`

Let us observe the first few terms.

`c_1=1/1c_0=1/{1!}c_0`

`c_2=1/2c_1=1/{2}cdot1/{1!}c_0=1/{2!}c_0`

`c_3=1/3c_2=1/3cdot1/{2!}c_0=1/{3!}c_0`
.
.
.
`c_n=1/{n!}c_0`

Hence, the solution is

`y=sum_{n=0}^infty1/{n!}c_0x^n=c_0sum_{n=0}^infty{x^n}/{n!}=c_0e^x`,

where `c_0` is any constant.