Question

Use a Taylor Series solution, centered at zero to solve the initial-value problem below. (Find a 5th degree approximation only) dy/dx = x + y y(0) = 1

Answer 1

We're looking for a solution of the form

`y=\displaystyle\sum_(n=0)^\infty a_nx^n=a_0+a_1x+a_2x^2+\cdots`

Given that
`y(0)=1`, we would end up with
`a_0=1`.

Its first derivative is

`y'=\displaystyle\sum_(n=0)^\infty na_nx^(n-1)=\sum_(n=1)^\infty na_nx^(n-1)=\sum_(n=0)^\infty(n+1)a_(n+1)x^n`

The shifting of the index here is useful in the next step. Substituting these series into the ODE gives

`\displaystyle\sum_(n=0)^\infty(n+1)a_(n+1)x^n-\sum_(n=0)^\infty a_nx^n=x`

Both series start with the same-degree term
`x^0`, so we can condense the left side into one series.

`\displaystyle\sum_(n=0)^\infty\bigg((n+1)a_(n+1)-a_n\bigg)x^n=x`

Pull out the first two terms (
`x^0` and
`x^1`) of the series:

`a_1-a_0+(2a_2-a_1)x+\displaystyle\sum_(n=2)^\infty\bigg((n+1)a_(n+1)-a_n\bigg)x^n=x`

Matching the coefficients of the
`x^0` and
`x^1` terms on either side tells us that

`\begin{cases}a_1-a_0=0\\2a_2-a_1=1\end{cases}`

We know that
`a_0=1`, so
`a_1=1` and
`a_2=1`. The rest of the coefficients, for
`n\ge2`, are given according to the recurrence,

`(n+1)a_(n+1)-a_n=0\implies a_(n+1)=(a_n)/(n+1)`

so that
`a_3=\frac{a_2}3=\frac13`,
`a_4=\frac{a_3}4=\frac1{12}`, and
`a_5=\frac{a_4}5=\frac1{60}`. So the 5th degree approximation to the solution to this ODE centered at
`x=0` is

`y\approx1+x+x^2+\frac{x^3}3+(x^4)/(12)+(x^5)/(60)`