Question

[infinity]

Substitute y(x)= Σ 2 anx^n and the Maclaurin series for 6 sin3x into y' - 2xy = 6 sin 3x and equate the coefficients of like powers of x on both sides of the equation to n= 0. Find the first four nonzero terms in a power series expansion about x = 0 of a general
n=0
solution to the differential equation.

У(Ñ)= ___________

Answer 1

Recall that

`\sin(x)=\displaystyle\sum_(n=0)^\infty(-1)^n(x^(2n+1))/((2n+1)!)`

Differentiating the power series series for y(x) gives the series for y'(x) :

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

Now, replace everything in the DE with the corresponding power series:

`y'-2xy = 6\sin(3x) \implies`

`\displaystyle\sum_(n=0)^\infty (n+1)a_(n+1)x^n - 2\sum_(n=0)^\infty a_nx^(n+1) = 6\sum_(n=0)^\infty(-1)^n((3x)^(2n+1))/((2n+1)!)`

The series on the right side has no even-degree terms, so if we split up the even- and odd-indexed terms on the left side, the even-indexed
`(n=2k)` series should vanish and only the odd-indexed
`(n=2k+1)` terms would remain.

Split up both series on the left into even- and odd-indexed series:

`y'(x) = \displaystyle \sum_(k=0)^\infty (2k+1)a_(2k+1)x^(2k) + \sum_(k=0)^\infty (2k+2)a_(2k+2)x^(2k+1)`

`-2xy(x) = \displaystyle -2\left(\sum_(k=0)^\infty a_(2k)x^(2k+1) + \sum_(k=0)^\infty a_(2k+1)x^(2k+2)\right)`

Next, we want to condense the even and odd series:

• Even:

`\displaystyle \sum_(k=0)^\infty (2k+1)a_(2k+1)x^(2k) - 2 \sum_(k=0)^\infty a_(2k+1)x^(2k+2)`

`=\displaystyle \sum_(k=0)^\infty (2k+1)a_(2k+1)x^(2k) - 2 \sum_(k=0)^\infty a_(2k+1)x^(2(k+1))`

`=\displaystyle a_1 + \sum_(k=1)^\infty (2k+1)a_(2k+1)x^(2k) - 2 \sum_(k=0)^\infty a_(2k+1)x^(2(k+1))`

`=\displaystyle a_1 + \sum_(k=1)^\infty (2k+1)a_(2k+1)x^(2k) - 2 \sum_(k=1)^\infty a_(2(k-1)+1)x^(2k)`

`=\displaystyle a_1 + \sum_(k=1)^\infty (2k+1)a_(2k+1)x^(2k) - 2 \sum_(k=1)^\infty a_(2k-1)x^(2k)`

`=\displaystyle a_1 + \sum_(k=1)^\infty \bigg((2k+1)a_(2k+1) - 2a_(2k-1)\bigg)x^(2k)`

• Odd:

`\displaystyle \sum_(k=0)^\infty 2(k+1)a_(2(k+1))x^(2k+1) - 2\sum_(k=0)^\infty a_(2k)x^(2k+1)`

`=\displaystyle \sum_(k=0)^\infty \bigg(2(k+1)a_(2(k+1))-2a_(2k)\bigg)x^(2k+1)`

`=\displaystyle \sum_(k=0)^\infty \bigg(2(k+1)a_(2k+2)-2a_(2k)\bigg)x^(2k+1)`

Notice that the right side of the DE is odd, so there is no 0-degree term, i.e. no constant term, so it follows that
`a_1=0`.

The even series vanishes, so that

`(2k+1)a_(2k+1) - 2a_(2k-1) = 0`

for all integers k ≥ 1. But since
`a_1=0`, we find

`k=1 \implies 3a_3 - 2a_1 = 0 \implies a_3 = 0`

`k=2 \implies 5a_5 - 2a_3 = 0 \implies a_5 = 0`

and so on, which means the odd-indexed coefficients all vanish,
`a_(2k+1)=0`.

This leaves us with the odd series,

`\displaystyle \sum_(k=0)^\infty \bigg(2(k+1)a_(2k+2)-2a_(2k)\bigg)x^(2k+1) = 6\sum_(k=0)^\infty (-1)^k (x^(2k+1))/((2k+1)!)`

`\implies 2(k+1)a_(2k+2) - 2a_(2k) = (6(-1)^k)/((2k+1)!)`

We have

`k=0 \implies 2a_2 - 2a_0 = 6`

`k=1 \implies 4a_4-2a_2 = -1`

`k=2 \implies 6a_6-2a_4 = \frac1{20}`

`k=3 \implies 8a_8-2a_6 = -\frac1{840}`

So long as you're given an initial condition
`y(0)\\eq0` (which corresponds to
`a_0`), you will have a non-zero series solution. Let
`a=a_0` with
`a_0\\eq0`. Then

`2a_2-2a_0=6 \implies a_2 = a+3`

`4a_4-2a_2=-1 \implies a_4 = \frac{2a+5}4`

`6a_6-2a_4=\frac1{20} \implies a_6 = (20a+51)/(120)`

and so the first four terms of series solution to the DE would be

`\boxed{a + (a+3)x^2 + \frac{2a+5}4x^4 + (20a+51)/(120)x^6}`